ARM Module 1

7/27/2019 ARM Module 1

http://slidepdf.com/reader/full/arm-module-1 1/65

UNITAR e-Learning course on Advance Risk ManagementModule 1: Review of Basic Knowledge: Statistics, Basic Matrix Algebra Operations and Mathematics of Finance

www.unitar.org/pft0

e-Learning Course on

Advanced Risk Management

Module 1: Review of Basic Knowledge:

Statistics, Basic Matrix Algebra

Operations and Mathematics of Finance

Author: Dr. E. Cosio-Pascal

United Nations Institute for Training and Research




www.unitar.org/pft1

MODULE 1

GOALS

• The course aims at providing the student with medium to advanced level knowledge

of financial risk management and its applications to the field of finances.

• In order to be able to apply the risk management tools, it is necessary to have a

minimum background on quantitative fields like linear algebra, calculus, statistics

and mathematics of finance.

• Module 1 of the course aims to putting all participants at the same operational level

in these fields.

LEARNING OBJECTIVES

By the end of this module you will be able to:

• Construct histograms and frequency tables from empirical data;

• Compute and interpret major statistical distributions and their parameters used in

risk management;

• Solve systems of simultaneous equations using determinants;

• Compute basic matrix algebra operations;

• Compute the inverse of a matrix and finding its Eigenvalues and Eigenvectors;

• Calculate simple and compound interest applying different rules for counting days in

the financial year;

• Calculate annuities certain and solve generic problems concerning annuities;

• Calculate and build different types of amortisation tables using annuities.

Module 1: Goals and Objectives




www.unitar.org/pft2

• ABBREVIATIONS AND ACRONYMS

• STATISTICS

o Statistical Parameters used in Risk Management

o Theoretical Distributions that are used toAnalyse Risk

o Statistical Tools that are used in Risk Analysis

• MATRIX ALGEBRA

o Basic Concepts

o Operations with Matrices

• MATHEMATICS OF FINANCE

o Basic Concepts

o Annuities-Certain and Amortisation Schedules

• BIBLIOGRAPHY

• ARTICLES OR SHORT PUBLICATIONS

• BOOKS

Table of Content




www.unitar.org/pft3

ALCO: Asset Liability Management Committee.

ALM: Asset Liability Management.

ATM: Average time to maturity.

CaT: Cost-at-Risk.

CBDMS: Computer Based Debt Management System.

COSO: Committee of Sponsoring Organisations of the Tradeway Commission.

CIR: Cox-Ingersoll-Ross Model.DMO: Debt Management Office.

DOD: Debt Outstanding and Disbursed, nominal value of outstanding debt.

EC: Economic Capital

EL: Expected Loss.

GAO: US Government Accountability Office (previous Accounting Office).

INTOSAI: International Organisation of Superior Audit Institutions.

IMF: International Monetary Fund.

JCR: Japan Credit Rating Agency.

MPL: Maximum Probable Loss

NII: Net interest income.

NW: Net worth.

PE Ratio: Share’s Price per Earnings Ratio.

RAROC: Economic capital and risk-adjusted return on capital.

ROE: Return on Equity.

SAI: Superior Audit Institution

SDR: Special Drawing Rights, unit of account of the IMF.

S & P: Standard & Poor’s.

UL: Unexpected Loss

USD: United States Dollar.

VaR: Value-at-Risk.

Abbreviations and Acronyms




www.unitar.org/pft4

1.1. Probabili ty Densit ies

1.1.1 Construction of Histograms and Percentages from Empirical Data

Statistics is a science that finds its roots on empirical observation:the number of times that a certain events are repeated. This isobviously linked to the units in which the observed events aremeasured: it can belong to the set of integer numbers or real

numbers. For instance, if we are dealing with loans, bonds or securities, they are measured in integer numbers, as it is impossible,at least theoretically, to issue half a financial instrument. Now, if weare measuring the worth of each financial instrument, this measurebelongs to real numbers, because this is measured in monetary unitswhich, for practical purposes, can be seen as a continue variable.

Statistical calculations are highly facilitated by the use of computers, and it is assumed in thiscourse that the e-Students have a computer and that are literate using a calculationspreadsheet, and it is highly recommended to reproduce all the numerical examples on aspreadsheet in order to assure the understanding of the different concepts that will bepresented all along the course.

Let us take an example for presenting graphically some numeric data. Let assume that thecash flow payments on the debt of a given debtor agency is given in Table 1.1.1. Table 1.1.1contains four loans which the three first ones are reimbursed in equal annual instalments, andthe last one is a bullet payment at five years. It is assumed that the payments are made endof period, i.e. end of each year, for principal and interest payments.

Table 1.1.1 will allow explaining how a weight average is calculated: for the average rate of interest and the average maturity the weights that are used are the total cash flow for eachone of the loans. It has to be noted that very often the face value or the debt outstanding anddisbursed (DOD), in other words the stock of outstanding debt, may be used instead. In thepresent case we are using the cash flow because it takes into account in the allocatedweights the interest payments that are not necessarily negligible.

If we denote by iw the weight associated to value iv , then the weighted average iscalculated as

1:

(1.1.1)

n

i i

i 1n

i

i 1

w v

wa

w

=

=

=∑

∑

1A barred letter denoting a variable means the weighted average of that variable.

1. Statistics




www.unitar.org/pft5

Table 1.1.1

Loans' Cash Flow = principal + interest

1,000 of Monetary Units

Year Loan 1 Loan 2 Loan 3 Loan 4 TotalFace V. 1,000.00 2,500.00 1,750.00 2,000.00 7,250.00

Rate 10.00% 8.50% 7.50% 9.50% 8.74%

Maturity 10 5 20 5 9.31

2009 200.00 712.50 218.75 190.00 1,321.25

2010 190.00 670.00 212.19 190.00 1,262.19

2011 180.00 627.50 205.63 190.00 1,203.13

2012 170.00 585.00 199.06 190.00 1,144.06

2013 160.00 542.50 192.50 2,190.00 3,085.00

2014 150.00 0.00 185.94 0.00 335.94

2015 140.00 0.00 179.38 0.00 319.38

2016 130.00 0.00 172.81 0.00 302.812017 120.00 0.00 166.25 0.00 286.25

2018 0.00 0.00 159.69 0.00 159.69

2019 0.00 0.00 153.13 0.00 153.13

2020 0.00 0.00 146.56 0.00 146.56

2021 0.00 0.00 140.00 0.00 140.00

2022 0.00 0.00 133.44 0.00 133.44

2023 0.00 0.00 126.88 0.00 126.88

Total 1,440.00 3,137.50 2,592.19 2,950.00 10,119.69

Total cash flow for the period 10,119.69

The average interest rate r is calculated as the weighted average of the individual ratesweighted by the face value, the result is then:

4

i i

i 14

i

i 1

w r 1,000.00 10% 2,500.00 8.5% 1750.00 7.5% 2000.00 9.5%

r 8.74%7,250.00

w

=

=

× + × + × + ×= = =

∑

∑

Similarly, the average maturity is calculated as2:

4

i ii 1

4

i

i 1

w m 1, 000.00 10 2,500.00 5 1,750.00 20 2,000.00 5m 9.31

7,250.00w

=

=

× + × + × + ×= = =∑∑

Once the average parameters of the data are determined, it is also important to have an ideaof the distribution of the data as well. This is undertaken by means of a graphicrepresentation of the data. In this particular case, in which we are dealing with a debt servicecash flow, this is called the “Redemption Profile” when it shows only the amortisation of capital or “Cash Flow Profile” when interest payments are also included, as is the case inTable 1.1.1. This is represented in the form of a histogram that is represented in Figure 1.1.1.

2 The reader should not confuse this concept with the risk indicator Average Time to Maturity (ATM)

that will be explained latter in the course.




www.unitar.org/pft6

Figure 1.1.1Loans Cash Flow Profile

0.00

500.00

1,000.00

1,500.00

2,000.002,500.00

3,000.00

3,500.00

2 0 0 9

2 0 1 0

2 0 1 1

2 0 1 2

2 0 1 3

2 0 1 4

2 0 1 5

2 0 1 6

2 0 1 7

2 0 1 8

2 0 1 9

2 0 2 0

2 0 2 1

2 0 2 2

2 0 2 3

Years

1 , 0

0 0 M o n e t a r y

U n i t s

Loan 1

Loan 2

Loan 3

Loan 4

Total

Figure 1.1.1 is extremely useful for the debt manager as it shows very clearly wherematurities are bunching and can become a risk. In this specific case, Figure 1.1.1 shows thatyear 2013 can present this risk.

In order to complement the analysis shown in Figure 1.1.1 it is also useful to transform thedata into percentages in order to obtain more information. In our present case, thepercentage will be calculated in respect to the total cash flow in the period: 10,119.69. This isto be found in Table 1.1.2.

It is also useful to produce a graphic representation of the relative figures of debt service, inorder to assess the same risk as it was assessed in Figure 1.1.1 but now with relative figures,i.e. percentages. This is represented in Figure 1.1.2 which is the graphic representation of

figures of Table 1.1.2.

It can be seen in Figure 1.1.2 that the bunching of maturities is confirmedand that practically 30.5 percent, i.e. almost one third, of the total cash flowhas to be paid in year 2013. This is obviously a serious liquidity and rollover risk indicator for the debt or the risk manager, as this bunching of maturitiesimplies a heavy draw on resources in order to pay the debt service in thatyear, and either we could have not the liquid resources available or wecould have difficulties in rolling over the debt falling due in that year.

It is useful to show with formulae how Table 1.1.2 is calculated. In fact thiswill also be useful to recall how a double sum operates. If we denote the cash flow of loan j at

year i ijcf , then the percentage of total cash flow to be honoured in year i due to loan j is

denoted ij p , that is calculated in our case as:

(1.1.2)ij

ij 15 4

ij

i 1 j 1

cf p

w= =

=

∑∑




www.unitar.org/pft7

Table 1.1.2

Cash Flow of Loans in Percentage of total Cash Flow

Year Loan 1 Loan 2 Loan 3 Loan 4 Total

2009 1.98% 7.04% 2.16% 1.88% 13.06%

2010 1.88% 6.62% 2.10% 1.88% 12.47%2011 1.78% 6.20% 2.03% 1.88% 11.89%

2012 1.68% 5.78% 1.97% 1.88% 11.31%

2013 1.58% 5.36% 1.90% 21.64% 30.49%

2014 1.48% 1.84% 3.32%

2015 1.38% 1.77% 3.16%

2016 1.28% 1.71% 2.99%

2017 1.19% 1.64% 2.83%

2018 1.58% 1.58%

2019 1.51% 1.51%

2020 1.45% 1.45%

2021 1.38% 1.38%2022 1.32% 1.32%

2023 1.25% 1.25%

Figure 1.1.2Loans Cash Flow Profile in Percentage

0.00%

5.00%

10.00%

15.00%

20.00%

25.00%

30.00%

35.00%

2 0 0 9

2 0 1 0

2 0 1 1

2 0 1 2

2 0 1 3

2 0 1 4

2 0 1 5

2 0 1 6

2 0 1 7

2 0 1 8

2 0 1 9

2 0 2 0

2 0 2 1

2 0 2 2

2 0 2 3

Years

P e r c e n t a g e o f t o t a l

d e b t

Loan 1

Loan 2

Loan 3

Loan 4

Total

Formula (1.1.2) should not present any difficulty, in fact the first index i, denotes the order of the row in the table: if we fix the year, for instance for 2009 which is the first row, i.e. i=1, wehave summing on j, i.e. on the columns j for i=1, we calculate the cash flow corresponding toall four loans in 2009:

4

1j

j 1

w 200.00 712.50 218.75 190.00 1, 321.25=

= + + + =∑

And obviously in our case:

15 4

ij

i 1 j 1

w 10,119.69= =

=∑∑




www.unitar.org/pft8

The reader should note that for obtaining the above result, first we fix the row to obtain thecash flow by year, and then we sum the result of each year over the rows, i.e. fifteen years.

Therefore, for instance, the first percentage of year 2009, year 1, for Loan 1, j=1, in Table1.1.2 is obtained as:

1111 15 4

ij

i 1 j 1

cf 200 p *100 1.98%

10,119.69w

= =

= = =

∑∑

Obviously, the ratio has to be multiplied by 100 in order to express values in percentage. Thereader should note that obviously:

15 4

ij

i 1 j 1

p 100%= =

=∑∑

1.1.2 Cumulative Histograms

To complete the analysis, the calculation of cumulated percentages of the cash flow impliedby the debt service can be compiled. This is represented in Table 1.1.3, where the cumulatedamounts have to add up to 100 percent as it is shown in the last column of the table.

Table 1.1.3 is extremely instructive as it shows the relative percentage of the total cash flowfor the period paid until a given year. For instance, Table 1.1.3 adds the valuable informationthat with the present redemption profile, we would have paid 79.2 percent of the total due

cash flow by the end of 2013. This is to say that almost 80 percent of the required cash flowfor fifteen years has to be paid in the first five years. This is also a valuable risk indicator.These calculations can also be represented graphically, which is represented in Figure 1.1.3.

Figure 1.1.3 also shows that by 2013 around 80 percent of the total cash flow of the periodwould have been paid. This implies a heavy service in the first third of the period comparedto the total horizon of fifteen years.

Formulae for calculating Table 1.1.3 are as follows:

1

11 i1

i 1

CP p 1.98%=

= =

∑




www.unitar.org/pft9

Table 1.1.3

Cumulated Cash Flow of Loans in Percentage

Year Loan 1 Loan 2 Loan 3 Loan 4 Total

2009 1.98% 7.04% 2.16% 1.88% 13.06%

2010 3.85% 13.66% 4.26% 3.76% 25.53%2011 5.63% 19.86% 6.29% 5.63% 37.42%

2012 7.31% 25.64% 8.26% 7.51% 48.72%

2013 8.89% 31.00% 10.16% 29.15% 79.21%

2014 10.38% 12.00% 82.53%

2015 11.76% 13.77% 85.68%

2016 13.04% 15.48% 88.68%

2017 14.23% 17.12% 91.50%

2018 18.70% 93.08%

2019 20.21% 94.60%

2020 21.66% 96.04%

2021 23.04% 97.43%2022 24.36% 98.75%

2023 25.62% 100.00%

Figure 1.1.3Cumulated Loans Redemption Profile in Percentage

0.00%

20.00%

40.00%

60.00%

80.00%

100.00%

120.00%

2 0 0 9

2 0 1 0

2 0 1 1

2 0 1 2

2 0 1 3

2 0 1 4

2 0 1 5

2 0 1 6

2 0 1 7

2 0 1 8

2 0 1 9

2 0 2 0

2 0 2 1

2 0 2 2

2 0 2 3

Years

C u m u l a t e d p e r c e

n t a g e s

Loan 1

Loan 2

Loan 3

Loan 4

Total

Where 11CP stands for the cumulated percentage for the first year of the first loan, similarly,

21CP which is the cumulated percentage for the second year of the first loan is calculated as:

2

21 i1

i 1

CP p 1.98% 1.88% 3.85%=

= = + =∑




www.unitar.org/pft10

1.2. Statistical Parameters used in Risk Management

Before application of statistics, there is necessary to recall some

basic notions which are basic for assessment of risk: economiccapital and risk-adjusted return on capital (RAROC). Economiccapital gives a common framework for measurement of risk, as wellas other applications as to calculate the amount of equity capital that,for instance, a bank should hold. In the banking activity RAROC hasbecome the standard for measuring risk-adjusted profitability, i.e. itallows the comparison of profitability of different transactions. In order to explain all theseconcepts, this section will start with an introduction to economic capital and the relationshipbetween capital, risk and probability of default

3.

1.2.1 Examples of Risk Loss Distribution

The difference between asset’s values minus liability’s values is called capital. This is arelative straightforward concept; however, in reality the value of assets and liabilities varies ondaily basis affecting the value of capital, which has an incidence in the capacity of aninstitution to honour its debts. Therefore, there is a direct relationship between the amount of capital that an institution holds, the amount of risk it takes and the probability of theinstitution’s defaulting.

In order to illustrate this, consider an example in which the activity is to set up a bank. Inorder to do so, let us assume that there are 5 million that are put by a group of investors thatare eager to share profits (or losses), the shareholders, as their participation is throughpurchasing of shares. In addition to the equity that is put by the shareholders, the new bankborrows 95 million from investors that want a relative safe return on their investment: they donot share profits or losses; they are paid on a fixed income basis spelled out on the terms of the loan. Therefore, the bank has 100 million to operate (5 million in shares + 95 millionborrowed). Let us assume that the rate of interest that the 95 million loan bears is 5 percentper annum. In that case, at the end of the year the bank has to pay to its creditors is 99.8million:

95 1.05 99.75× =

The bank in order to make a profit has to invest the total of resources 100 million at a rate of return higher than 5 percent. Let us assume that the bank can buy corporate bonds that pay6 percent interest. In that case, if none of the corporations that have issued those bondsdefaults, the bank will give to the shareholders the difference between the income of corporate bonds minus the payment to the creditors of the 95 million. In order to measure theprofitability of the operation for the shareholders, we calculate the Return on Equity (ROE),expressed in percentage, and which is defined as:

(1.2.1) t t 1t

t 1

E EROE 100

E

−

−

−= ×

Where:

tROE : Return on Equity in period t

tE : Equity at end of period t

t 1E − : Equity at beginning of period t

3 RAROC will be dealt with in the following Module, in this section the examples would be bound to

losses on assets and the related Economic Capital (EC) to minimise the probability of loss.





In our example, ROE is calculated as follows:

The bank receives 106 million and has to pay to creditors 99.75, thus there is now an equityend of period of 6.25 million. ROE is of 25 percent for the period:

1 01

0

E E 6.25 5.00ROE 100 100 25.00%E 5.00− −= × = × =

In order to obtain the above result it is necessary that all corporations paid back the bank’sinvestment. But what would happen if there is a percentage of default on the investedvalues? For instance, what would happen if only 99 percent of the investment is repaid? In

that case, (106 0.99) 99.75 5.19× − = and ROE would fall to 3.8 percent:

1 01

0

E E 5.19 5.00ROE 100 100 3.80%

E 5.00

− −= × = × =

If at the end of the period the losses may be even worst if the percentage of default increases.If the percentage of default increases to 2 percent, then ROE decreases to 17.40− , i.e. there

is a loss for the shareholders of 17.4 percent. And if there is a percentage of default of 10percent among the issuers of the corporate bonds, the shareholders will loose more than thevalue of their equity and the bank goes out of business.

In order to analyse and assess the probability of bankruptcy of the bank, it is possible toelaborate different scenarios. This is shown in Table 1.2.1.

Table 1.2.1Results of Ten Possible Equity Variation Scenarios for the Bank:

Equity 5 and Borrowed Funds 95

Scenario % of LoansRepaid

AssetValueEnd of Period

EquityEnd of Period

Return onEquity

(ROE) in %

1 84.00% 89.04 0.00 -100.00%

2 90.00% 95.40 0.00 -100.00%

3 95.00% 100.70 0.95 -81.00%

4 96.00% 101.76 2.01 -59.80%

5 96.50% 102.29 2.54 -49.20%

6 97.00% 102.82 3.07 -38.60%

7 97.50% 103.35 3.60 -28.00%

8 98.00% 103.88 4.13 -17.40%

9 99.00% 104.94 5.19 3.80%

10 100.00% 106.00 6.25 25.00%

As we do not have any inference on the frequency on which each occurrence may arrive, letus assume that all the ten have the same probability to occur. Under that assumption, it ispossible to plot the possible outcomes in a histogram, making intervals of occurrences. Thedata are summarised in Table 1.2.2.





Table 1.2.2Frequency of Occurrences:

Asset Variation Scenarios End of Period

Interval of Asset

Value End of Period

Frequency of Occurrence

89<x<95 1

95<x<100 1

100<x<101 1

101<x<103 3

103<x<105 2

105<x<106 2

The histogram using data of Table 1.2.2 is represented in Figure 1.2.1.

Figure 1.2.1Frequency of Occurrences:


0

1

2

3

89-95

95-100100-101

101-103

103-105

105-106

The histogram in Figure 1.2.1 gives a rough indication of the probability distribution for the

asset value end of period. It shows, for example, that there is a 20 percent probability that theasset value will be less than 100 at end of period, i.e. that there is 20 percent chances thatthe bank will go bankrupt.

The conclusion stated in the above paragraph is obtained by taking the number of cases inwhich the asset value is less than 100 and divided by the totality of cases in the scenario:

2100 20%

10× =

This would be more evident if we calculate the cumulated frequencies implied by the tenscenarios. This is summarised in Table 1.2.3.





Table 1.2.3Cumulated Frequency of Occurrences:


Interval of AssetValue End of Period

Cumulated Frequency of Occurrence

89<x<95 1

x<100 2

x<101 3

x<103 6

x<105 8

x<106 10

The cumulated histogram plotted with the data shown in Table 1.2.3 are shown in Figure1.2.2.

Figure 1.2.2Cumulated Frequency of Occurrences:


0

1

2

3

4

5

6

7

8

9

10

89<x<95

x<100

x<101

x<103

x<105

x<106

But one would be interested in knowing what would happen if the bank would operate with alarger share of equity. Let us see how these scenarios and the final result would be affectedin case the bank would operate with an initial equity of 10 and would be borrowing only 90instead of 95.

In that case, the same ten scenarios are represented in Table 1.2.4, that shows that ROEwould be less important in case of non-default by creditors. However, the probability of thebank going bankrupted decreases from 20 to 10 present, which is a substantial increase insecurity margin.





Table 1.2.4Results of Ten Possible Equity Variation Scenarios for the Bank:

Equity 10 and Borrowed Funds 90

Scenario % of LoansRepaid

AssetValue Endof Period

EquityEnd of Period

Return onEquity (ROE)

1 84.00% 89.04 0.00 -100.00%

2 90.00% 95.40 0.90 -91.00%

3 95.00% 100.70 6.20 -38.00%

4 96.00% 101.76 7.26 -27.40%

5 96.50% 102.29 7.79 -22.10%

6 97.00% 102.82 8.32 -16.80%

7 97.50% 103.35 8.85 -11.50%

8 98.00% 103.88 9.38 -6.20%

9 99.00% 104.94 10.44 4.40%

10 100.00% 106.00 11.50 15.00%

In reality, there are an infinite number of possible values for the asset value end of period,which implies that instead of working with an histogram it would be more practical to useprobability density functions for which there are tables in order to build our tests andconfidence intervals. This will be dealt with in the next section.

The example given above shows that there is a close relationship between initial capital, therisk taken and the probability that the bank would going bankrupted. Economic capital

4is one

of the most important variables because it provides a unique framework to measure risks witha single metric

5. Therefore, economic capital (EC) is the net value the bank must have at

beginning of period to ensure that there is only a “small” probability of defaulting within thatperiod. The “net value” is the value of assets minus liabilities, as it was said before. The“small” probability is the probability that that corresponds to the bank’s credit rating. For instance, an A-rated financial institution is assumed to have less than 0.1 percent probabilityof default within a given year.

It is impossible to actually observe the probability of default of a single bank, because anysingle bank would either default or not default. However, by looking at the average defaultrate of all banks in a given rating grade, it is possible to link the credit rating to a probability of default. This is illustrated in Table 1.2.5

6.

4Economic capital is the amount of capital—assessed on a realistic basis—which a firm requires tocover the risks that it is running such as market, credit and operational risks. It is the amount of moneywhich is needed to secure survival in a worst case scenario. Firms and financial services regulatorsshould then aim to hold risk capital of an amount at least equal to economic capital. Typically,economic capital is calculated by determining the amount of capital that the firm needs to ensure that

its realistic balance sheet stays solvent over a certain time period with a pre-specified probability.Therefore, economic capital is often calculated as value at risk. The concept of economic capitaldiffers from regulatory capital in the sense that regulatory capital is the mandatory capital theregulators require to be maintained while economic capital is the best estimate of required capital thatfinancial institutions use internally to manage their own risk and to allocate the cost of maintaining

regulatory capital among different units within the institution.5 Metric is used with the meaning of “standard of measurement”.6

Example taken from Marrison (2002).





Table 1.2.5Correspondence between Credit Rating and

Annual Probability of Default

S&P Rating Probability of Default

AAA 0.01%

AA 0.04%

A 0.12%

BBB 0.50%

BB 3.00%

B 11.00%

CCC 28.00%

Default 100.00% Linking the examples given above with the definition of capital, it is clear that theshareholder’s equity is a good shield against default, because—as shown in the numericalexample—the larger the equity the smaller the probability of default. In fact, EC is the amountthat shareholders must invest into the bank at beginning of period, so that the bank can carryout its planned business and maintain its targeted credit rating.

1.2.2 Economic Capital for Different Types of Risks

For the credit risk of lending operations, as the examples given in the previous section, therequirement of EC depends on the probability distribution of the losses as it was illustratedwith the numerical examples. But let us define first what is understood by “credit risk”:

Credit Risk

It is the risk of non-performance by borrowers on loans or other financialassets or by a counterparty in financial contracts.

Actually, the examples given in the previous section were dealing with this kindof risk, as we were taking into account the probability of non-repayment, i.e. thepercentage of non-performing bonds. Therefore, if the reader understood thecalculations explained in that section, there would be any difficulty inunderstanding the formal mathematical manipulations of this concept.

There are three key variables to describe the risk loss distribution: the expected loss (EL), theunexpected loss (UL) and the maximum probable loss (MPL). The statistical interpretation of these statistics will be deal with later, for the moment let us say that EL is the mean of the

loss and UL is the standard deviation. The bank should expect to loose, on average, EL per period and the MPL is the bound of a confidence interval that implies that there is a very smallprobability that the losses that may arrive should be larger than MPL.

The value of the purchased bonds at the end of period will depend on the coupon rate, on onehand, and on the other hand on the percentage of bonds that default. This relationship canbe expressed as:

(1.2.2) ( ) ( )t t 1 AA A 1 i 1−= + − δ

And

(1.2.3) ( )t t 1 DD D 1 i−= +





Where:

tA : Value of assets end of period t

t 1A − : Value of assets beginning of period t

tD : Value of debt end of period t

t 1D − : Value of debt beginning of period t

Ai : Coupon rate for period t

δ : Percentage of the bonds that default in period t

Di : Interest rate paid on debt for period t

The economic capital remaining at the end of the year is the value of the assets minus therepayable debt. This is expressed as:

(1.2.4) t t tEC A D= −

Where:tEC : Economic capital at end of period t

tD : Due debt, including interest, at end of period t

Replacing (1.2.2) and (1.2.3) into (1.2.4) we can write:

(1.2.5) ( ) ( ) ( )t t 1 A t 1 DEC A 1 i 1 D 1 i− −= + − δ − +

Given that the bank’s goal is to avoid bankruptcy, the minimum value of tEC that it can afford

is zero, i.e. the maximum bearable loss without incurring bankruptcy would be equal to theequity end of period; but actually, this precisely the maximum possible loss (MPL) that the

bank could incur. Therefore, the maximum value—in order to avoid bankruptcy—that δ can

take in (1.2.5) should be equal to MPL expressed as a percentage of t 1A − , that will be

denoted as MLPδ , where:

(1.2.6) MLP

t 1

MPL

A −

δ =

Then, we can write (1.2.5) as:

( ) ( ) ( )t 1 A MPL t 1 D0 A 1 i 1 D 1 i− −= + − δ − +

And then to obtain the value of t 1D − :

(1.2.7)( ) ( )

( )t 1 A MPL

t 1

D

A 1 i 1D

1 i

−−

+ − δ=

+

But the required capital beginning of period is:

(1.2.8) t 1 t 1 t 1EC A D− − −= −





Replacing (1.2.7) into (1.28) and rearranging terms, we can write:

(1.2.9)( ) ( )

( )D A MPL A

t 1 t 1

D

i i 1 iEC A

1 i− −

− + δ +=

+

In (1.2.9), ( )D Ai i− represents the debt interest that must be paid minus the interest earned

from assets, i.e. the smaller this difference the smaller the required initial amount of capital.

MLPδ represents the “worst-scenario” loss of assets, therefore, ( )MPL A1 iδ + implies that a

high risk increases the required initial amount of capital, and the other way around. The term

( )D1 i+ in the denominator means that the highest Di the highest the required initial amount

of capital.

The expression in (1.2.9) is quite complex, however, making some hypotheses it can be

simplified. Let assume that Ai is the average interest needed in order to cover EL. Then, we

can write:

(1.2.10) A D D

t 1

ELi i i

A −

= + = + α

Where

t 1

EL

A −

α =

Replacing (1.2.10) into (1.2.9) we find:

(1.2.11) D

t 1 t 1 MPLD

1 i

EC A 1 i− −

⎛ ⎞+ + α

= δ − α⎜ ⎟+⎝ ⎠

But actually, if we expect α to be small, and then we can write:

( )Dt 1 t 1 MPL t 1

D t 1 t 1

1 i MPL EL1 EC A A

1 i A A− − −

− −

⎛ ⎞+ + α≈ ⇒ ≈ δ − α = −⎜ ⎟+ ⎝ ⎠

Then, we conclude that:

(1.2.13) t 1EC MPL EL− ≈ −

(1.2.13) is a very intuitive logical conclusion: the economic capital is the difference betweenthe maximum possible loss, MPL, minus the expected loss, EL. In fact, we can deduct from

(1.2.9) the exact value of t 1EC − in similar terms as in (1.2.13):

( ) ( )( )

( )( )

( )( )

D A MPL A A D A

t 1 t 1 t 1 t 1

D t 1 D D

i i 1 i 1 i i iMPLEC A A A

1 i A 1 i 1 i− − − −

−

− + δ + + −= = −

+ + +

(1.2.14)( )( )

( )( )

A D A

t 1 t 1

D D

1 i i iEC MPL A

1 i 1 i− −

+ −= −

+ +





If the rate on the bonds, Ai , is such that it equals the rate needed to honour the borrowed

funds, Di , (1.2.14) is equal to MPL . However, the economic capital takes already into

account the expected loss, EL , and it has to be subtracted to MPL , in which case (1.2.14)shows exactly the same result as (1.2.13).

Market Risk

Market risk is associated with changes in market prices, such as interest rates, exchangerates, and commodity prices. Therefore, market risks are associated with trading andinvesting, in a general sense.

Market risks require also an economic capital (EC) to play the role of a contingent reserve.The required economic capital can be considered to be the amount of money that theshareholders put in reserve at the beginning of the period so that the trading operation cancarry out its strategy and maintain the desired debt rating.

The trading of a financial institution will imply taking market risks through carrying out a set of

investment strategies within predefined limits. The profitability of each strategy has aprobability distribution. In fact, this is associated to the technique called value-at-risk, whichwill be dealt with in the third module, and we would need to determine the probabilitydistribution and to calculate the lower bound of the set of maximum losses.

For sake of simplicity and to help the reader to understand concepts before going into thedetail of calculations, let us assume that we have the distribution, a normal curve, and that wehave calculated the lower bound of the set of maximum losses. This is represented in Figure1.2.3.

Figure 1.2.3 represents the distribution of profits, which is symmetric in relation to the averageprofits, i.e. we have the same probability of making more or less than this average: slightlyless than 50 percent on each side, because the probability of obtaining the average profit is

not nil.

Figure 1.2.3Lower Bound of the Set of Maximum Losses for Market Risk

Figure 1.2.3 illustrates that there is a probability ω over a period that the loss will be worst

than W. This is expressed in mathematical notation as:

[ ]P profit W< = ω





Where P stands for density probability function. The example would be enough for thereader to understand the methodology, let us add that the techniques of value-at-risk wouldallow as to find the value for W. Once W is estimated, then the determination of the requiredeconomic capital is straightforward too, the simplest way to cope with this risk is to invest W—

or if we want to increase safeness, a multiple of W—at the risk-free rate in the market, f i , like

in a savings account deposit at the beginning of period. In that case the required EC iscalculated as follows:

(1.2.15)

f

WEC

1 i=

+

Implying that at the end of the period the reserve is enough to cover the risk:

(1.2.16) ( )f EC 1 i W+ =

The reader should notice that (1.2.15) and (1.2.16) assume a very simple investment

decision: investing at the risk-free rate. However, alternative investment assumptions may betaken into account in the analysis in order to calculate EC.

Operational RiskOperational risks include a range of risks, such as errors in the various stages of executingand recording transactions; inadequacies or failures in internal controls or in systems andservices; reputational risk; legal risk; security breaches; or natural disasters that affectbusiness activity.

Conceptually, the calculation of the required economic capital for covering operational risks issimilar to the one used for market risks, however, the probability distribution may have adifferent shape. The most challenge issue in order to estimate the required economic capitalfor operational risks is to find the data—accurate enough—in order to be able to estimate the

characteristics of the probability distribution for dealing with this kind of risk.





1.3. Theoretical Distributions that are used to Analyse Risk

1.3.1 Mean, Variance, Standard Deviation, Skew and Kurtosis

The properties of numerical observations, like those presented in the lastsection, can be described as random variables that follow a specificprobability distribution. The properties of the random variable can bequantified in terms of specified parameters, namely mean, variance,standard-deviation, skew and kurtosis.

We have to bear in mind, however, that there are two different conceptsregarding these parameters. The first concept is the set of parametersthat describe the actual underlying process that produces the random

results, which we may identify as the theoretical distribution that describes the observedresults.

The second concept is the set of parameters that we can calculate from our sample, thereforeproducing estimations or estimates of mean, variance, standard-deviation, skew andkurtosis. These estimates will approach with a certain degree of confidence the “true”theoretical values, but they are not necessarily exactly equal to those.

In the real world, we cannot know the true probabilistic distribution, we can only observe andcalculate results obtained from samples of the underlying statistical process, and then usethese as the best estimates to understand the underlying process. We will use Greek lettersto denote the theoretical values and Latin letters to denote the estimates of those parameters.

Mean of a Distribution

The mean represents the centre of gravity of a distribution, i.e. the weighted observations are“equally” distributed below and above the mean value. The mean is, normally, denoted by theGreek letter μ . This letter denotes, in general, the theoretical mean value of a probability

distribution. Assuming that f(x) is the probability density function for x, and f(x)dx is the

probability of x falling in the range dx, we can write:

(1.3.1) xf(x)dx+∞

−∞μ = ∫

This would imply the knowledge of the real probabilistic distribution f(x), fact that in realityseldom happens. This is because we have only limited information, for instance a sample of a larger population. Therefore, one of the statistician tasks is to, first, estimate theparameters of a distribution, and second, to deduce the type of the distribution he is dealingwith. This may sound quite difficult; however, we will see that this is a manageable problem.

As we calculated in Section 1.1 the average rate of interest and the average maturity of a setof loans, we calculate in the same way an estimation of the value of μ . This is expressed in

(1.3.2):

(1.3.2)

m mi

i i i

i 1 i 1

n 1x x n x

n n= =

= =∑ ∑

Where:

m

i

i 1

n n=

= ∑





The meaning of (1.3.2) is that we calculate an estimation of the mean μ using a sample of n

observations, of which the value ix has been observed in times and we have m clusters of

same observed values. Actually, in

nis an estimation of the probabilityf(x)dx for ix to fall in

the range dx, i.e. to be observed in times out of n in our sample. The formula expressed in

(1.3.2) is called an estimator of μ . When this estimator is applied to values and x takes a

numerical value, this value is called an estimate or estimation of μ .

It can be shown, that (1.3.2) is an unbiased estimator of μ 7. This means, roughly, that x

tends in probability to μ , and is expressed as:

(1.3.3) ( )E x = μ

Formula (1.3.3) reads that the mathematical expectation of x equals μ . This result justifies

the application of (1.3.2) for estimating the theoretical value μ of a distribution.

Let us take the asset values of Table 1.2.4, where in =1 for all ix , as each value has been

observed only once:

( )1

x 89.04 95.40 100.70 101.76 102.29 102.82 103.35 103.88 104.94 106.00 101.0210

= + + + + + + + + + =

The value x 101.02= means that the “weights” of the values observed are in equilibrium

below and above this value. If we look at Figure (1.2.1) we can see that the value 101.02 isat the beginning of the class 101-103, and intuitively we can see that at the left and the rightof this value the weights, which are represented by the surface of the rectangles on the figure,

are evenly distributed.

Variance and Standard Deviation of a Distribution

The standard deviation gives a measure of the degree to which the random results may bedistributed away from the mean. This is a key statistics, in particular for risk measurement,because it gives a way to measure how different the outcomes might be from the desiredresult, i.e. to avoid a certain loss.

In order to measure how far the results may be away from the mean, we have to take thedifferences of the observed values minus the estimation of the mean. However, if we do thatwith the values as such, given that the mean is the centre of gravity, the sum of differencesequals zero:

n

i

i 1

(x x) 0=

− =∑

Therefore, in order to measure the dispersion around the mean we have to avoid minus singsin the differences. We obtain this by taking the square value of the differences andcalculating the mean of these differences. The statistics calculated in this way is calledvariance and for the theoretical distribution is calculated as follows:

7 The demonstration of this statment is beyond the aim of our course. For the student it is enough to

know that the proposed estimator is an unbiased estimator of the theoretical mean.





(1.3.4) ( )22 x f (x) dx

+∞

−∞⎡ ⎤σ = − μ⎣ ⎦∫

The estimator of the variance is:

(1.3.5) ( )2n2 i

i 1

x xs

n 1=

−=−∑

Where n is defined as in (1.3.2). It can be demonstrated that2s is an unbiased estimator of

2σ , and this is expressed as:

(1.3.6) ( )2 2E s = σ

Formula (1.3.6) reads that the mathematical expectation of 2s equals

2σ . This result justifies

the application of (1.3.5) for estimating the theoretical value2σ of a distribution.

The standard deviation is defined as the square root of the variance:

(1.3.7)2σ = σ

Because (1.3.6), it is considered that (1.3.5) can be utilised as an estimator of the standarddeviation by taking the square root:

(1.3.8)( )

2n

i

i 1

x xs

n 1=

−=

−∑


observed only once. It is more practical to calculate the estimation of the variance with acalculation table. The calculations are shown in Table 1.3.1.

Table 1.3.1Estimation of the Variance and Standard Deviation

ScenarioAsset Value

End of Period

( )2

ix x−

1 89.04 143.472 95.40 31.56

3 100.70 0.10

4 101.76 0.55

5 102.29 1.62

6 102.82 3.25

7 103.35 5.44

8 103.88 8.19

9 104.94 15.38

10 106.00 24.82

Total 234.38





Where x 101.02= and hence:

2 234.38s 26.04

10 1= =

−and s 26.04 5.10= =

Skew of a Distribution

The skew is a measure of the asymmetry of a distribution. As far as risk measurement isconcerned, it tells us whether the probability of winning is similar to the probability of losing.The smaller the absolute value of the skew, the smallest the asymmetry of the distribution,hence higher the probability of observe even cases, i.e. the same probability of losing or winning the same amount. The skew is defined as follows:

(1.3.9) ( )3

ixf x dx

∞

−∞

⎡ ⎤− μ⎛ ⎞ς = ⎢ ⎥⎜ ⎟σ⎝ ⎠⎢ ⎥⎣ ⎦∫

The estimator of (1.3.9) is:

(1.3.10)( ) ( )

3ni

i 1

n x xz

n 1 n 2 s=

⎡ ⎤−⎛ ⎞= ⎢ ⎥⎜ ⎟− − ⎝ ⎠⎢ ⎥⎣ ⎦∑

Where n is defined as in (1.3.2). It can be demonstrated that z is an unbiased estimator of ς , and this is expressed as:

(1.3.11) ( )E z = ς

Formula (1.3.11) reads that the mathematical expectation of z equals ς . This result justifiesthe application of (1.3.11) for estimating the theoretical value ς of a distribution.


observed only once. It is more practical to calculate the estimation of the skew with acalculation table. The calculations are shown in Table 1.3.2.

Table 1.3.2Estimation of the Skew

ScenarioAsset Value

End of Period

3

i

x x

s

−⎛ ⎞⎜ ⎟⎝ ⎠

1 89.04 -12.931

2 95.40 -1.334

3 100.70 0.000

4 101.76 0.003

5 102.29 0.015

6 102.82 0.044

7 103.35 0.095

8 103.88 0.176

9 104.94 0.454

10 106.00 0.930Total -12.547





Where x 101.02= and s 5.10= , hence:

( )10

z 12.547 1.7439 8

= − = −×

Kurtosis of a Distribution

The kurtosis is useful describing extremes events, i.e. losses that are so bad that canhappen only with very little probability. However, they may happen. Let us assume that thisbad event has a probability of 1 to 1000 to happen. In order to understand the application of kurtosis, let us take two hypothetical portfolios, both with the same mean, standard deviationand skew, but with different kurtosis. Every 1000 days, the portfolios might be expected tosuffer a “bad” loss. In these extreme events, the portfolio with the higher kurtosis may beexpected to suffer worse losses than the one with lower kurtosis.

The kurtosis is defined as follows:

(1.3.12)

4x

f(x) dx∞

−∞

⎡ ⎤− μ⎛ ⎞κ = ⎢ ⎥⎜ ⎟σ⎝ ⎠⎢ ⎥⎣ ⎦∫

The estimator of (1.3.12) is:

(1.3.13)( )

( ) ( ) ( )

4ni

i 1

n n 1 x xk

n 1 n 2 n 3 s=

⎡ ⎤+ −⎛ ⎞= ⎢ ⎥⎜ ⎟− − − ⎝ ⎠⎢ ⎥⎣ ⎦∑

Where n is defined as in (1.3.2). It can be demonstrated that k is an unbiased estimator of

κ , and this is expressed as:

(1.3.14) ( )E k = κ

Formula (1.3.14) reads that the mathematical expectation of k equals κ . This result justifiesthe application of (1.3.14) for estimating the theoretical value κ of a distribution.


observed only once. It is more practical to calculate the estimation of the kurtosis with acalculation table. The calculations are shown in Table 1.3.3.





Table 1.3.3

Estimation of the Kurtosis

ScenarioAsset Value End

of Period

4

ix x

s

−⎛ ⎞

⎜ ⎟⎝ ⎠

1 89.04 30.350803

2 95.40 1.468791

3 100.70 0.000015

4 101.76 0.000447

5 102.29 0.003860

6 102.82 0.015547

7 103.35 0.043606

8 103.88 0.098926

9 104.94 0.348870

10 106.00 0.908339Total 33.239204

Where x 101.02= and s 5.10= , hence:

( )10 11

k 33.239204 7.259 8 7

×= =

× ×

1.3.2 Normal Distribution

The Normal Distribution, also known as the Gaussian Distribution8

having a bell shape, isthe most commonly used distribution to describe random changes in market risk factors, such

as exchange rates, interest rates and equity prices.

This distribution is very commonly used because the Central Limit Theorem, which statesthat if the number of elements in our sample grows infinitely, then the distribution will tend gobe normally distributed. In real life, however, “infinity” may be quite small and the theorem isapplicable very often. The probability distribution of random variable x normally with meanμ and standard deviation σ is:

(1.3.15)

( )2

2

x

21f (x) e

2

⎡ ⎤−μ⎢ ⎥−

σ⎢ ⎥⎣ ⎦=πσ

The cumulative distribution being:

(1.3.16)

( )2

2

x

21f (x)dx e dx 1

2

⎡ ⎤−μ⎢ ⎥−

∞ ∞ σ⎢ ⎥⎣ ⎦

−∞ −∞= =

πσ∫ ∫

8Because the German mathematician E. Gauss introduced this distribution in the 18

thcentury.





The probability of x taking a value between a and b, where a<b, P(a x b)< < , is calculated

as:

(1.3.17)

( )2

2

x

b 2

a

1e dx P(a x b)

2

⎡ ⎤−μ⎢ ⎥−

σ⎢ ⎥⎣ ⎦ = < <πσ ∫

In practice it is not necessary to calculate the definite integral given in (1.3.17) because thereare tables for different values of the density distribution that are of simple use.

If a random variable x has a normal distribution with mean μ and standard deviation σ , this is

expressed synthetically with the following notation:

(1.3.18) x ~ N(µ,σ)

The tables for the normal distribution are calculated for a random variable with mean 0 andstandard deviation 1. Therefore we have to make a variable transformation to use the tables,

but this is not a serious constraint.

The kurtosis of a normal distribution is constant and equals 3. This distribution is socommonly used that it is defined “excess kurtosis” the value found in the sample minus 3.Distributions that have a kurtosis larger than 3, are said to have leptokurtosis . If we denote

the excess kurtosis δ , then it is calculated as:

(1.3.19)

( )2

2

x4

21 x3 e dx 3

2

⎡ ⎤−μ⎢ ⎥−

∞ σ⎢ ⎥⎣ ⎦

−∞

⎡ ⎤− μ⎛ ⎞⎢ ⎥δ = κ − = −⎜ ⎟⎢ ⎥σπσ ⎝ ⎠

⎢ ⎥⎣ ⎦∫

It can be shown that d is an unbiased estimator of δ , i.e. that ( )E d = δ , where d is

calculated as:

(1.3.20)( )

( ) ( ) ( )

( )

( ) ( )

24ni

i 1

n n 1 n 1x xd 3

n 1 n 2 n 3 s n 2 n 3=

⎡ ⎤+ −−⎛ ⎞= −⎢ ⎥⎜ ⎟− − − − −⎝ ⎠⎢ ⎥⎣ ⎦∑

For the numerical example of the previous section, where we found that k 7.25= , the

application of (1.3.20) gives:

29

d 7.25 3 2.918 7= − × =×

This result shows that the kurtosis of our distribution is very large and in consequence, of losses larger than expected have to be taken into account.





1.3.3 Log-Normal Distri bution

When dealing with variables that cannot take negative values, like interest rates or stock andcommodity prices, the Log-Normal Distribut ion is preferred to the normal distribution. If the

random variable is log-normally distributed, then the logarithm of the variable will be normallydistributed:

1.3.4 Beta Distribution

The Beta Distribution is useful in cases like the credit-risk losses, whichare typically highly skewed. The formula of the Beta distribution is a

complex mathematical expression; however, its calculation is available inmost spreadsheets applications. It only requires two parameters, α and

β , to define the shape of the distribution. They are calculated and

estimated as follows:

( )2

2

1μ − μα = − μ

σand

( )( )

2

2

11

μ − μβ = + μ −

σ

The estimators for α and β are defined as follows:

(1.3.21) ( )2

2x 1 xa xs −= − and ( ) ( )

2

2x 1 x b x 1s−= + −

1.4. Statist ical Tools that are used in Risk Analysis

1.4.1 Confidence Intervals, Confidence Levels and Percentiles

Confidence Intervals provide tools to use probability distributions to assess the possibility of future events to materialise. If we assume a normal distribution, a confidence interval is given

by (1.3.17), i.e. the probability P(a x b)< < of x taking a value between a and b, where a<b.

However, in risk management, we are more concerned by Confidence Levels rather thaninterval levels. This is because we are dealing with extreme cases, as in Section 1.2.2 when

calculating the “minimum of the worst loss” represented in Figure 1.2.3. Actually, Figure 1.2.3represents a normal distribution, and the confidence level W with a probability of ω to occur

is calculated as:

[ ]

( )2

2

x

W 21P profit W e dx

2

⎡ ⎤−μ⎢ ⎥−

σ⎢ ⎥⎣ ⎦

−∞< = = ω

πσ ∫

If we assume that x ~ N(0,1), the confidence levels are shown in Table 1.4.1.





Table 1.4.1Confidence Levels for x ~ N(0,1)

Confidence Level Probability %

-1.00 15.90

-1.64 5.00

-1.96 2.50

-2.00 2.30

-2.32 1.00

-2.50 0.60

-3.00 0.10

The most used confidence level is –2.32, which implies a probability of 1 percent, i.e. aprobability of 99 percent of a favourable outcome. Using a N(0,1) distribution is not a major problem, as all variable x following a normal distribution can be reduced to a N(0,1) through achange of origin and scale: subtracting the mean to all observations and dividing the results

by the standard deviation.

The Percentile is the inverse of the confidence level, e.g. the 1%percentile is –2.32 times thestandard deviation. This percentile, as its corresponding confidence level, is used very oftenas it implies a probability (or percentile) of 99 percent favourable cases.

1.4.2 Correlation and Covariance

In the preceding sections we have been analysing the behaviour of a single variable.However, very often we need to analyse the joint behaviour of a couple of variables. For instance, changes in interest rates and changes in commodity prices, or changes in interestrates and changes in exchange rates. These changes and their interrelation are measuredwith two statistics: the covariance and the correlation coefficient.

The covariance is calculated in a similar way as the variance. Notwithstanding, as we aredealing with two different variables, instead of taking the square value of the differencebetween the variable and its mean, we take the multiplication of the difference between thetwo variables and their means. The estimator of the covariance between two randomvariables, x and y, is expressed in (1.4.1).

(1.4.1) ( ) ( )n

xy i i

i 1

1s x x y y

n 1 =

= − −⎡ ⎤⎣ ⎦− ∑

Let us take a hypothetical example of the variation of loans volume of a bank in relation withthe rate of interest applied to the funds lent. This is summarised in Table 1.4.2.





Table 1.4.2Covariance between Rates of Interest and Volume of Loans

NºRate of Interest

Volumeof LoansMillionUSD

( )ix x− ( )iy y− ( ) ( )i ix x y y− −

1 0.030 159 -0.022500 31 -0.70

2 0.035 155 -0.017500 27 -0.48

3 0.040 140 -0.012500 12 -0.15

4 0.045 140 -0.007500 12 -0.09

5 0.050 138 -0.002500 10 -0.03

6 0.055 130 0.002500 2 0.01

7 0.060 120 0.007500 -8 -0.06

8 0.065 110 0.012500 -18 -0.22

9 0.070 100 0.017500 -28 -0.48

10 0.075 85 0.022500 -43 -0.96

Total 0.525 1,277 -3.173

Mean 0.053 128 Covariance -0.35

Table 1.4.2 illustrates the behaviour of the covariance: when one variable increases and theother decreases the covariance is negative, and the other way around as it is in our example.If the rate of interest increases there is a decrease in the volume of loans made by the bank.If the variation of one of the variables implies sometimes a variation with different sign, thecovariance tends to be small. In our case, the covariance is calculated as:

xy

3.173s 0.35

10 1

−= = −

−

However, we cannot tell when the covariance is large or small in order to draw a conclusionon the joint behaviour of the variables. This is because the size of the covariance is affectedby the units in which the variables are measured, in our case by million of dollars andpercentages, which are of very different dimension. In order to be able to draw a conclusionon the cause and effect of one variable on the other, the concept of Coefficient of Correlation has been introduced.

The Coefficient of Correlation is defined by normalising the covariance dividing it by theindividual standard deviations, by doing so we eliminate the effect of units of measure of thevariables on the statistics. The coefficient of correlation is expressed in (1.4.2).

(1.4.2)xy

xy

x y

σρ =σ σ

The unbiased estimator of (1.4.2) is calculated as follows:

(1.4.3)xy

xy

x y

sr

s s=

It can be shown that: xy1 r 1− ≤ ≤





Table 1.4.3Coefficient of Correlation between Rates of Interest and Volume of Loans

NºRate of Interest

Percentage

Volume of LoansMillion USD ( )

2

ix x− ( )2

iy y−

1 0.030 159 0.000506 980

2 0.035 155 0.000306 745

3 0.040 140 0.000156 151

4 0.045 140 0.000056 151

5 0.050 138 0.000006 106

6 0.055 130 0.000006 5

7 0.060 120 0.000056 59

8 0.065 110 0.000156 313

9 0.070 100 0.000306 767

10 0.075 85 0.000506 1,823

Total 0.002063 5,102Variance: 0.000229 567

Therefore, (1.4.3) allows drawing conclusions on the joint behaviour of the two variables:

• If xyr is close to 1, the variables are highly correlated and vary in the same sense

• If xyr is close to 1− , the variables are highly correlated and vary in contrary sense

• If xyr is close to 0, the variables are not correlated and vary independently the one to

the other

In order for the statistician to know when xyr is sufficiently close to 1, 1− or 0, confident

levels can be fixed using probabilities.

Table 1.4.3 shows the calculations for example in Table 1.4.2 in order to obtain the estimateof the correlation coefficient. The steps for calculating the estimate of the coefficient of correlation with our sample are:

x

0.002063s 0.015138

10 1= =

−and y

5,102s 23.809662

10 1= =

−

Therefore: xy

0.35r 0.97798

0.000229 567

−= = −

×

The above numerical result allows us to conclude that the variations of rate of interest have astrong effect on the volume of loans, and this effect is negative, i.e. if the rate of interestincreases, the volume of loans decreases, or if the rate of interest decreases the volume of loans increases. This last statement is important because it implies that:

xy yxr r =





Sum of Random Variables

Those results will allow us to deal with the measurement of the sum of random variables,which is a problem that we will come across very often. For example, the probability of losson a portfolio is the sum of the probabilities of loss of the instruments that compose theportfolio. Similarly, the probability of a trading loss over a year is the sum of the probabilitiesof trading losses on individual days in the year. Let us take as an example the case in which

our random variable y is the sum of two random variables, i i iz x y= + , then:

(1.4.4)2 2 2

i i i z x y xyz x y z x y s s s 2s= + ⇒ = + ⇒ = + +

The reader should be able to verify the result given in (1.4.4) without any difficulty. However,there is another way, which may be more practical for our purposes, to write the variance of zexpressed in (1.4.4) using (1.4.3):

(1.4.5)2 2 2

z x y xy x ys s s 2r s s= + +

In the case in which we have more than two variables in the sum composing another randomvariable, let us assume m variables, we can generalise the results obtained in (1.4.4) and(1.4.5) as follows:

(1.4.6)

m m m m2

i i i z ij i j

i 1 i 1 i 1 j 1

z x z x s r s s= = = =

= ⇒ = ⇒ =∑ ∑ ∑∑

One common and useful application of (1.4.6) is when the correlation between the variables

ix is zero, i.e. they are independent. This is the assumption made for daily changes in

market variables, for instance, for calculating the variance of the loss over multiple days, for which we assume that market variations are independent from one day to another. Making

this assumption, the variance of the loss over K days is the sum of the variances of each onethe losses for each one of the K days:

(1.4.7)2 2 2 2

K 1 2 K s s s ... s∑ = + + +

If we further assume that:2 2 2

1 2 K s s ... s= = = , i.e. that the variance of the loss is the same for

each day, then (1.4.7) becomes:

(1.4.8)2 2

K K K K s Ks s s K ∑ ∑= ⇒ =

The relation expressed in (1.4.8) shows that the standard deviation of the loss over K days,under the hypotheses made, is the standard deviation of the loss over one day, multiplied bythe square root of K. This relationship is very often used in the management of risk oncumulative losses over multiple days, i.e. to predict how bad the cumulative losses over Kdays might be.





1.4.3 Random Time Series Equations

Many phenomena that are needed to observe for risk management, like interest rates or

equity prices, the value in one time period depends on the value in previous periods. Thisfield of risk management requires a solid knowledge of statistical and econometric tools;therefore, we will limit this section to general notions of the subject.

For instance, one model for stock prices is to say that the price is random walk in which thestock price, SP, on any given date is equal to the previous day’s price plus a random number

t x :

(1.4.9) t t 1 tSP SP −= + x

An assumption often made is that the random number is drawn from a normal distribution,multiplied by the required standard deviation:

If ɛt ~ N(0,1), then: t t t t 1 ts T SP SP s Tε − ε= ε Δ ⇒ = + ε Δ x

Here sε is the standard deviation of the variation of the stock price over one period, and TΔ

is the number of time steps between each period. If sε would be the daily standard deviation

and the time steps days, then TΔ = 1. If sε was the daily standard deviation and the time

steps years, then TΔ = 250, the number of trading days in a year.

In practice, the random walk of equities is closer to being a geometric process, i.e. as thestock price increases, the size of the random changes also increases. This geometricprocess is included in the model expressed in (1.4.10):

(1.4.10) t t 1 t 1 tSP SP SP s T− − ε= + ε Δ

The model can be further refined by adding a mean expected growth rate η :

(1.4.11) t t 1 t 1 t 1 tSP SP SP T SP s T− − − ε= + ηΔ + ε Δ

(1.4.11) is not appropriate for describing the evolution of interest rates. Interest rates havenot a long-term expected growth, but rather over long periods then tend to return to a set level

called the long-term interest rate average mr . There are many models for describing random

interest rates process. The Cox-Ingersoll-Ross (CIR) model:

(1.4.12) t t 1 m t 1 rt r t 1r r c(r r ) s r − − −= + − + ε

Where tr is the rate of interest at time t, c is the rate of convergence towards mr : if t 1r − is

grater than mr the term m t 1c(r r )−− becomes negative, moving down the rates of interest.

The square root of t 1r − scales the disturbances, so that when rates are low, the disturbance

will be low. There is, however, a possibility of creating negative interests rates.





2.1. Basic Concepts

2.1.1 Where does the Concept of Matrix come from?

Matrices are just tables of numbers or variables, like the ones presentedin Section 1, e.g. Tables 1.1, 1.2 and 1.3. As in those cases, rows andcolumns are numbered in order to locate each element in the matrix withthe indexes, i for the row and j for the column. This means that, as the

examples presented with the cash flows the element in the matrix ije ,

where e stands for “element”, means that it is located in the i th row andthe j

thcolumn.

2.1.2 Basic Matrix and Vectors Operations: Addi tions and Subtractions

Matrices form a non-commutative algebra. This means that not all operations arecommutative like in the real numbers algebra. An operation × is commutative if:

a b b a× = ×

The above equality is not necessarily true for matrices, and the reader should be veryattentive to this fact. In this section we will be giving the rules for matrix operations without

mathematical demonstrations, because this is a course on risk management and not on linear algebra.

Matrices have the four basic operations: addition, subtraction, multiplication and division. Aset of numbers, in order to form an “Algebra”, have a “neutral” element for a specificoperation. In the case of real numbers, the neutral element for addition is zero, because for any number r belonging to the set of real numbers:

r 0 r + =

Also, the inverse element of addition is the opposite of any number, that is, adding theopposite of any number to the number itself will yield the neutral element for the addition: 0.

For example, the opposite of r is r − , so r ( r) 0+ − = .

Subtraction is not commutative. For that reason, it is often helpful to look at subtraction as

addition of the minuend and the opposite of the subtrahend, that is: a b a ( b)− = + − . When

a subtraction is written as a sum, then the commutability of addition holds.

Also for the set of real numbers, the neutral element for multiplication is one, because for anynumber belonging to the set of real numbers:

r 1 r × =

2. Matrix Algebra





Also, the multiplicative inverse is the reciprocal of any number, that is, multiplying thereciprocal of any number by the number itself will yield the multiplicative identity:

11r r r 1

r

−× = × =

Division is not commutative, and as it is helpful to look at subtraction as addition, it is helpfulto look at division as multiplication of the dividend times the reciprocal of the divisor, that is:

1a 1a a b

b b

−= × = × . When a division is written as a product, it will obey all the properties of

multiplication.

An (n,m) matrix consists of a rectangular array with n rows and m columns, in which the

elements are either numbers or algebraic expressions. An example of a matrix (3 3)× is

given below:

(2.1.1)

11 12 13

21 22 23

31 32 33

a a a

a a a

a a a

⎛ ⎞

⎜ ⎟⎜ ⎟⎜ ⎟⎝ ⎠

It has to be noted that each element in the matrix has two indexes, the first one indicates the

row order number and the second the column order number. For instance 23a indicates the

element in second row, third column.

2.1.3 Systems of Simultaneous Equations and Determinants

Actually, the “invention” of matrices comes from a form of synthetic writing for dealing with

systems of simultaneous equations, as we will see in the next section. Let us take oneexample of three simultaneous equations with three unknowns.

(2.1.2)

3x y z 1

2x 3y 5z 2

x y z 6

+ − =

− + =

+ + =

Mathematicians found that it would be easier to manipulate these kinds of problems if theabove system is written as:

(2.1.3)

3 1 1 x 1

2 3 5 y 2

1 1 1 z 6

−⎛ ⎞⎛ ⎞ ⎛ ⎞

⎜ ⎟ ⎜ ⎟ ⎜ ⎟− =⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠ ⎝ ⎠

This is because it is easier to write than the original detailed system:

(2.1.4) AX B=

Where:

3 1 1

A 2 3 5

1 1 1

−⎛ ⎞⎜ ⎟= −

⎜ ⎟⎜ ⎟⎝ ⎠

;

x

X y

z

⎛ ⎞⎜ ⎟=

⎜ ⎟⎜ ⎟⎝ ⎠

; and

1

B 2

6

⎛ ⎞⎜ ⎟=

⎜ ⎟⎜ ⎟⎝ ⎠





The synthetic notation expressed in (2.1.4) means that in order to solve the system (2.1.2) we

need to find a matrix1A−

such that1A A I− = , where I is the neutral element for

multiplications of matrices, then the solution of the system of equations given in (2.1.2) wouldbe written as:

(2.1.5)1X A B−=

The matrix1A−

is called the inverse of A. The calculations for obtaining such a matrix will bedealt with in the following section. For the moment, we will be solving the system (2.1.2)using determinants.

The method, probably the most used to solve a system like the one in (2.1.2), is “bysubstitution”. This means that we find the value of one of the unknowns in one of theequations and replace its value into both remaining equations: this brings a system of threeequations and three unknowns to a system of two equations and two unknowns. Then, themethod is repeated: we obtain the value of one of the unknowns in function of the other in oneequation and we replace in the remaining one: we have then one equation with one unknownthat is solved. The value obtained for this unknown is replaced into the other equations andthe process continues until the three unknowns are determined.

Mathematicians have created a “short-cut” for this method—a “recipe” in fact—that will savework and diminish probability of calculation errors. This solving method is called “bydeterminants”. The determinant of a system of equations is the group of coefficients in theequations for each one of the unknowns. In fact matrix A contains the determinant of thesystem. The determinant of A defined by (2.1.2) is written as:

(2.1.6)

3 1 1

A 2 3 5

1 1 1

−

= −

In order to calculate A , let first see how a determinant of dimension (2 2)× , which is called

a “second-order” determinant, is calculated. For example, the second-order determinant in(2.1.7) is calculated below:

(2.1.7) ( )1

1 44

) 22

33

(2= − × = −×

The rule is that the diagonal “left to right”—in red—bears a “plus sign” and the diagonal “rightto left”—in blue—bears a “minus sign”. Therefore, the value of the second-order determinant

in (2.1.7) is equal to 2− . In order to calculate the third-order determinant defined in (2.1.6)following the method called “expanding a determinant by minors”

9, we have to decompose it

into second-order determinants, i.e. decomposing it in a series of determinants of a “minor”order. To do that, we have to choose a row or a column for finding the minors. Let us takethe 3

rdrow in (2.1.6) to find the minors. It comes:

(2.1.8)

3 1 11 1 3 1 3 1

A 2 3 53 5

1

12

15

12 3

11

−− −

= − = − +− −

9

We follow the method “expanding the determinant by minors”. There is another method by theFrench mathematician Gabriel Cramèr published in 1750, but which is valid only for determinants of a

maximum dimension of (3x3).





Each one of the minors is obtained by eliminating in the third-order determinant all theelements in the row and the column of the element. This is shown in (2.1.8) showing the

respective colours for each one of the minor’s element. Then the value of A is:

(2.1.9) ( ) ( ) ( )A 1 5 ( 1 3) 3 5 ( 1 2) 3 311 (11 2)= × × − − × − − × × − − × + × × − − ×⎡ ⎤ ⎡ ⎤ ⎡ ⎤⎣ ⎦ ⎣ ⎦ ⎣ ⎦

Therefore: A 26= −

The elements between parenthesis in (2.1.9)—which are in fact two-order determinants—arecalled the Cofactors of the of the original three-order determinant (or matrix) element. For instance, in our example, that we find the minors of the last row, the cofactor are:

Cofactor of 31a : ( )31

1 1C 1 5 ( 1 3) 2

3 5

−= = × − − × − =⎡ ⎤⎣ ⎦−


3 1C 3 5 ( 1 2) 17

2 5

−= − = − × − − × =⎡ ⎤⎣ ⎦


3 1C 3 3 (1 2) 11

2 3= = × − − × = −⎡ ⎤⎣ ⎦−

It has to be noted that the cofactor 32C bears a minus sign. Actually, there is a rule for

assigning the correct sign to the cofactors. If the sum of the indexes i+j is even, the cofactor sign is plus, if it is odd the sign of the cofactor is minus. In practice this is translated as:

i j

ij ijC ( 1) M+= −

Where ijM denotes the determinant minor of ija .

Determinants of higher order can be calculated by successive expansions of this type. Bychoosing rows of columns containing zeros, some terms can be eliminated. There arevarious rules for transforming a given determinant, which can be used to obtain a row or column most of whose elements are zeros. Determinants have many applications inmathematics and other fields, e.g., in the solution of simultaneous linear equations, which isgoing back to the system of simultaneous equations expressed in (2.1.2). The recipe to solve

(2.1.2) has the following steps:

1. Calculate the determinant of the system A . The determinant of the system has to

be different to zero for the system to have a solution10

2. Calculate the determinant of each one of the unknowns of the system3. The solution of the system for each unknown is the determinant of the unknown

divided by the determinant of the system

How to do step 1 has been explained above. Now we are going to deal with steps 2 and 3.

10 The solution of the system geometrically gives the point in three dimensions (x, y, z), where the three

equations intersect each other. If the determinant of the system is equal to zero, this means that at leasttwo of the equations are parallel, therefore they do not cross each other and there is no common point

(x, y, z) that satisfies the three equations simultaneously.





Step 2: calculate the determinant of the unknowns of the system¨

Each one of the columns of A , see (2.1.8), are composed by the coefficients of each one of

the unknowns in each one of the equations: column 1 is composed of the coefficientsmultiplying x , column 2 is composed of the coefficients multiplying y and column 3 is

composed of the coefficients multiplying z . The determinant of each one of the unknowns isdefined as the determinant of the system where the column corresponding to the unknownsearched is replaced by the vector B defined in (2.1.4). Then we have:

1

2

1 1

x 3 5 0

1 16

−

= − =

3 1

y 2 5

1

2

6

91

1 1

−

= = −

1

2

3 1

z 2 3 65

1 1 6

= − = −

In each one of the above determinants the coefficients of the concerned variables have been

replaced by vector B elements that are printed in red. The value of the determinant of eachvariable has been calculated using the same method as the one exposed in (2.1.9).

Step 3: calculate the value of the unknowns of the system:

x 0x 0

A 26= = =

−

y 91y 3.5

A 26

−= = =

−

z 65z 2.5

A 26

−= = =

−

Replacing the values into (2.1.2) it easily verified that the triplet ( )x 0;y 3.5;z 2.5= = =

satisfies the system11

.

11

The reader is invited to verify all the calculations. It has to be noted that it is easy to create atemplate in Excel in order to solve a third-order determinant by expanding it to second-order

determinants. For higher order determinants, there is also possible to programme an Excel spreadsheet.





2.2. Operations with Matrices

One of the important notions that we have to keep in mind when operatingwith matrices is the dimension of the matrix. The dimension of the matrix

is defined as the number of rows and the number of columns that composethe matrix. For instance, in the system written in (2.1.4) matrix A is of

dimension (3 3)× , matrices X and B are of dimension (3 1)× 12. This is

written as (3 3)×A , ( 3 1)×X , and ( 3 1)×B when it is necessary to indicate the

dimension of the matrices, but in general the dimension is omitted,however, we have to bear always in mind the dimension of the matrices we

are working with in order to avoid errors.

2.2.1 Adding and Transposing Matrices

For instance, 21a is the element of the matrix situated in the second row, first column, and in

general the elementij

a denotes the element situated in the ith

row and the jth

column. A

matrix can be multiplied by a constant, for instance:

(2.2.1)

11 12 13 11 12 13

21 22 23 21 22 23

31 32 33 31 32 33

a a a ka ka ka

k a a a ka ka ka

a a a ka ka ka

⎛ ⎞ ⎛ ⎞⎜ ⎟ ⎜ ⎟=⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

For instance, a practical example of (2.2.1) is:

1 1 1 5 1 5 ( 1) 5 1 5 5 5

5 2 0 1 5 2 5 0 5 ( 1) 10 0 51 1 1 5 1 5 1 5 ( 1) 5 5 5

− × × − × −⎛ ⎞ ⎛ ⎞ ⎛ ⎞⎜ ⎟ ⎜ ⎟ ⎜ ⎟

− = × × × − = −⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟ ⎜ ⎟− × × × − −⎝ ⎠ ⎝ ⎠ ⎝ ⎠

Matrices can be added, but they have to be of the same dimension, because each element of each matrix is added to the corresponding element of the other matrix, as shown in (2.2.2):

(2.2.2)11 12 11 12 11 11 12 12

21 22 21 22 21 21 22 22

a a b b a b a b

a a b b a b a b

+ +⎛ ⎞ ⎛ ⎞ ⎛ ⎞+ =⎜ ⎟ ⎜ ⎟ ⎜ ⎟+ +⎝ ⎠ ⎝ ⎠ ⎝ ⎠


3 2 4 0 3 4 2 0 7 2

4 1 2 3 4 ( 2) 1 3 2 4

− + − + −⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞+ = =⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟− + − +⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠

Finally, the transpose of a matrix (n m)×M is the (m n)× matrix formed by interchanging the

elements of the rows and columns of M. It is denoted by M′ . The jth

row of M′ is the transposeof the j

thcolumn of M and vice versa. An example is given below:

12 The reader should note that we use bold letters to denote matrices, in order to differentiate those from

single scalars or variables.





(2.2.3)

11 12

11 21 31

21 22

12 22 32

31 32

m mm m m

m m 'm m m

m m

⎛ ⎞⎛ ⎞⎜ ⎟= ⇒ = ⎜ ⎟⎜ ⎟⎝ ⎠⎜ ⎟

⎝ ⎠

M M


5 25 3 1

3 02 0 4

1 4

⎛ ⎞⎛ ⎞⎜ ⎟= ⇒ = ⎜ ⎟⎜ ⎟ ⎝ ⎠⎜ ⎟

⎝ ⎠

M M'

2.2.2 Dimension and Conformability of Matrices

The way to write the system of equations (2.1.2) in the form of (2.1.4) gives straightforwardthe rule for multiplications of matrices: in order to obtain the first equation of system (2.1.2)we need to multiply each element of the first row of matrix A by each element of the firstcolumn of matrix X13, and subsequently for the two other equations. The example allows usto define the rule for multiplying matrices: the pre-multiplier, i.e. the left matrix in theoperation, has to contain the same number of columns as the post-multiplier, i.e. the right

matrix in the operation. In our case this is true, because the dimension of A is (3 3)× , i.e.

three rows and three columns, and X is (3 1)× , i.e. three rows and one column. It has to be

noted that the result of the multiplication is B, which dimension is (3 1)× , i.e. the result has

the same number of rows that the pre-multiplier and the same number of columns as thepost-multiplier. Writing the dimension of each matrix as an index, we have:

(2.2.4) (3 3) (3 1) (3 1)× × ×=A X B

The operation that is expressed in (2.2.4) is non-commutative—because given the matricesdimensions—we can multiply AX, but not XA. If the pre-multiplier matrix has the samenumber of columns as the post-multiplier matrix has rows, it is said that the matrices areconformable. Therefore, the operation AX is possible because in that order, A first and X second, are conformable, whereas the operation XA is not conformable and thus impossibleto perform. In addition to that, the result of the multiplication has a dimension equal to thenumber of rows of the pre-multiplier matrix and to the number of columns of the post-multiplier matrix.

Let us take another example with three matrices, for instance M and N as multipliers and Q asthe result of the multiplication. Let us assume that M and N are conformable, in particular

they dimensions are respectively (3 3)× and (3 2)× . Then we can write, along what has

been explained, that:

(2.2.5) (3 3) (3 2) (3 2)× × ×=M N Q

That is, Q has the same number of rows as M and the same number of columns as N. Now,let us see how the detailed operations work. In fact they are very much as the operationimplicitly shown in (2.1.2):

13 In fact a matrix that contains only one row or one column is a “vector”, but the algebra of matrices is

applicable to vectors, making those a “type of matrix” for calculation purposes.





(2.2.6)

11 12 13 11 12 11 12

21 22 23 21 22 21 22

31 32 33 31 32 31 32

m m m n n q q

m m m n n q q

m m m n n q q

⎛ ⎞ ⎛ ⎞ ⎛ ⎞⎜ ⎟ ⎜ ⎟ ⎜ ⎟=⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠ ⎝ ⎠

Following the implicit rule of (2.1.2), i.e. that each of the elements of the first row of M ismultiplied by each of the elements of the first column of N, and their sum will give the firstelement in the first row and first column of Q. This is written as:

3

1h h1 11

h 1

m n q =

=∑

Writing i for the row and j for the column, then we can generalise and write the generalformula in order to calculate each element of Q:

(2.2.7)

3

ih hj ij

h 1

m n q =

=

∑

Let us take a numerical example in order to calculate ijq :

(2.2.8)

11 12

21 22

31 32

2 3 1 1 2 q q

6 1 2 2 2 q q

3 5 1 1 3 q q

−⎛ ⎞⎛ ⎞ ⎛ ⎞⎜ ⎟ ⎜ ⎟ ⎜ ⎟− =⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟ ⎜ ⎟− −⎝ ⎠ ⎝ ⎠ ⎝ ⎠

Let us take for instance 21q , element of Q. Following the rule expressed in (2.2.7), it comes

that:

21q ( 6) 1 1 2 2 1 2= − × + × + × = −

Repeating the operation for each one of the elements of Q, we obtain:

2 3 1 1 2 7 7

6 1 2 2 2 2 4

3 5 1 1 3 8 7

−⎛ ⎞⎛ ⎞ ⎛ ⎞⎜ ⎟⎜ ⎟ ⎜ ⎟= − = − −⎜ ⎟⎜ ⎟ ⎜ ⎟⎜ ⎟⎜ ⎟ ⎜ ⎟− − − −⎝ ⎠⎝ ⎠ ⎝ ⎠

Q

Variance-Covariance Matrix

One common application of matrixes in statistics is the calculation of the variance-covariance matrix. In Section 1.4.2 we defined the covariance between two randomvariables. This is written as:

( ) ( ) ( )

( ) ( ) ( )

( ) ( )

( ) ( )

( ) ( )

1 1

2

1 i n x xy

xy i i 2

1 i n xy y

n n

x x y y

x x x x x x s s1S x x y y

y y y y y y s sn 1

x x y y

⎛ − − ⎞⎜ ⎟⎜ ⎟ ⎛ ⎞⎛ − − − ⎞⎜ ⎟= =− − ⎜ ⎟⎜ ⎟ ⎜ ⎟− − −− ⎜ ⎟⎝ ⎠ ⎝ ⎠⎜ ⎟⎜ ⎟− −⎝ ⎠

M ML L

L LM M





Where xyS is the estimator of xyΣ , the theoretical variance-covariance matrix. xyS is the

product of two matrixes, the elements of each of them is the difference of the value taken bythe variable minus the estimation of its mean. In the case, in which we have two variables as

in Section 1.4.2, the dimension of the first matrix is ( )2 n× , where n is the number of

observations in our sample. The second matrix, which is the transpose of the first one, has adimension ( )n 2× . Therefore the dimension of xyS in our case is ( )2 2× , and is a

symmetric square matrix. The matrix is symmetric because xy yxs s= . In the case in which

we deal with m random variables in our sample, the dimension of xyS is ( )m m× , which is

also a symmetric square matrix.

2.2.3 Inverse of a Matrix

The system of equations written in matrix form in (2.1.4):

=AX B

Means that we need to find a matrix1−

A such that1− =A A I , where I is the neutral element

for multiplications of matrices, then the solution of the system of equations given in (2.1.2)would be written as in (2.1.5):

1−=X A B

Matrix1−

A is called the Inverse Matrix of A . The calculation of 1−

A is very similar to thecalculations for solving the system (2.1.2). We will give the steps to follow in order to inversea matrix without giving a formal mathematic demonstration.

Inverse of a matrix

The inverse of a matrix is the transposed cofactors matrix divided by the determinant of thematrix. Therefore, a matrix can be inversed only if its dimension is square—it has the samenumber of rows as number of columns—and if its determinant is different from zero.

Let us take matrix A of system (2.1.2) to show the steps in order to calculate1−

A .

3 1 1

2 3 5

1 1 1

−⎛ ⎞⎜ ⎟= −⎜ ⎟⎜ ⎟⎝ ⎠

A

Step 1: Calculate the Determinant of AThis has been calculated in Section 2.1, in particular in (2.1.9) and the result that was foundis:

26= −A

We have to recall that the necessary and sufficient condition for a system of equations tohave a solution is that:

0≠A





Step 2: Calculate the Matrix of Cofactors

This is similar to the calculation of a three-order determinant, but we need to calculate thecofactors for ALL the elements of A, not only one row or column as in (2.1.8) and (2.1.9). Also, we have to respect the rules for the sign of the cofactor given in Section 2.1:

(2.2.9)i j

ij ijC ( 1) M+= −

For instance the cofactor 11C of element 11a 3= is calculated as follows:

( ) ( )1 1 2

11

3 5C ( 1) ( 1) ( 3) 1 5 1 8

1 1

+ −= − = − − × − × = −⎡ ⎤⎣ ⎦

For cofactor 21C of element 21a 2= :

( ) ( )2 1 3

211 1C ( 1) ( 1) 1 1 ( 1) 1 21 1

+ −= − = − × − − × = −⎡ ⎤⎣ ⎦

Repeating the operation for each one of the elements of A, we find the matrix of cofactors

AC of A:

11 12 13

A 21 22 23

31 32 33

C C C 8 3 5

C C C 2 4 2

C C C 2 17 11

−⎛ ⎞ ⎛ ⎞⎜ ⎟ ⎜ ⎟= = − −⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟− −⎝ ⎠ ⎝ ⎠

C

Step 3: Transpose the Matrix of Cofactors

The rule for transposing matrices has been given in (2.2.3). The transpose of matrix AC is

matrix'

AC formed by interchanging the elements of the rows and columns of AC :

11 21 31

'

A 12 22 32

13 23 33

C C C 8 2 2

C C C 3 4 17

C C C 5 2 11

− −⎛ ⎞ ⎛ ⎞⎜ ⎟ ⎜ ⎟= = −⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟− −⎝ ⎠ ⎝ ⎠

C

Step 4: Divide the Transpose of the Matrix of Cofactors by the Determinant of A

We have finally that:

1 '

A

0.3077 0.0769 0.07691

0.1154 0.1538 0.6538

0.1923 0.0769 0.4231

−

−⎛ ⎞⎜ ⎟= = − −⎜ ⎟⎜ ⎟−⎝ ⎠

A CA

Taking equality (2.1.5):

1−=X A B





We can now find the values of the unknowns of system (2.1.2) using1−

A :

x 0.3077 0.0769 0.0769 1 0

y 0.1154 0.1538 0.6538 2 3.5

z 0.1923 0.0769 0.4231 6 2.5

−⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟= − − =⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟

⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟−⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠

The reader should have already verified that those values satisfy system (2.1.2)

We can also find the third-order matrix that is the multiplicative identity:

1− =AA I

Then it comes:

3 1 1 0.3077 0.0769 0.0769 1 0 0

2 3 5 0.1154 0.1538 0.6538 0 1 01 1 1 0.1923 0.0769 0.4231 0 0 1

− −⎛ ⎞⎛ ⎞ ⎛ ⎞⎜ ⎟ ⎜ ⎟ ⎜ ⎟

= − − − =⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟ ⎜ ⎟−⎝ ⎠ ⎝ ⎠ ⎝ ⎠I

The reader should verify that the matrix I is the multiplicative identity for all third-order matrix.

2.2.4 Eigenvalues and Eigenvectors of a Matrix

Given a square matrix A , an Eigenvalue λ and its associated Eigenvector V are, by

definition, a pair obeying the relation14

:

(2.2.10) = λA V

Any non-zero vector V satisfying (2.2.10) is called and Eigenvector of A belonging to the

Eigenvalue λ . Equivalently, (2.2.10) can be written as:

(2.2.11) ( ) 0− λ =A I V

Where I is the identity matrix of the same dimension as A . Therefore, matrix λI is a

diagonal matrix where the diagonal are the Eigenvalues of A . However there is a conditionfor this which is related to is an extremely useful theorem for risk management, in particular for Monte Carlo simulation that reads as follows.

Theorem on Existence and Properties of Eigenvalues and Eigenvectors

A squared matrix(n n)×A is similar to a diagonal matrix Λ if and only if A has n

linearly independent Eigenvectors. In this case, the diagonal elements of Λ are the

corresponding Eigenvalues, and1−=Λ B AB , where B is the matrix whose columns

are the Eigenvectors of A .

The demonstration of this theorem is beyond the aim of this course; therefore, we will accept

it without demonstration. It has to be noted that the property1−=Λ B AB implies that matrix

14 The terms Characteristic Value and Characteristic Vector, or Proper Value and Proper Vector are

sometimes used instead of Eigenvalue and Eigenvector.





A can be diagonalised, where Λ is the diagonal. Then A has an extremely usefuldiagonal factorisation:

(2.2.12)1−=A BΛB

Using this factorisation, the algebra of A reduces to the algebra of the diagonal matrix Λ ,which can be deducted in a relatively easier manner.

For instance, let us take a matrix (2 2)×A as follows:

4 2

3 1

⎛ ⎞= ⎜ ⎟−⎝ ⎠

A

Step 1: Finding the Eigenvalues of A

Applying (2.2.11) we deduct that the determinant of ( )− λA I should be zero—because the

vector V is not nil—then:

( )( )

4 2

3 1

− λ⎛ ⎞− λ = ⎜ ⎟− + λ⎝ ⎠

A I

Hence:

( )

4 20

3 1

− λ− λ = = ⇒

− + λA I

2 3 10 0λ − λ − =

This shows that in order to find the Eigenvalues of A we have to solve a polynomial of

second degree, and in general, calculating the Eigenvalues of a matrix of dimension ( )n n×

is equivalent to finding the n roots of an n-degree polynomial. In our example, the polynomial

is easily factorised as: ( )( )5 2 0λ − λ + = , thus the Eigenvalues of A are: 15λ = and

22λ = − . It has to be noted that in cases where factorisation is less obvious, the roots can

be found by applying the quadratic equation formula:

2 b b 4ac

2a

− ± −λ = , where:

2a b c 0λ + λ + = .

Step 2: Finding the Eigenvectors of A

In order to find an Eigenvector of A belonging to the Eigenvalue 15λ = , we have to subtract

Eigenvalue 5 down the diagonal of A to find the matrix:

( )1

4 5 2 1 2

3 1 5 3 6

− −⎛ ⎞ ⎛ ⎞− λ = =⎜ ⎟ ⎜ ⎟− − −⎝ ⎠ ⎝ ⎠

A I





And to solve a system of equations such that:

( ) 11

1 1

12

v1 20 0

v3 6

− ⎛ ⎞⎛ ⎞− λ = ⇒ =⎜ ⎟⎜ ⎟−⎝ ⎠⎝ ⎠

A I V

Or

(2.2.13)11 12

11 12

v 2v 0

3v 6v 0

− + =

− =

The system (2.2.13) has only one free equation, because if we multiply the first equation by

3− we find the second equation. Hence, the system has several non-zero solutions, for

example the couple 2 and 1:

1

2

1

⎛ ⎞= ⎜ ⎟

⎝ ⎠V

1V is an Eigenvector of A belonging to the Eigenvalue

15λ = .

In order to find an Eigenvector of A belonging to the Eigenvalue2

2λ = − , we proceed as

before: subtracting 2− down the diagonal of A , and we find the system of equations in

order to find the elements of 2

V :

(2.2.14)12 22

12 22

6v 2v 0

3v v 0

+ =

+ =

Therefore:

2

1

3

−⎛ ⎞= ⎜ ⎟

⎝ ⎠V

2V is an Eigenvector of A belonging to the Eigenvalue 2

2λ = − .

Step 3: Factorisation and Diagonalisation of A

Now we have to find a matrix B that satisfies1−=Λ B AB :

11 12

12 22

v v 2 1

v v 1 3

−⎛ ⎞ ⎛ ⎞

= =⎜ ⎟ ⎜ ⎟⎝ ⎠⎝ ⎠B And so

3 1

7 71

1 2

7 7

−

−

⎛ ⎞

= ⎜ ⎟⎜ ⎟⎝ ⎠B

And we can verify that the theorem applies:

3 1

1 7 7

1 2

7 7

4 2 2 1 5 0

3 1 1 3 0 2

−

−

⎛ ⎞ −⎛ ⎞⎛ ⎞ ⎛ ⎞= = ⎜ ⎟ =⎜ ⎟⎜ ⎟ ⎜ ⎟⎜ ⎟ − −⎝ ⎠⎝ ⎠ ⎝ ⎠⎝ ⎠

Λ B AB

Unfortunately, this method has some limitations. A general polynomial of order n> 4 cannot besolved by a finite sequence of arithmetic operations and radicals. Therefore, generalEigenvalue algorithms are expected to be iterative. But fortunately, there are many software

packages that contain algorithms for calculation of Eigenvalues and Eigenvectors.





Symmetric Matrices, Eigenvalues and Eigenvectors

An important property of the factorisation of matrices concern symmetric matrices. Asymmetric matrix is a square matrix that is equal to its transpose, and it happens thatcovariance and correlation matrices are symmetric, because the covariance and thecorrelation between x and y are exactly equal to the covariance and the correlation between yand x. The properties of symmetric matrices facilitate the calculations for finding theEigenvalues and Eigenvectors of this type of matrices.

Theorem on Symmetric Matrices, Eigenvalues and Eigenvectors

A square (n n)×A symmetric matrix has n linearly independent Eigenvectors. Moreover, these

Eigenvectors can be chosen such that they are linearly independent to each other. Thus a

symmetric matrix A can be decomposed as '=A BΛB , where B is an orthogonal matrixwhich columns are the Eigenvectors of A, and Λ is real and diagonal, where the diagonal iscomposed of the Eigenvalues of A.

Definition of an Orthogonal Matrix

An orthogonal matrix is such a matrix that its transpose is equal to its inverse:1' ' '−= ⇒ = =B B B B BB I

Therefore, in the case of symmetric matrices—covariance and correlation matrices—thecalculations are highly simplified, as for obtaining the inverse of the Eigenvectors matrix itsuffices to transpose it.

The covariance matrix—or correlation matrix—factorisation is necessary for Monte Carlo

simulations, as matrix B keeps the correlation structure among the different variables andthis is very important for the consistency of simulation scenarios.





3.

3.1. Basic Concepts

The payment of interest as reward for the use of capital is anestablished part of economic life. It may be regarded as a reward paidby the borrower, who is given the use of capital, to the creditor, theowner of the capital. In financial theory both capital and interest areexpressed in terms of monetary units.

In practice the interest which it has been agreed will be paid for the useof capital is payable at stated intervals of time. The rate of interest

which operates during one on these intervals is found to by dividing the amount of interestagreed to be paid in relation to the capital invested. The formal definition of a rate of interestis:

RATE OF INTEREST: The amount contracted to be paid in one unit interval of time for each unit of capital invested

The general financial practice is to make the unit interval of time a year, and this unfortunatelytends to induce the preconceived idea that rates of interest must be annual rates. It is indeedcustomary to describe rates of interest “per annum” when in fact they are not.

Effective Rate of Interest

The effective rate of interest during any period is the one that satisfies the definition of rate of interest given above. Let us state that if one period the rate of interest applied is “i”, then theeffective rate of interest is “i” as long as at the end of the period, one unit of capital hasaccrued from 1 to 1+i.

Nominal Rate of Interest

The common financial practice is to express rates of interest per annum, even though theinterest is paid more than frequently than once per year. For instance, a US Treasury bond ispaid on quarterly basis and in spite of this the rate is said to be “per annum”. In those cases,actually, what really happens is that the rate expresses as per annum is called nominal rate and the effective rate is calculated as the nominal rate divided by the number of times the

rates accrues in the year. The nominal rate of interest, which is annual, is denoted(m )

iif it

accrues m times a year. The effective rate of interest is the one applied to each accrual

period and denoted

(m )ii

m= . The interest payment may be made several times per annum,

let us say “p” times. Four cases can be distinguished:

The case where m=p The case where m>p and m/p is and integer The case where m<p and p/m is an integer The case where m/p or p/m is not integer

For sake of simplicity, in this course the normal case would be m=p, i.e. that the payment datecoincides with the accrual periodicity of the interest rate. Consequently, in this course the

3. Mathematics of Finance





nominal rate of interest will be expressed as(p) (m)i i= and the effective rate of interest will be

calculated as

(p) (m)i ii

p m= = .

3.1.1 Simple Interest and Number of Day Rules in the Financial Year

The number of days, which is the time unit of account for calculation of interest, takes aparamount importance in Mathematics of Finance as there are different working hypothesesin different parts of the world, and therefore for applicable to different currencies. Thecounting of days is used mainly in short-term transactions

15, though not necessarily limited to

those.

Simple Interest Future Value

Let us assume that the annual nominal rate of interest to be applied on “s” monetary unitsover a certain number of days is “i”. Then the interest accrued on “s”, let us say 15 days in ayear, would be “S” and is calculated as follows:

(3.1.1)i 15

S s 1365

×⎛ ⎞= +⎜ ⎟⎝ ⎠

S is said to be calculated by simple interest by opposition to compoundinterest that will be dealt with in the next section.

Table 3.1.1 shows the different conventions for counting the days for interest calculationsaround the world. The convention to apply has to be very clearly defined in order to calculateaccurately the corresponding accrued interest.

Table 3.1.1

Convention of Number of Days in Interest Calculations

Type Vale Description

yb_US 0 US (NASD) 30/360 - As with the European 30/360 yb_EU, with theadditional provision that if the end date occurs on the 31st of a month itis moved to the 1st of the next month if the start date is earlier than the30th.

yb_Act 1 Uses the exact number of elapsed days between the two dates, as wellas the exact length of the year.

yb_Act360 2 Uses the exact number of elapsed days between two dates butassumes the year only have 360 days

yb_Act365 3 Uses the exact number of elapsed days between two dates butassumes the year always has 365 days

yb_EU 4 European 30/360 - Each month is assumed to have 30 days, such thatthe year has only 360 days. Start and end dates that occur on the 31stof a month become equal to the 30th of the same month.

The first column of Table 3.1.1 shows the notation for the convention utilised. The secondcolumn shows the value of the parameter that is assigned to the convention when usingcomputerised algorithms and the third column describes the application of the convention.For instance, as the calculation is expressed in (3.1.1). The convention applied could beyb_Act if the year is not leap, or if it is it could be yb_Act365. If yb_Act360 or yb_EU isapplied, then (3.1.1) would be calculated as:

15 A transaction is considered short-term if it is carried on a maximum of one year: 365 days or 366 for

lap years. In this course, any transaction above one year will be considered as a long-term transaction.





(3.1.2)i 15

S s 1360

×⎛ ⎞= +⎜ ⎟⎝ ⎠

It has to be noted that the difference in calculations between some of the conventions would

appear only when the accrual period is larger than one calendar month. Let us take anumerical example for calculating (3.1.2) assuming s=100 and a nominal rate of interest of 5percent.

0.05 15S 100 1 100.21

360

×⎛ ⎞= + =⎜ ⎟⎝ ⎠

The reader is encouraged to make the same calculations assuming the yb_Act convention,and he will find that (3.1.2) gives a higher interest amount for the same rate and the sameperiod than (3.1.1) for periods larger than one month.

Simple Interest Present Value

The calculation expressed in (3.1.1) and (3.1.2) imply a forward vision, i.e. the future value Sof an invested amount s. If the problem we are facing is the opposite, for instance whatamount s should we invest today—present value—to obtain an amount of S in 15 days, thiswould be calculated as:

(3.1.3)

1S i 15

s S 1i 15 365

1365

−×⎛ ⎞= = +⎜ ⎟×⎛ ⎞ ⎝ ⎠+⎜ ⎟

⎝ ⎠

Let us take a numerical example for calculating (3.1.3) assuming that we want to know what

amount s we have to invest today in order to obtain S=100 in 15 days assuming a nominalrate of interest of 5 percent.

10.05 15

s 100 1 99.79365

−×⎛ ⎞= + =⎜ ⎟

⎝ ⎠

3.1.2 Compound Interest: Future and Present Values

Compound interest means that interest is re-invested and becomes part of principal16

. Thereare also two different, but complementary, ways of look at compound interest. One is forwardlooking, i.e. calculating future values, and the other one is calculating present values, i.e.

calculating the today’s value of future amounts of money.

Future Value

Future value allows us to calculate the future worth of and investment made today, or in aseries of deposits from today to the maturing date. Let us start calculating the future value of an invested amount of s monetary units at a rate of i per period at date 0 and maturing at date1.

(3.1.4) 1S s is s(1 i)= + = +

16In all the sections dealing with composed interest we assume that the days convention is yb_EU.





At end of period 1, (3.1.4) gives the future value: 1S . Assuming the amount is re-invested for

another period, a second period, the future amount at end of period 2 is calculated as:

(3.1.5)2

2S s(1 i) s(1 i) i s(1 i)= + + + × = +

If the future value at the end of each period is re-invested until the end of period n, then theformula to calculate the future value maturing at the end of the n period is:

(3.1.6)n

nS s(1 i)= +

The calculation of the future value using formula (3.1.6) is called compound interest byopposition to simple interest that was deal in Section 3.1.1. Let us take a numerical example:how much is the future value in 10 periods of an amount s=100 at an effective rate of interestof 5 percent per period:

10

10S 100(1 0.05) 162.89= + =

It has to be noted that the future value has a forward approach: by investing a known amounttoday the calculation gives the future value after n periods.

Present Value

Present value allows us to calculate the present worth of an amount to be obtained at a givenfuture date, or of a series of amounts from today to the maturing date. Let us start calculatingthe present value, at date 0, of an amount of “a” monetary units at a rate of i per period at

date 1. We want to find the value 0A to be invested at a rate “i” that will accrue amount “a” at

the end of the period. Following (3.1.4), we can write:

(3.1.7) 0A (1 i) a+ =

From (3.1.7), we find the amount 0A that invested at beginning of period will accrue to the

amount “a” at end of period:

(3.1.8)1

0

aA a(1 i)

(1 i)

−= = ++

If we increase the number of periods, proceeding as in the developments shown in (3.1.5)and (3.1.6), we find the general formula for present value:

(3.1.9) n0 n

aA a(1 i)(1 i)

−= = ++

For the numerical example let us find what amount we have to invest today in order to obtain

a= 0A 100 monetary units in 10 periods at a compound rate of interest of 5 percent. The

calculation applying (3.1.9) gives:

10

0A 100(1 0.05) 61.39−= + =





Comparing Future Value and Present Value

Taking formulae (3.1.6) and (3.1.9) it is clearly seen that future value has a forward vision andpresent value a backward vision:

• In the case of future value, we invest an amount “s” to obtain nS at the end of nperiods; therefore future value has a forward vision because brings the value fromdate 0 to date n in the future.

• In the case of present value, we calculate how much we need to invest at date 0, 0A ,

in order to obtain an amount of “a” in n periods; therefore present value has abackward vision, because brings the value from date n, in the future, to date 0, today.





3.2. Annuities-Certain and Amortisation Schedules

3.2.1 Annuities-Certain

An annuity is a series of payments at fixed periods of time continuingduring the existence of a given status, for instance, during the lifetimeof a person, or to a widow as long she has children under a certainage, or more simply, for a fixed period of time, independent of anycontingency. In this last case, the series of payments is calledannuity-certain. These series of payments are concerned by thetheory of compound interest as seen in the previous paragraph. However, we will restrict our exposition to annuities-certain as defined above. For sake of simplicity, we will restrict our annuities-certain analysis to the convention yb_EU, i.e a year with 360 days and 12 monthswith 30 days each. The analysis of annuities-certain can be extended to other conventions,

however, it happens very seldom.

Future Value of Annuities-Certain

Let us assume that we commit, at date 0, to deposit at the end of each period an amount of 1monetary unit during n periods at a compound rate of interest “i”. In that case, the futurevalue of the annuity will be calculated as:

(3.1.10)n 1 n 1

nS (1 i) (1 i) ... (1 i) 1− −= + + + + + + +

Wheren

S denotes the future value of an annuity of 1 per annum—payable for a term certain

of n periods—the symboln

being used to denote a fixed period of time, in this case n

periods. The series in the right side of (3.1.10) is the sum of the first n terms of a geometrical

series. The calculation of the sum, nZ , of the first n terms of a geometrical series, assuming

that the geometrical reason is q, is calculated as follows:

(3.1.11)0 1 2 n 1 2 n 1

nZ q q q ... q 1 q q ... q − −= + + + + = + + + +

Multiplying nZ by q, we obtain:

(3.1.12)2 n

nqZ q q ... q = + + +

And subtracting (3.1.11) from (3.1.12):

(3.1.13)

nn

n n n

q 1qZ Z q 1 Z

q 1

−− = − ⇒ =

−

In (3.1.10) q (1 i)= + , therefore applying the result obtained in (3.1.13) to (3.1.10):

(3.1.14)

n

n

(1 i) 1S

i

+ −=

It has to be noted that (3.1.14) is applied when the payment is made end of period. In casethe payment is made beginning of period, which in most of the cases of future value





calculations happens. In that case, there is, for each of the payments one more period for theinterest to accrue. In that case the future value is simple calculated as:

(3.1.15)n n

S (1 i)S= +&&

Wheren

S&& denotes the future value of a series of n payments of one monetary unit during n

periods, made beginning of period, and accruing at an effective rate of interest i.

The formula in (3.1.14) is for an effective rate i. When dealing with nominal rates, i.e. a

nominal rate(p )i for a payment of one monetary unit p times a year, (3.1.14) shall be

calculated as:

(3.1.16)

np(p)

(p )np

i1 1

pS

i

p

⎛ ⎞+ −⎜ ⎟

⎝ ⎠=

Let us take a numerical example. We want to calculate the future value S of a series of payments of 100 monetary units during 10 years made semi-annually beginning of period at anominal rate of interest of 8 percent. In that case:

(p )

n 10

p 2

i 0.08

s 100

=

=

=

=

Therefore, the calculation to be made is:

20

20 20

(1 0.04) 1S 100S 100(1 0.04)S 100(1 0.04) 3,096.92

0.04

+ −= = + = + =&&

It has to be noted that the fact of paying beginning of period will result in a higher accruedfuture value, as each payment has one more period to accrue. To show this statement let usproceed to do the same calculations as before but with payments end of period:

20

20

(1 0.04) 1S 100S 100 2,977.81

0.04

+ −= = =

Present Value of Annuities-CertainIn this case, the question that we wish to answer is what amount do we have to invest today,

na , in order to obtain a series of payments of one monetary unit during n periods, payable

end of period, at a given effective rate of interest “i”. This is calculated as follows:

(3.1.17)n

1 2 n 1 2 n 1(1 i) (1 i) .. (1 i) (1 i) 1 (1 i) .. (1 i)− − − − − − −⎡ ⎤= + + + + + + = + + + + + +⎣ ⎦a

The series shown between brackets in the right side of (3.1.17) is the sum of the first n terms

of a geometrical series where1q (1 i)−= + . In this case, in order to calculate the value of this

sum, it is more practical to subtract (3.1.12) from (3.1.11):





(3.1.18)

nn

n n n

1 q Z qZ 1 q Z

1 q

−− = − ⇒ =

−

Thus, applying (3.1.18) to the sum between brackets in (3.1.17):

(3.1.19)

n2 n 1

1

1 (1 i)1 (1 i) .. (1 i)

1 (1 i)

−− − −

−

− ++ + + + + =

− +

Replacing (3.1.19) into (3.1.17) we have:

(3.1.20)

n

n

n 11

1 1

1 (1 i)

(1 i) 1

1 (1 i) (1 i)(1 i)

1 (1 i) (1 i)

−− −−

− −

− +

+ −

⎡ ⎤− + += + = ×⎢ ⎥− + +⎣ ⎦

a

Simplifying (3.1.20) we obtain the final result:

(3.1.21)n

n1 (1 i)i

−− +=a

As in the case of future value, the formula in (3.1.21) is for an effective rate i. When dealing

with nominal rates, i.e. a nominal rate(p )i for a payment of one monetary unit p times a year,

(3.1.21) shall be calculated as:

(3.1.22)np

(p )np

(p )

i1 (1 )

p

i

p

−− +=a

Let us take a numerical example. We want to calculate the present value A of a series of payments of 100 monetary units during 10 years made semi-annually end of period at anominal rate of interest of 8 percent. In that case:

(p )

n 10

p 2

i 0.08

a 100

=

=

=

=

Therefore, the calculation to be made is:

20

201 (1 0.04)A 100 100 1,359.03

0.04

−− += = =a

The interpretation of the above result is that we need 1,359.03 monetary units today, at aprevailing nominal rate of interest of 8 percent per annum, in order to receive 100 monetaryunits at end of semi-annual periods during 10 years.

Obviously, if the series of payments is to be made beginning of period, then the present valueis calculated as follows:

(3.1.23) n n(1 i)= +&&a a





Wheren

&&a denotes the present value of a series of n payments of one monetary unit during n

periods, made beginning of period, and accruing at an effective rate of interest i. This type of annuity-certain is called an “annuity-due”.

It has to be noted that by paying beginning of period, the required amount will decrease by

one monetary unit, as it is de facto subtracted from the initial amount. The series of paymentwill also have one period less to accrue interest. Actually, (3.1.23) can be written as:

(3.1.24)n n 1

1−

= +&&a a

This means that by using a present value of an annuity-certain paid beginning of period, fromthe practical point of view, we are making a down payment that is subtracted from our initial

investment and we obtain a shorter annuity:n 1−

a . The needed investment will be smaller

than in the case of payments end of period. In order to show this point let us calculate thesame numerical example as above under the hypothesis of payments beginning of period.

n

20

1 (1 0.04)A 100 100 (1 0.04) 1,413.390.04

−

− += = × + =&&a

3.2.3 Types of Problems Concerning Annui ties-Certain

There are four types of problems concerning annuities-certain:

1. Given s or a, i and n find S or A2. Given S or A, i and n find s or a3. Given (S, s) or (A, a) and i find n4. Given (S, s) or (A, a) and n find i

First Type of Problem

The first type of problem was given and solved as numerical examples in the previoussection. In this section we will be solving the other three types of problems by givingnumerical examples of the solution.

Second Type of Problem: Future Value

Given S, i and n find s: What is the amount s to be deposited at the end of each semester inorder to obtain 1,000.00 monetary units in ten years at a nominal rate of interest of 8 percent?

The solution is given by (3.1.16):

20

20 20

(1 0.04) 1 1,000.001,000.00 sS s s 33.580.04 (1 0.04) 1

0.04

+ −= = ⇒ = =⎛ ⎞+ −⎜ ⎟⎝ ⎠

Second Type of Problem: Present Value

Given A, i and n find a: If we deposit today 1,000.00 monetary units, what amount “a” paidsemi-annually during ten years at a nominal rate of interest of 8 percent will be obtain?






20

20

20

1 (1 0.04) 1,000.001,000.00 a a a 73.58

0.04 1 (1 0.04)

0.04

−

−

− += = ⇒ = =

⎛ ⎞− +⎜ ⎟

⎝ ⎠

a

Third Type of Problem: Future Value

Given S, s and i find n. We wish to obtain 1,000.00 monetary units at the end of the accrualperiods by investing 50.00 monetary units at the end of each period. Given that the effectiverate of interest is 4 percent, in how many periods will we obtain the desired future value?


nn

n

(1 0.04) 1 1,000.00 0.041,000.00 50.00S 50.00 1 (1 0.04)

0.04 50.00

+ − ×= = ⇒ + = +

Therefore:

However, the calculations can be carried forward more accurately. In fact, the value for nshown above is rounded to the nearest entire number but actually its exact value is

n 14.9866= . In that case, once we got this value, we can deduct that in order to have theexact desired amount we have to make 14 deposits of 50 monetary units at the end of eachone of the periods, plus a smaller payment at the end of the 15

thperiod. This last payment is

calculated as follows:

( ) ( )14 14

1,000.00 50.00S 1 0.04 x x 1,000.00 50.00S 1 0.04= + + ⇒ = − +

x 1, 000.00 50.00 18.29191 1.04 48.82= − × × =

The answer to our problem is that under the hypothesis made on the nominal rate of interest,in order to obtain 1,000.00 monetary units we need to make 14 deposits of 50.00 monetaryunits end of period, and a last deposit of 48.82 monetary units at the end of the 15

thperiod.

Third Type of Problem: Present Value

Given A, a and i find n: If we deposit today 1,000.00 monetary units, and we wish to obtain anamount of 50.00 monetary units paid at the end of each period. Given that the effective rateof interest is 4 percent, during how many periods will we get the periodical payment?


n

nn1 (1 0.04) 1,000 0.04

1,000 50 50 (1 0.04) 10.04 50

−−− + ×

= = ⇒ + = −a





Therefore:

As in the same type of problem for FV, we can find a more accurate answer to this type of problem. In fact, the exact value of n is n=41.0354. This means that the amount of 1,000monetary units deposited initially will allow us to get 40 payments of 50 and one more of aslightly different value. This is calculated as follows:

( ) ( ) ( )40 40

41 411,000 50 x 1 0.04 x 1 0.04 1,000 50

−= + + ⇒ = + −a a

( )41x 1.04 1,000 989.64 51.73= × − =

The answer to our problem is that under the hypothesis made on the nominal rate of interest,the 1,000.00 monetary units deposited initially, will allow us to get 50.00 monetary units endof period during 40 periods, and a last payment of 51.73 monetary units at the end of the 41

st

period.

Fourth Type of Problem: Future Value

Given S, s and n find i. We wish to obtain 1,000.00 monetary units at the end of the 10accrual periods by investing 75.00 monetary units at the end of each period. At what effectiverate of interest do we have to invest in order to obtain the desired future value?


10 10

10

(1 i) 1 1,000 (1 i) 11,000.00 75S 75

i 75 i

+ − + −= = ⇒ =

Therefore:10(1 i) 1

13.3333i

+ −=

The above equation in i is difficult to solve by applying algebra. It is more practical to proceedby iterations and interpolation. We need a value of i that will give a value smaller than13.3333, but close to it, and a value of i that will give a value larger than 13.3333, but close toit. Proceeding by iterations we find:

10(1 0.062) 113.30525

0.062

+ −= and

10(1 0.0625) 113.33657

0.0625

+ −=

With these values we can find a closer value of i by interpolation:

i 0.062 0.0625 0.062

13.3333 13.30525 13.33657 13.30525

− −=

− −

( ) ( )13.3333 13.30525 0.0625 0.062i 0.062 0.0624

13.33657 13.30525

− × −= + =

−

The desired rate is i=6.24%





The reader should notice that it might be that we cannot invest at exactly the rate that weneed, for instance in practice it might be difficult to find a nominal rate i=6.24%. However, wemay find a rate i=6.25. In that case, the problem can be adjusted for the last payment to bedifferent to the others as we did with the third type of problem solution.

Fourth Type of Problem: Present ValueGiven A, a and n find i: If we deposit today 1,000.00 monetary units, and we wish to obtain anamount of 150.00 monetary units paid at the end of each period during ten periods. At whateffective rate of interest do we have to invest in order to obtain the desired series of payments?


10

10 101 (1 i) 1,000 1 (1 i)1,000 150 150

i 150 i

− −− + − += = ⇒ =a

Therefore:

101 (1 i)6.6667

i

−− +=

As in the previous case, the above equation in i is difficult to solve by applying algebra. It ismore practical to proceed by iterations and interpolation. We need a value of i that will give avalue smaller than 6.6667, but close to it, and a value of i that will give a value larger than6.6667, but close to it. Proceeding by iterations we find:

101 (1 0.0725)6.943128

0.0725

−− += and

101 (1 0.082)6.649969

0.082

−− +=

With these values we can find a closer value of i by interpolation:

i 0.0725 0.082 0.0725

6.6667 6.943128 6.649969 6.943128

− −=

− −

( ) ( )6.6667 6.943128 0.082 0.0725i 0.0725 0.0811

6.649969 6.943128

− × −= + =

−

The desired rate is i=8.11%

Here the same remark as in the case of FV is applicable: it might be that we cannot invest atexactly the rate that we need, for instance in practice it might be difficult to find a nominal ratei=8.11%. But, we may find a rate i=8.125. In that case, the problem can be adjusted for thelast payment to be different to the others as we did with the third type of problem solution.However, in this case we may have more flexibility, as this may be applicable to purchases of bonds, in which case the desired rate can be interpreted and calculated as the yield of our investment. This will be deal with in the next module.

3.2.4 Amor tisation of Loans and Annuities-Certain

Formulae (3.1.21) and (3.1.22) is the present value of a series of payments of one monetary

unit per annum for n periods, and means that if a purchaser investsn

a he will secure a yield

of at the rate i on his capital during n periods. The payment he receives is one per annum,and it must be clear that he cannot regard this payment purely as income, or interest,





because if he did not reinvest any part of it, then at the end of n periods when the annuityceased he would find that ne had none of his original capital left. Hence each payment of onemonetary unit must comprise:

• Interest on the capital invested, and

• A portion representing a repayment of principal

In the above paragraph the transaction has been considered as the purchase of a given

series of payments for the price of n

a . It would lead to exactly the same result if the

transaction was described as a loan of n

a to be repaid by n periodical instalments of one

monetary unit. From the point of view of the purchaser, or the lender, as the case may be,

the transaction is still the investment of capital sum of n

a . From the point of view of the

vendor, or the borrower, the transaction is in each case an undertaking to provide a series of

n periodical payments of one monetary unit in return for the cash payment of n

a at the

outset.

Let us calculate an amortisation table using an annuity-certain: we borrow 10 million monetaryunits at a nominal rate of 8 percent per annum to be repaid in semi-annual instalments duringfive years. The loan is fully disbursed on 15 October 2009.

First, we need to calculate the amount of the annuity payment. This is calculation is achievedapplying (3.1.22):

10

101 (1 0.04)10,000,000 a a

0.04

−− += =a

Hence:

10

10,000,000a 1,232,909.44

1 (1 0.04)0.04

−= =

⎛ ⎞− +⎜ ⎟⎝ ⎠

Therefore, we have to pay ten semi-annual instalments of 1,232,909.44 in order to

reimburse the 10 million including the interest at a nominal interest rate of 8 percent. Now wecan construct the amortisation schedule for the loan, which appears in Table 3.2.1.

Table 3.2.1Annuity-Certain end of Period: Loan Amortisation Schedule

Period Date Outstanding Principal Interest Annuity

0 15-Oct-09 10,000,000.001 15-Apr-10 9,167,090.56 832,909.44 400,000.00 1,232,909.44

2 15-Oct-10 8,300,864.74 866,225.82 366,683.62 1,232,909.44

3 15-Apr-11 7,399,989.88 900,874.85 332,034.59 1,232,909.44

4 15-Oct-11 6,463,080.03 936,909.85 295,999.60 1,232,909.44

5 15-Apr-12 5,488,693.79 974,386.24 258,523.20 1,232,909.44

6 15-Oct-12 4,475,332.10 1,013,361.69 219,547.75 1,232,909.44

7 15-Apr-13 3,421,435.94 1,053,896.16 179,013.28 1,232,909.44

8 15-Oct-13 2,325,383.94 1,096,052.01 136,857.44 1,232,909.44

9 15-Apr-14 1,185,489.85 1,139,894.09 93,015.36 1,232,909.44

10 15-Oct-14 0.00 1,185,489.85 47,419.59 1,232,909.44

TOTALS 10,000,000.00 2,329,094.43 12,329,094.43





As indicated, the payment is constant. However, the shares that are devoted to pay interestand principal vary. Let us analyse how Table 3.2.1 is built: the disbursement is made on 15October and the first instalment on 15 April, six months later. The share corresponding to theinterest payment is calculated by applying the effective interest rate per period, 4 percent, tothe outstanding amount beginning of period:

10,000,000.00 0.04 400,000.00× =

And therefore, the share devoted to repay principal is:

1,232,909.44 400,000.00 832,909,.44− =

The outstanding amount beginning of period on 15 April 2010 is calculated as:

10,000,000.00 832,909,.44 9,167,090.56− =

The remaining cells in Table 3.2.1 are calculated following the three steps described above.

Table 3.2.2Annuity-Certain beginning of Period: Loan Amortisation Schedule

Period Date Outstanding Principal Interest Annuity

0 15-Oct-09 8,717,774.18

1 15-Apr-10 7,784,259.33 933,514.85 348,710.97 1,282,225.82

2 15-Oct-10 6,813,403.88 970,855.45 311,370.37 1,282,225.82

3 15-Apr-11 5,803,714.21 1,009,689.67 272,536.16 1,282,225.82

4 15-Oct-11 4,753,636.96 1,050,077.25 232,148.57 1,282,225.82

5 15-Apr-12 3,661,556.62 1,092,080.34 190,145.48 1,282,225.82

6 15-Oct-12 2,525,793.06 1,135,763.56 146,462.26 1,282,225.82

7 15-Apr-13 1,344,598.96 1,181,194.10 101,031.72 1,282,225.82

8 15-Oct-13 116,157.10 1,228,441.86 53,783.96 1,282,225.82

9 15-Apr-14 0.00 116,157.10 4,646.28 120,803.38

10 15-Oct-14 0.00

TOTALS 8,717,774.18 1,660,835.77 10,378,609.95

Table 3.2.1 has been constructed following (3.1.21) and (3.1.22). But how would the tablelook if we use (3.1.23) or (3.1.24), an annuity-due, instead? Using these latest formulaemean that the payment is beginning of period, and we would be making a de facto down

payment equal to the annuity payment calculated above, 1,232,909.44 , that has to besubtracted from the original loan amount, 10,000,000.00 , because at date 0, i.e. at 15 April

2009, no interest has accrued. This is represented in Table 3.2.2, where the initialdisbursement is:

10,000,000.00 1,232,909.44 8,717,774.18− =

Besides the above difference, Table 3.2.2 is calculated exactly as Table 3.2.1. Comparingthe two tables, it is clear that those formulae (3.1.23) and (3.1.24) are related, namely:

n n(1 i)= +&&a a and

n n 11

−= +&&a a





Grace Period: Deferred Annuity-Certain

If the first payment of the annuity-certain is not to be made at the end of the first period, but isto be deferred for m periods from the present time, the annuity is called to be a deferred

annuity to distinguish from an immediate annuity, and the symbol used isn

m a . In the case

where the deferred annuity represents the repayment of a loan, the deferral period is calledgrace period.

Table 3.2.3Deferred Annuity-Certain: Loan Amortisation Schedule

Period Date Outstanding Principal Interest AnnuityPresentValue.

0 15-Oct-09 10,000,000.00

1 15-Apr-10 10,000,000.00 400,000.00 400,000.00 384,615.38

2 15-Oct-10 10,000,000.00 400,000.00 400,000.00 369,822.49

3 15-Apr-11 10,000,000.00 400,000.00 400,000.00 355,598.54

4 15-Oct-11 10,000,000.00 400,000.00 400,000.00 341,921.68

5 15-Apr-12 10,000,000.00 400,000.00 400,000.00 328,770.84

6 15-Oct-12 9,167,090.56 832,909.44 400,000.00 1,232,909.44 974,386.24

7 15-Apr-13 8,300,864.74 866,225.82 366,683.62 1,232,909.44 936,909.85

8 15-Oct-13 7,399,989.88 900,874.85 332,034.59 1,232,909.44 900,874.85

9 15-Apr-14 6,463,080.03 936,909.85 295,999.60 1,232,909.44 866,225.82

10 15-Oct-14 5,488,693.79 974,386.24 258,523.20 1,232,909.44 832,909.44

11 15-Apr-15 4,475,332.10 1,013,361.69 219,547.75 1,232,909.44 800,874.46

12 15-Oct-15 3,421,435.94 1,053,896.16 179,013.28 1,232,909.44 770,071.60

13 15-Apr-16 2,325,383.94 1,096,052.01 136,857.44 1,232,909.44 740,453.46

14 15-Oct-16 1,185,489.85 1,139,894.09 93,015.36 1,232,909.44 711,974.48

15 15-Apr-17 0.00 1,185,489.85 47,419.59 1,232,909.44 684,590.85

TOTALS 10,000,000.00 2,329,094.43 12,329,094.43 8,219,271.07

The value of an annuity-certain for n periods deferred m periods, i.e. with a grace period of mperiods, may be considered either as the present value of a the single sum which representsthat value of the annuity at the moment it is entered upon: m periods form the present, or asthe difference between the present value of an annuity-certain for (m+n) periods les thepresent value of the payments in the first m periods which are not received:

(3.1.25) ( )n n n m m

mm 1 i

+

−= + = −a a a a

In the numerical example above:

1010,000,000.00 a= a

Therefore, if we introduce a grace period or a deferral of two years and a half, the value of theannuity would be:

( )5

1 0.04 10, 000, 000.00 8, 219, 271.07−

+ × =





The amortisation table for this case is shown in Table 3.2.3. A new column has been addedin Table 3.2.3 that is the sum of the present value of the annuity payments. It is clearly shownthat the value, we could say market value, of such an operation is 8,219,271.07 monetaryunits and not 10 million, as the face value would induce. This means that from the point of view of the purchaser, or the lender, the opportunity cost for this operation would be thedifference between the present value of the annuity and the face value:

10,000,000.00 8,219,271.07 1,780,728.93− =

This amount is precisely the present value of the first five payments of interest during thegrace or deferral period:

5h

h 1

(1 0.04) 400,000,00 1,780,728.93−

=

+ × =∑

From the point of view of the borrower, or the vendor, this operation implies an opportunitygain which is exactly equal to the opportunity cost of the purchaser, or the lender. This is in

fact the result of the formulae expressed in (3.1.25), which numerical values are:

( ) ( )10 10

5 55 1 0.04 1 0.04 10,000,000.00 8,219,271.07

− −= + = + × =a a

And:

10 15 55 10,000,000.00 1,780,728.93 8,219,271.07= − = − =a a a





Articles or Short Publications

Bergström, Pål, and Anders Holmlund, (2000), A Simulation Model Framework for

Government Debt Analysis. Swedish National Debt Office, 30 November, Stockholm.

Blommestein, Hans J., (2005), Overview of Advances in Risk Management of GovernmentDebt, OECD, Paris.

Cassard, Marcel, and David Folkers-Landau, (1997), Risk Management of Sovereign Assetsand Liabilities, IMF Working Paper WP/97/166, December, Washington.

____________________________________, (1997), Sovereign Debt: Managing the Risks.

IMF, Finance and Development, December, Washington.

Claessens, Stijn, (2004), Taking Stock of Risk Management Techniques for Sovereigns.Sovereign Debt Project, Initiative for Policy Dialogue, University of Columbia and UnitedNations, October, New York.

Cosío-Pascal, Enrique, (2007), Manual on Debt and Risk Indicators, Inter-AmericanDevelopment Bank, Latin-American and Caribbean Debt Group, January, New York andWashington.

Currie, Elizabeth, and Antonio Velandia, (2002), Risk Management of Contigent LiabilitiesWithin a Sovereign Asset-Liability Framework. The World Bank, January,Washington.

IMF, PDRD, (2000), Debt- and Reserve-Related Indicators of External Vulnerability, 23March, Washington.

Storkey, Ian, (2001), Sovereign Debt Management: A Risk Management Focus. The Financeand Treasury Professional of Australia, May, Melbourne.

Velandia, Antonio, (Without date), A Risk Quantification Model for Public Debt Management.The World Bank, Washington.

World Bank and IMF, (2001), Guidelines for Public Debt Management, Washington, March.

World Bank and the IMF, (2002), Guidelines for Public Debt Management: AccompanyingDocument, Washington, November.

World Bank and IMF, (2003), Amendments to the Guidelines for Public Debt Management,Washington, November.

World Bank and IMF, (2007), Strengthening Debt Management Practices: Lessons fromCountry Experiences and Issues Going Forward, 27 March, Washington.

World Bank and IMF, (2007), Strengthening Debt Management Practices—Lessons fromCountry Experiences and Issues Going Forward: Background Paper , 27 March, Washington.

Bibliography




Books

Anderson, Torben Juul, (1993), Currency and Interest Rates Hedging. New York Institute of Finance, New York.

Crouhy, Michael, Dan Galai and Robert Mark, (2006), The Essentials of Risk Management.McGraw Hill, New York.

Hoercher, Karen A., (2006), Essentials of Managing Treasury, John Wiley & Sons, Hoboken,New Jersey.

Hull, John C. (2006), Options, Futures and Other Derivatives. Pearson-Prentice Hall, NewJersey.

Ludwig, Mary S. (1993), Understanding Interest Rates Swaps. McGraw Hill, New York.

Marrison, Chris, (2002), The Fundamentals of Risk Measurement. McGraw Hill, New York.

Piga, Gustavo, (2001) Derivatives and Public Debt Management. International SecuritiesMarket Association and Council on Foreign Relations, Zürich and New York.

Pitts, Mark and Frank J. Fabozzi, (1990), Interest Rates, Futures and Options. Probus,Chicago.

Soler Ramos, José A., Kim B. Staking, Alfonso Ayuso Calle, Paulina Beato, Emilio BotínO’Shea, Miguel Escrig Meliá and Bernardo Carrasco, (2000), Financial Risk Management: APractical Approach for Emerging Markets. John Hopkins, Inter-American Development Bank,Grupo Santander and Caribbean Development Bank, Washington.

Wheeler, Graeme, (2004), Sound Practice in Government Debt Management. The WorldBank, Washington.

Documents

ARM Module 1