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Lecture skeleton for Statistics and Business Finance based on Quantitative Investment Analysys
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Topic 1:
Time Value of Money
(Part I)
Associate Professor Ishaq Bhatti
La Trobe Business School
E-Mail: [email protected]
Slides have been drafted by the La Trobe University, School of Business based on DeFusco et al (2007)
Statistics for Business and Finance
Chapter 1 Time Value of Money
1.2
1. INTRODUCTION
What is the time value of money?
Is $1 today equal to $1 tomorrow?
Would you agree to pay $500 to a friend and receive $500 back 1 year from now?
Would you agree to pay $500 to a friend and receive $1000 back 1 year from now?
Time Value of Money
1.3
2. INTEREST RATES
Would you agree to receive $9,500 now and pay $10,000 now?
What if you receive $9,500 now and pay $10,000 one year from now?
Time Value of Money
1.4
2. INTEREST RATES
An interest rate is a rate of return that reflects the relationship between
differently dated cash flows.
How much is your return in the previous example?
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Time Value of Money
1.5
2. INTEREST RATES
Interest rate may be referred to as:
Required rate of return: The minimum rate of return an investor must receive in
order to accept the investment.
Discount rate: The rate we use to discount future cash flows.
Opportunity cost: Value that investors forgo by choosing a particular course of
action.
Time Value of Money
1.6
3. FUTURE VALUE OF A SINGLE CASH FLOW
You invest $9,500 now and receive $10,000 one year from now.
This $10,000 includes the initial $9,500 plus $500 interest on that.
Future Value is equal to the Present Value of the investment plus interest on the investment.
rPVFV
PVrPVFV
1
Time Value of Money
1.7
3. FUTURE VALUE OF A SINGLE CASH FLOW
Now what if you invest that money for one more year?
Hence, formula for future value of a single cash flow after N periods is:
21
11
rPVFV
rPVrrPVFV
NrPVFV 1
Time Value of Money
1.8
Example 1: FV of a Lump Sum
An institution promises to pay you a lump sum, six years from now at an 8% annual interest
rate, if you invest $2,500,000 today.
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Time Value of Money
1.9
3.1 The Frequency of Compounding
Some investments pay interest more than once a year.
Financial institutions often quote an annual interest rate.
If your bank states the annual interest rate is 8%, compounded monthly, how much is the
monthly interest rate?
Time Value of Money
1.10
3.1 The Frequency of Compounding
With more than one compounding period per year, the future value formula can be
expressed as:
mN
s
m
rPVFV
1
Time Value of Money
1.11
Example 2: FV of a Lump Sum with Monthly Compounding
An investment has a six-year maturity and annual quoted interest rate is 8% compounded
monthly. FV if you invest $2,500,000 is:
Time Value of Money
1.12
3.2 Continuous Compounding
If the number of compounding periods per year becomes infinite, then interest is said to
compound continuously.
Formula for FV of a sum in N years with continuous compounding is:
7182818.2in which
e
PVeFVNrs
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Time Value of Money
1.13
Example 3: FV of a Lump Sum with Continuous
Compounding
An investment has a six-year maturity and annual quoted interest rate is 8% compounded
continuously. FV if you invest $2,500,000 is:
Time Value of Money
1.14
3.3 Stated and Effective Rates
In all of the examples 1-3 stated interest rate and maturity were 8% and 6 years. But they all
have different future values:
PV = $2,500,000
FV Compounded annually = $3,967,186
FV Compounded Monthly = $4,033,755
FV Compounded Continuously = $4,040,186
Time Value of Money
1.15
3.3 Stated and Effective Rates
Examples 1-3 illustrate that when interest rate is compounded, monthly or continuously,
effective rate is more than the stated rate.
The effective annual rate is calculated as:
Or for continuous compounding:
1 1 mateInterest RPeriodicEAR
1 sreEAR
Time Value of Money
1.16
Example 4: Effective Annual Rate
An investment has a six-year maturity and annual quoted interest rate is 8%. What is the EAR if interest rate is compounded monthly? What if it is compounded continuously:
For continuous compounding:
8%r
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Time Value of Money
1.17
4. FUTURE VALUE OF A SERIES OF
CASH FLOWS
Common terms used in this topic:
Annuity: a finite set of sequential cash flows
Ordinary annuity: first cash flow occurs one year from now
Annuity due: first cash flow occurs immediately (at t=0)
Perpetuity: an infinite set of cash flows beginning one year from now.
Time Value of Money
1.18
4. FUTURE VALUE OF A SERIES OF
CASH FLOWS
Consider an ordinary annuity paying 5% annually. Suppose we have 5 annual deposits
of $100 starting one year from now.
We are interested in FV of this ordinary annuity.
Now: t=0 2 1 3 4 5
$100 $100 $100 FV= ?
$100 $100
Time Value of Money
1.19
4. FUTURE VALUE OF A SERIES OF
CASH FLOWS
Total FV is equal to the sum of FV of every single payment:
FV of 1st payment :
FV of 2nd payment :
FV of 3rd payment :
FV of 4th payment :
FV of 5th payment :
105.01100$ FV 205.01100$ FV 305.01100$ FV 405.01100$ FV 505.01100$ FV
Time Value of Money
1.20
4. FUTURE VALUE OF A SERIES OF
CASH FLOWS
Total FV is equal to the sum of FV of every single payment:
But if all the payments are equal, we can arrive at a general annuity formula:
r
rAFV
N11
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Time Value of Money
1.21
4. FUTURE VALUE OF A SERIES OF
CASH FLOWS
Hence, In this example we have:
Time Value of Money
1.22
5. PRESENT VALUE OF A SINGLE CASH FLOW
Present value is the discounted value of a future cash flow.
PV of a lump sum can be found through the following equation:
N
NrFV
rFVPV
1
1
1
Time Value of Money
1.23
Example 5: PV of a Lump Sum
An institution promises to pay you $100,000 in six years with an 8% annual interest rate. How
much should they invest today to have this
money at the end of 6th year?
Time Value of Money
1.24
5.1 The Frequency of Compounding
With more than one compounding period per year, the present value formula can be
expressed as:
mN
s
m
rFVPV
1
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Time Value of Money
1.25
Example 6: PV of a Lump Sum with Monthly Compounding
A company must make a $5 million payment 10 years from now. How much should they
invest today if the annual interest rate is 6%,
compounded monthly?
Time Value of Money
1.26
6. PRESENT VALUE OF A SERIES OF
CASH FLOWS
Consider an ordinary annuity paying 5% annually. Suppose we have 5 annual deposits
of $100 starting one year from now.
We are interested in PV of this ordinary annuity.
Now: t=0 2 1 3 4 5
$100 $100 $100
PV= ? $100 $100
Time Value of Money
1.27
6. PRESENT VALUE OF A SERIES OF
CASH FLOWS
Total PV is equal to the sum of PV of every single payment:
PV of 1st payment :
PV of 2nd payment :
PV of 3rd payment :
PV of 4th payment :
PV of 5th payment :
105.01100$ PV 205.01100$ PV 305.01100$ PV 405.01100$ PV 505.01100$ PV
Time Value of Money
1.28
6. PRESENT VALUE OF A SERIES OF
CASH FLOWS
Total PV is equal to the sum of PV of every single payment:
But if all the payments are equal, we can arrive at a general annuity formula:
r
rAPV
N11
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Time Value of Money
1.29
6. PRESENT VALUE OF A SERIES OF
CASH FLOWS
Hence, In this example we have:
Time Value of Money
1.30
Present Value of an Infinite Series of Equal
Cash Flows
Present value of an infinite series of equal cash flows can be calculated as:
r
APV
Time Value of Money
1.31
Example 6: An Infinite Series of Equal Cash Flows
A type of British government bonds pays $100 per year in perpetuity. What would it be worth
today if interest rate were 5%?
Time Value of Money
1.32
Thank You!
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Topic 2:
Time Value of Money (Part II) Discounted Cash Flow Applications
Associate Professor Ishaq Bhatti
La Trobe Business School
E-Mail: [email protected]
Slides have been drafted by the La Trobe University, School of Business based on DeFusco et al (2007)
Statistics for Business and Finance
Chapter 2 Discounted Cash Flow Applications
1. INTRODUCTION
Key Time Value of Money concepts: NPV and IRR
Making investment decision
Portfolio return measurement
Calculation of money market yields
Discounted Cash Flow Applications
2.1 Net Present Value (NPV)
Invest or Not?
You first need to know the present value of the cash flows
Second, you need to know how much the project cost you
If the project costs less than the PV of the cash flows you will invest; otherwise you will not
Discounted Cash Flow Applications
2.1 Net Present Value (NPV)
NPV is a method for choosing among alternative investments. The Net
Present Value is the present value of
cash inflows, minus the present value of
cash outflows.
0
1 0(1 ) (1 )
N Nt t
t tt t
CF CFNPV CF
r r
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Discounted Cash Flow Applications
Example 1: NPV
A project generates $100m next year, $150m
in year 2 and $120m in year 3. If the project
costs $310m and you can finance the
investment with an opportunity cost of 10%,
will invest in the project?
Discounted Cash Flow Applications
Example 1: NPV
What if the opportunity cost is 5%?
Would you invest in this project?
Discounted Cash Flow Applications
2.2 Internal Rate of Return (IRR)
Internal Rate of Return is the rate of return that makes the NPV equal to zero
IRR can be calculated using financial software or financial calculators, or trial and error method!
0)1(
...)1()1( 2
2
1
10
N
N
IRR
CF
IRR
CF
IRR
CFCFNPV
Discounted Cash Flow Applications
2.2 Internal Rate of Return (IRR)
Investment decision is to invest if IRR is more than the interest rate (opportunity cost) and not invest if IRR is
lower than the interest rate.
If discount rate is more than IRR, NPV will be negative and if interest rate is lower that IRR, NPV will be positive
Therefore, NPV and IRR will always lead to the same decision
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Discounted Cash Flow Applications
Example 2: IRR
A project generates $100m next year, $150m
in year 2 and $120m in year 3. If the project
costs $310m. What is the internal rate of
return? Is it 8.34%, 9.12% or 10.27%?
Discounted Cash Flow Applications
Example 2: IRR
Will you invest if the opportunity cost is 10%?
What if the interest rate is 5%?
Discounted Cash Flow Applications
2.3 Problems with IRR rule
IRR and NPV always lead to the same invest/not invest decision, but sometimes they rank the projects differently
If the scale of the projects differs
If projects have different timing of future cash flows
If there is a conflict between IRR and NPV, we should follow NPV as it reflects the real change in investors wealth
Discounted Cash Flow Applications
2.3 Problems with IRR rule
IRR and NPV always lead to the same invest/not invest decision, but sometimes they rank the projects differently
If the scale of the projects differs
If projects have different timing of future cash flows
If there is a conflict between IRR and NPV, we should follow NPV as it reflects the real change in investors wealth
If the sign of the cash flows changes more than once, we may get more than one IRR.
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Discounted Cash Flow Applications
3. Portfolio Return Measurement
Holding Period Return (HPR), the fundamental concept
Return that an investor earns over a specified holding period
0
101
P
DPPHPR
0
1
1
is the initial investment
is the price at the end of the period
is the cash paid by the investment
P
P
D
Discounted Cash Flow Applications
3.1 Money Weighted Rate of Return
IRR is called Money Weighted Rate of Return because it depends on timing and the dollar value of cash flows
IRR is not a good measure for investment managers
Usually, client decides when and how much to invest or withdraw
An evaluation tool should only judge the investment manager only for his own decisions, not for the clients
Discounted Cash Flow Applications
3.2 Time Weighted Rate of Return
The preferred measurement tool in investment management industry
Measures the compound rate of growth of each $1 of initial investment over the period
Does not depend on the dollar value of the investment
Not affected by withdrawals or additions to the portfolio
Discounted Cash Flow Applications
3.2 Time Weighted Rate of Return
Calculation of Time Weighted Rate of Return
Price the portfolio before any additions or withdrawals, breaking the period into subperiods
Calculate the HPR for each subperiod
Take the geometric mean of the calculated Holding Period Returns (HPR)
n1 2 nTWRR= (1+r ) (1+r ) ... (1+r ) 1
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Discounted Cash Flow Applications
Example 3: Money Weighted Rate of Return
At t = 0, an investor buys one share at $200. At time t = 1, he purchases an additional
share at $225,
At the end of Year 2, t = 2, he sells both shares for $235 each.
During both years, the share pays a per-share dividend of $5.
Calculate the Money Weighted Rate of Return
Discounted Cash Flow Applications
Example 3: Money Weighted Rate of Return
t = 0 t = 1 t = 2
$200 $225 $10 2 x $235 $5
Discounted Cash Flow Applications
Example 4: Time Weighted Rate of Return
t = 0 t = 1 t = 2
$200 $225 $10 2 x $235 $5
Discounted Cash Flow Applications
Example 4: Time Weighted Rate of Return
First period:
t = 0 t = 1 t = 2
$200 $225 $10 2 x $235 $5
P0=$200 P1=$225
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Discounted Cash Flow Applications
Example 4: Time Weighted Rate of Return
t = 0 t = 1 t = 2
$200 $225 $2 x 5 2 x $235 $5
P0=$225 P1=$235
Discounted Cash Flow Applications
Example 4: Time Weighted Rate of Return
n1 2 nTWRR= (1+r ) (1+r ) ... (1+r ) 1
t = 0 t = 1 t = 2
$200 $225 $2 x 5 2 x $235 $5
Discounted Cash Flow Applications
4. Money Market Yields
Consider two 1-year bonds of a company:
One with a $100 face value and $10 coupon at maturity;
The other with a $110 face value but no coupon
Are they selling at the same price today?
What is the interest of the second bond?
Many short-term debts (one-year maturity or less) pay no explicit coupon but they are sold at a discount.
Pure discount instruments
Discounted Cash Flow Applications
4. Money Market Yields
Pure discount instruments such as T-bills are quoted on a Bank Discount basis, rather than
on a price basis:
rBD: annualized yield on a Bank Discount basis
D: dollar discount = face value purchase price
F: face value
T: actual number of days remaining to maturity
360: bank convention of the number of days in a year
tF
DrBD
360
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Discounted Cash Flow Applications
Thank You!
Topic 3:
Statistical Concepts and Market
Returns
Associate Professor Ishaq Bhatti
La Trobe Business School
E-Mail: [email protected]
Slides have been drafted by the La Trobe University, School of Business based on DeFusco et al (2007)
Statistics for Business and Finance Chapter 3
Statistical Concepts and Market Returns
2.59
INTRODUCTION
Statistical methods provide a powerful set of tools for analyzing data.
Descriptive statistics includes basics of describing and analyzing data.
We explore four properties of return distributions: Where the returns are centered (central tendency)
How far returns are dispersed from their center (dispersion)
Whether the distribution of returns is symmetrically shaped or not (skewness)
Whether extreme outcomes are likely (kurtosis)
Statistical Concepts and Market Returns
2.60
Descriptive and Inferential Statistics
Descriptive statistics is to summarize a small set of data (sample) effectively to describe the important aspects of a larger dataset (population)
Statistical inference is to make forecasts. estimations or judgments about a larger dataset (population) from a smaller group (sample) actually observed.
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Statistical Concepts and Market Returns
2.61
Frequency Distributions
The simplest way of summarizing data is the frequency distribution.
A frequency distribution is a tabular display of data summarized into a
relatively small number of intervals.
Statistical Concepts and Market Returns
2.62
Example: Frequency Distributions
Banna Insurance Pty Ltd - 4 sales incentive
programs A, B, C and D. 40 Salespeople
asked for their opinion of the preferred
program.
Fill the following table:
B A D C A C D B D B
D D B A D B D A D C
D B C D A D B D B C
B A D B A B A C D B
Statistical Concepts and Market Returns
2.63
Example 1: Frequency Distributions
Program Frequency Relative Frequency Cumulative
Frequency Cumulative Relative
Frequency
A
B
C
D
Statistical Concepts and Market Returns
2.64
Histogram & Bar Chart
Histogram consist of adjacent rectangles whose bases are marked off by class width and their heights are
proportional to the frequencies they possessed.
Bar chart is special form of histograms where bars are not adjacent and the data have been grouped into a
frequency distribution.
The following two slides display the bar chart of absolute and relative frequency distributions of example
1. Similarly you can compute histogram of ASX200
returns; Text page
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Statistical Concepts and Market Returns
2.65
Bar Chart of Banna Insurance Example
Statistical Concepts and Market Returns
2.66
Bar Chart of Banna Insurance Example
Statistical Concepts and Market Returns
2.67
Measures of Central Tendency
Arithmetic Mean:
The (arithmetic) mean is the sum of the observations divided by the number
of observations.
The population mean is given by
The sample mean looks at the arithmetic average of the sample of data:
The mean return of ASX200 is 0.70%
NXN
i
i
1
nXXn
i
i
1
Statistical Concepts and Market Returns
2.68
Example: Arithmetic Mean
Find the mean of 46, 54, 42, 46, 32:.
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Statistical Concepts and Market Returns
2.69
Median
Median is the value of the middle item of a set of items, sorted by ascending or descending
order.
If n is an odd number it occupies the (n+1)/2 position.
If n is an even number it is the average of items in the n/2 and (n+2)/2 positions.
Unlike the mean, the median is not affected by a few large observations.
Statistical Concepts and Market Returns
2.70
Example: Median
Find the median of 46, 54, 42, 46, 32:.
Statistical Concepts and Market Returns
2.71
Mode
The mode is the most frequently occurring value in a distribution.
A distribution can have more than one mode or even no mode.
Stock returns or other data from continuous distribution may not have a modal outcome but we often find the modal interval (intervals) Which internal in our Banna Insurance example is the
modal interval? (Hint: see the Histogram in slide 9)
Statistical Concepts and Market Returns
2.72
Example: Mode
Find the mode of 46, 54, 42, 46, 32:.
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Statistical Concepts and Market Returns
2.73
Weighted Mean
The weighted mean allows us to place greater importance on different observations.
For example, we may choose to give larger companies greater weight in our computation of an
index. In this case, we would weight each
observation based on its relative size.
The arithmetic mean is a special case where each observation is given the same weight.
i
i
n
i
iiw wXwX 1 where,1
Statistical Concepts and Market Returns
2.74
Geometric Mean
The geometric mean is most frequently used to average rates of change over time
or to compute the growth rate of a
variable.
which can also be calculated by
n, . . . , , i XXXXG in
n 21for 0 with ,]...[/1
21
n
i
iXn
G1
ln1
ln
Statistical Concepts and Market Returns
2.75
Geometric Mean Return
Geometric mean requires all data are positive It cannot be applied to observations with negative data like return.
The geometric mean return allows us to compute the average return when there is
compounding.
1)1(
)1)...(1)(1)(1(1
1
1
1
321
TT
t tG
TTG
RR
RRRRR
Statistical Concepts and Market Returns
2.76
Quartiles and Percentiles
If your lecturer tells you that your exam mark is in top 10%, what does it mean?
Median divides the data in half. The dataset can be also divided into:
Quartiles; Quintiles; Deciles; Percentiles
To find the value of yth percentile with n observations:
first organize data in ascending order
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Statistical Concepts and Market Returns
2.77
Quartiles and Percentiles
then the location of the yth percentile is at
Ly = (n + 1) * y%
however, some software package like Excel uses location Ly = [(n - 1)*y%] +1
if Ly is an integer, the value of yth percentile, Py, is equal to the observation at Ly
if Ly is not an integer, it can be written as Ly = k.d, where . is the decimal point, and
Py =Vk + .d (Vk+1 - Vk),
where Vk and Vk+1 are the values of kth and (k + 1)th
observations.
Statistical Concepts and Market Returns
2.78
Example: Percentiles
Find the values of 25th and 40th percentiles of the 46, 54, 42, 46, 32:
Note, median = P50, = Q2 = D5, P75 = Q3, , here Q and D denote Quartile
and Decile.
Statistical Concepts and Market Returns
2.79
Measures of Dispersion
One of simplest measures of dispersion is the range, which is the difference between
the maximum and minimum values in a
dataset:
Range = Maximum value Minimum value.
Statistical Concepts and Market Returns
2.80
Mean Absolute Deviation
Mean Absolute Deviation (MAD) measures the average distance that each observation
is from the mean:
nXXMADn
i
i
1
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Statistical Concepts and Market Returns
2.81
Population Variance and Standard
Deviation
Variance measures the average squared deviation from the mean:
Because the variance is not in the same units as the mean, sometimes we prefer the standard deviation, the
square root of variance, which is in the same units as
the mean
NXN
i
i
1
22 )(
NXN
i
i
1
2)(
Statistical Concepts and Market Returns
2.82
Sample Variance and Standard Deviation
Sample Variance and standard deviation slightly differ from population variance and
standard deviation:
)1()(
1
22
nXXsn
i
i)1()(
1
2
nXXsn
i
i
Statistical Concepts and Market Returns
2.83
Example: MAD and Standard Deviation
Find the MAD for 46, 54, 42, 46, 32:
iX iX XX iX X
Statistical Concepts and Market Returns
2.84
Example: MAD and Standard Deviation
Find Std Dev. for 46, 54, 42, 46, 32:
iX iX XX 2
iX X
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Statistical Concepts and Market Returns
2.85
Semivariance
Often times, observations above the mean are good the variance is not a good measure of risk. Semivariance looks at
the average squared deviations below the
mean:
where n* is the number of observations
greater than the average of observations.
XX
i
in
XX
allfor *
2
)1(
)(
Statistical Concepts and Market Returns
2.86
Coefficient of Variation
The coefficient of variation is the ratio of the standard deviation to their mean value.
measure of relative dispersion
can compare the dispersion of data with different scales
What is the coefficient of variation of the dataset in previous example?
XsCV
Statistical Concepts and Market Returns
2.87
Sharpe Ratio
The Sharpe ratio is the ratio of mean excess return to riskan application of mean and standard deviation analysis.
Risk averse investors who make decisions only in terms of mean and standard deviation prefer
portfolios with larger Sharpe ratios:
Assuming monthly risk-free interest rate is 0.3%, the Sharpe ratio of ASX200 is 0.121.
p
Fp
hs
RRS
Statistical Concepts and Market Returns
2.88
Skewness
Skewness measures the symmetry of a distribution.
A symmetric distribution has a skewness of 0.
Positive skewness indicates that the mean is greater than the median (more than half the deviations from
the mean are negative)
Negative skewness indicates that the mean is less than the median (less than half the deviations from
the mean are negative)
3
1
3)(
)2)(1( s
XX
nn
nS
n
i i
K
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Statistical Concepts and Market Returns
2.89
Graphic Illustration of Skewness
Statistical Concepts and Market Returns
2.90
Sample Excess Kurtosis
Kurtosis measures how peaked the distribution is relative to the normal distribution.
Using the sample excess kurtosis formula
KE 0 is called Mesokurtic, which means the distribution is normally distributed
KE > 0 is called Leptokurtic, which means the distribution is more peaked.
KE < 0 is called Platykurtic means the is less peaked.
)3)(2(
)3(3)(
)3)(2)(1(
)1( 2
4
1
4
nn
n
s
XX
nnn
nnK
n
i i
E
Statistical Concepts and Market Returns
2.91
Leptokurtic: Fat Tailed
Statistical Concepts and Market Returns
2.92
Thank You!
FIN5SBF
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Topic 4:
Probability Concepts
Associate Professor Ishaq Bhatti
La Trobe Business School
E-Mail: [email protected]
Slides have been drafted by the La Trobe University, School of Business based on DeFusco et al (2007)
Statistics for Business and Finance
Chapter 4 Probability Concepts
4.94
Uncertainty and Probability
A random variable is a quantity whose outcomes are uncertain.
An event is a specified set of outcomes.
Probability: the likelihood or chance that something is the case or will happen
The probability of any event, E, is a number between .
The sum of the probabilities of any set of mutually exclusive & exhaustive events equals.
Probability Concepts
4.95
Uncertainty and Probability
A random variable is a quantity whose outcomes are uncertain.
An event is a specified set of outcomes.
Probability: the likelihood or chance that something is the case or will happen The probability of any event, E, is a number between 0 and 1: 0 P(E)
1 The sum of the probabilities of any set of mutually exclusive &
exhaustive events equals 1.
Probability of an event A is equal to:
Probability Concepts
4.96
Example
An experiment is conducted in which a coin is tossed three times - the uppermost face recorded on each toss. Draw the tree diagram.
First throw Second throw Third throw
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Probability Concepts
4.97
Example
If A is the event of throwing 2 heads and 1 tail in any order, then:
A=
and P(A) =
Probability Concepts
4.98
Uncertainty and Probability
Mutually exclusive events are those only one of which can occur at a time.
Exhaustive events are the events that cover all possible outcomes.
How to estimate probability? An empirical probability is estimated by relative frequency of occurrence
based on historical data
A subjective probability is one drawing on personal or subjective judgment.
A priori probability is one based on logical analysis rather than on observation or personal judgment.
A priori or an empirical probability is also called objective probability.
Probability Concepts
4.99
Example
A die is thrown. Determine the probability of obtaining a number 2 or a 5.
AB
S
Probability Concepts
4.100
Example
A die is thrown. Determine the probability of obtaining a multiple of 2 or a multiple of 3.
A
B
S
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Probability Concepts
4.101
Example
A die is tossed twice. Determine the probability of obtaining a 3 on the first toss and number > 5 on the second toss.
A
B
S
Probability Concepts
4.102
Example
A die is tossed twice. Determine the probability of obtaining a 3 on the first toss and a total of 5 on both tosses.
A
B
S
Probability Concepts
4.103
Multiplication and Addition Rules for
Independent Events
When two events are independent, the joint probability is the product of two probabilities
Consequently, addition rule becomes
Example: What is the probability of tossing two heads in a row?
Probability Concepts
4.104
Independent Events
Two events are independent if and only if
)B(P)A|B(Ply equivalentor )A(P)B|A(P
FIN5SBF
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Probability Concepts
4.105
Multiplication Rule for Probability
The joint probability can be found using the multiplication rule:
On the other hand, if joint probability P(AB) and unconditional probability P(B) are known,
the conditional probability is
)B(P)B|A(P)AB(P
0P(B) ,)B(P
)AB(P)B|A(P
Probability Concepts
4.106
Unconditional and Conditional
Probabilities
Unconditional or marginal probability answers question, What is the probability of event A.
Conditional probability answers the question, What is the probability of event A, given that event B occurs.
Joint probability answers the question, What is the probability of both events A and B
happening.
Probability Concepts
4.107
Example
A die is tossed twice. Determine the probability of obtaining a total of 5 on both tosses if a 3 is obtained on the first toss.
A BS
3,1 3,3 3,4 3,5 3,6
3,2
2,3 1,4 4,1
Probability Concepts
4.108
Example:
A sample of Business Degree evening students was surveyed in order to investigate the relationship between age and marital status. The results of
the survey are tabled below:
Answer the following questions
Marital Status
S M
Age
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Probability Concepts
4.109
Example:
What is the sample size?
Are the events being single and being < 30 independent?
( 30)P Age
( )P Being Single
( AND < 30)P Being Single
( GIVEN < 30)P Being Single
Probability Concepts
4.110
The Total Probability Rule
)()|()()|(
)()()( .1
CC
C
SPSAPSPSAP
ASPASPAP
)()|(...)()|()()|(
)(...)()()( .2
2211
21
nn
n
SPSAPSPSAPSPSAP
ASPASPASPAP
eventsor scenarios exhaustive
and exclusivemutually are ,...,S where 21 nSS
where S is an even and SC is the even not-S or the
complement of S
Probability Concepts
4.111
Expected Value
The expected value of a random variable is the probability weighted average of the
possible outcomes of the random variable.
n
i
ii
nn
XXP
XXPXXPXXPXE
1
2211
)(
)(...)()()(
Probability Concepts
4.112
Variance
The variance of a random variable is the expected value of squared deviations from
the random variables expected value:
Note, a better notation is
2 2 2 2( ) {[ ( )] } ( ) ( )X E X E X E X E X
n
i
ii XEXXPX1
22 )]()[()(
2
X
FIN5SBF
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Probability Concepts
4.113
Variance
(i) Var (c) = 0
(ii) Var (cX) = c2 Var (X)
(iii) Var (c + X) = Var (X)
(iv) Var (X + Y) =Var (X) + Var (Y), if X and Y are independent
=Var (X) + Var (Y) + 2 COV (X,Y) if X and Y
are not independent
Probability Concepts
4.114
Standard Deviation
Standard deviation is the positive square root of variance.
2
Probability Concepts
4.115
Example
A random variable, X, has the following probability distribution:
xi
2P( xi)X=xi P(xi) or
P(X=xi)
xiP(xi)
0 0.1
1 0.6
2 0.3
Total 1.0
Probability Concepts
4.116
Example
Find the following:
E(X)
E(X2)
E(Y), if Y = aX cX2, where a = 1, c = 2
E(W), if W = (d c) X + a, where d = 8
Var(X)
Var(W)
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Probability Concepts
4.117
Conditional Expected Value
A conditional expected value is the expected value of a random variable X
given an event or scenario S, is denoted
E(X|S):
Total probability rule for expected value
)()|(...)()|()()|()|( 2211 nn SPSXESPSXESPSXESXE
nn XSXPXSXPXSXPSXE )|(...)|()|()|( 2211
Probability Concepts
4.118
Conditional Variance
Since variance is the expected value of random variable
we can define conditional variance
accordingly
SSXEXESXVar |)]|([)|( 2
2)]([ XEXE
Probability Concepts
4.119
Portfolio Expected Return and Variance
Investment diversification and portfolio
Portfolio is the profile of the investment
If there are n assets and you invest wi (i =1, 2, .., n) portion of your wealth in asset i, then the investment
portfolio is (w1, w2 , wn). The portfolio return is
where Ri is the return of asset i.
Modern portfolio theory often uses expected return as the measure of reward and the
variance of returns as a measure of risk.
nnp RwRwRwR ...2211
Probability Concepts
4.120
Properties of Expected Value
The expected value of a constant times a random variable equals the constant times
the expected value of the random variable
Expected value of the sum of random variables is equal to the sum of expected
values of the random variables:
)(...)()(
)...(
2211
2211
nn
nn
REwREwREw
RwRwRwE
)()( iiii REwRwE
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Probability Concepts
4.121
Calculation of Portfolio Expected Return
Given a portfolio with n securities, the expected return on the portfolio is a
weighted average of the expected
returns on the component securities.
)(...)()()( 2211 nnP REwREwREwRE
Probability Concepts
4.122
Covariance and Correlation
)])([(),( jjiiji ERRERRERRCov
)()(),( jijiij RRRRCov
Covariance of two random variables is defined as
Correlation coefficient of two random variables is defined as
Probability Concepts
4.123
Interpretation of Return Covariance
If the covariance is 0, the returns on the assets are unrelated.
If the covariance is negative (positive), when the returns on one asset is above its expected
value, the returns of the other asset tend to be
below (above) its expected value; i.e the two
returns tends to move in the same (opposite)
direction.
The covariance of a random with itself is its own variance.
Probability Concepts
4.124
Interpretation of Correlation Coefficient
Correlation is a scaled covariance that falls between -1 and +1.
A correlation of +1 means the variables are perfectly positively correlated.
A correlation of -1 means the variables are perfectly negatively correlated.
A correlation of 0 means the variables are uncorrelated.
FIN5SBF
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Probability Concepts
4.125
Portfolio Variance
Unlike portfolio expected return, portfolio variance is not a weighted average of the
variances of the securities in the portfolio.
To compute portfolio variance, we need to incorporate the interaction between each
pair of variables (correlation or
covariance).
Probability Concepts
4.126
Portfolio Variance
Portfolio variance for a two-security portfolio.
Portfolio variance for an n-security portfolio.
211221
2
2
2
2
2
1
2
1
2121
2
2
2
2
2
1
2
1
2
2
),(2)(
wwww
RRCovwwwwRP
),()(1 1
2
n
i
n
j
jijiP RRCovwwR
Probability Concepts
4.127
Thank You!
Topic 5:
Common Probability Distributions
Associate Professor Ishaq Bhatti
La Trobe Business School
E-Mail: [email protected]
Slides have been drafted by the La Trobe University, School of Business based on DeFusco et al (2007)
Statistics for Business and Finance
Chapter 5
FIN5SBF
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Common Probability Distributions
5.129
Random Variables and Distributions
A probability distribution specifies the probabilities of the possible outcomes of a random variable.
There are two types of random variables:
A discrete random variable can take on at most a countable number of possible values;
A continuous random variable takes infinitely many values, on an interval, say between [0, 1] or (-, +)
Common Probability Distributions
5.130
Random Variables and Distributions
For discrete random variable, the probability function specifies the probability that the random
variable takes on a specific value.
For continuous variable,
The probability density function p(x) specifies the probability density the random variable takes on the
value x or the approximate probability the random
variable takes on values around x of a unit length.
The cumulative distribution function P(x) gives the probability that the random variable is less than or equal
to x.
Common Probability Distributions
5.131
Bernoulli Random Variable
Sometimes a random variable can only take on two values, success or failure. This is referred to
as a Bernoulli random variable.
A Bernoulli trial is an experiment that produces only two outcomes.
Y = 1 for success and Y = 0 for failure.
p1)0Y(P)0(pp)1Y(P)1(p
Common Probability Distributions
5.132
Binomial Distribution
A binomial random variable X is defined as a number Bernoulli trials.
The probability of x successes out of n trials is
The mean and variance of B(n,p) are:
= np
2 = np(1-p).
n21 YYYX
xnxxnx ppxxn
npp
x
nxXPxp
)1(
!)!(
!)1()()(
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Common Probability Distributions
5.133
Binomial Distribution
General notations (pages 166-168):
n factorial: n! n(n-1)(n-2)1 and 0! 1
Combination (x choices out of n options)
Binomial distribution assumes
The probability, p, of success is constant for all trials
The trials are independent
!)!(
!
xxn
n
x
nCrn
Common Probability Distributions
5.134
Example: Binomial Distribution
Flipping a fair coin: Probability of head = 50% and probability of tail = 50%
If you flip three coins in a row, what is the probability you have two heads and one tail?
So, can we answer the question?
Common Probability Distributions
5.135
Example: Binomial Distribution
Three customers enter a clothing store. The probability that a customer will make a purchase p(s) is 0.30. investigate the probability distribution.
Common Probability Distributions
5.136
Example: A Binomial Model of Stock
Price Movements
If the probability of stock price moving up is 60% and down is 40%, what is the probability that the stock price goes up in exactly two years?
Find the probability of an upward movement in the first two years followed by a fall in the price in the third year.
What is the probability that the price goes down at least twice?
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Common Probability Distributions
5.137
Example: A Binomial Model of Stock
Price Movements
Common Probability Distributions
5.138
Continuous Uniform Distribution
Probability density function (pdf) and cumulative distribution function (cdf) of uniform distribution on [a, b] are:
The mean and variance of a continuous distribution:
Mean: E(X)= = (a+b)/2
Variance: Var (x) = 2 = (a+b)2/12
otherwise 0
for 1
)(bxa
abxf
1
for
for 0
)(
bfor x
bxaab
ax
ax
xF
Common Probability Distributions
5.139
Normal Distribution
Random variable X follows a normal distribution with mean and variance 2 (X ~ N(, 2)) if it has a probability density function as:
There is not a closed-form cdf for a normal distribution and we have to use a table of cumulative probabilities for a normal distribution
A normal distribution can be determined using its mean and variance.
x
xxf for
2
)(exp
2
1)(
2
2
Common Probability Distributions
5.140
Normal Distribution
A normal distribution has a skewness of 0 (it is symmetric)
its mean, median and mode (slightly abuse the term) are equal
It has a kurtosis of 3 (or excess kurtosis of 0).
A linear combination of two or more normal random variables is also normally distributed
This property is vary useful to determine the distribution of portfolio return, given each assets return.
FIN5SBF
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Common Probability Distributions
5.141
Graphic Illustration of Two Normal
Distributions
Common Probability Distributions
5.142
Units of Standard Deviation
Common Probability Distributions
5.143
Units of Standard Deviation
Approximately 50 percent of all observations fall in the interval (2/3).
Approximately 68 percent of all observations fall in the interval .
Approximately 95 percent of all observations fall in the interval 2.
Approximately 99 percent of all observations fall in the interval 3.
Common Probability Distributions
5.144
Confidence Intervals for Values of a Normal
Random Variable X
We expect
90 percent of the values of X to lie within the interval
95 percent of the values of X to lie within the interval
99 percent of the values of X to lie within the interval
These intervals are called 90%, 95% and 99% confidence intervals for X.
s96.1X
s65.1X
s58.2X
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Common Probability Distributions
5.145
Standard Normal Distribution
Standard normal distribution has a mean of zero and a standard deviation of 1.
If X is a normal random variable that X ~ N(, 2), then Z follows the standard normal distribution (i.e., Z ~ N(0, 1)), if:
XZ
Common Probability Distributions
5.146
Example: Normal Distribution
Find the following probabilities: (Z table is available in the next slide)
1. P (0 z 1.4)=
2. P (0 z 1.46)=
3. P (-1.5 z 1.5)=
Common Probability Distributions
5.147
Common Probability Distributions
5.148
Example: Normal Distribution
A portfolio has an estimated mean return of 12% and standard deviation of return of 22%.
What is the probability that portfolio return will exceed 20%?
What is the probability of that portfolio return will be between 5.5% and 20%?
What is the returns 90% confidence interval?
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Common Probability Distributions
5.149
Application: Safety-First Rule
Roy demonstrates that if portfolio return, RP, is normal then minimizing the probability of RP falling below, RL,
require maximizes
PLP RRE /])([SFRatio
Common Probability Distributions
5.150
Application: Safety-First Rule
You are managing an $800,000 portfolio for an investor whose objective is long-term growth. But she may want to liquidate $30,000
at the end of a year. If that need arises, she hopes the liquidation of
$30,000 would not invade the initial capital of $800,000. If return on
the portfolio is 8% with 3% std deviation:
To protect the initial investment, portfolio managers rank the investents based on the SFRatios
Common Probability Distributions
5.151
Application: Safety-First Rule (Cont.)
There are three investment alternatives : A B C
Expected annual return (%) 25 11 14
Standard deviation (%) 27 8 20
What is the shortfall level (RL)?
According to safety-first criterion, which of the three allocations is the best?
What is the probability that the return on the safety-first optimal portfolio will be less than the shortfall level?
Common Probability Distributions
5.152
Lognormal Distribution
The lognormal distribution is widely used for modeling asset prices.
A random variable Y follows a lognormal distribution if and only if X = lnY is normally distributed.
Note,
If X has mean and variance 2, then Y have mean exp( + 0.5 2) and variance exp(2 + 2)[exp(2) 1].
XY e
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Common Probability Distributions
5.153
Two Lognormal Distributions
Common Probability Distributions
5.154
Thank You!
Topic 6:
Sampling and Estimation
Associate Professor Ishaq Bhatti
La Trobe Business School
E-Mail: [email protected]
Slides have been drafted by the La Trobe University, School of Business based on DeFusco et al (2007)
Statistics for Business and Finance
Chapter 6 Sampling and Estimation
6.156
Sampling
In statistics we are often interested in obtaining information about the value of some parameters of a population.
To obtain this information we usually take a small subset of the population and try to draw some conclusions from this sample.
A sampling plan is the set of rules used to select a sample.
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Sampling and Estimation
6.157
Simple Random Sampling
A simple random sample is a subset of a larger population created in such a way that each element
of the population has an equal probability of being
selected.
Sampling and Estimation
6.158
Stratified Random Sampling
Stratified random sampling occurs when the population is divided into subpopulations
(strata) and a simple random sample is drawn
from each strata.
It guarantees that population subdivisions of interests are represented in the sample.
It generates more accurate estimates (smaller variance) than simple random sampling
Sampling and Estimation
6.159
Types of Sample Data
Cross-sectional data represent observations over individual units at a point in time;
Time series data is a set of observations on a variables outcomes in different time periods;
Panel data have both time-series and cross-sectional aspects and consist of observations
through time on a single characteristics of
multiple observational units.
Sampling and Estimation
6.160
Sampling Error and Statistic
Sampling error is the difference between the observed value of a statistic and the quantity it is intended to estimate.
Sampling distribution of a statistic is the distribution of all the distinct possible values that the statistic can assume when computed from samples of the same size randomly drawn from the same population.
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Sampling and Estimation
6.161
Example: Distribution of Sample Mean
Suppose we have a 'population' of 5 elements with values 1, 2, 3, 4, 5
What are the average and standard deviation of this population?
Now, consider all possible samples of size 3 to provide a point estimate of the population mean, and find the average of each
sample.
Sampling and Estimation
6.162
Example: Distribution of Sample Mean
Now, consider all possible samples of size 3 to provide a point estimate of the population mean, and find the average of each
sample. What is the average and standard deviation of the new
distribution? x Possible Samples, Size 3
1, 2, 3
1, 2, 4
1, 2, 5
1, 3, 4
1, 3, 5
1, 4, 5
2, 3, 4
2, 3, 5
2, 4, 5
3, 4, 5
Sampling and Estimation
6.163
Shape of the Sampling Distribution of Sample Mean
If n is large enough (>30) the sampling distribution of will be a normal distribution regardless of the distribution
type exhibited by the population.
If the population distribution is normal the sampling distribution of will be normal regardless of the sample
size.
X
X
Sampling and Estimation
6.164
Standard Error of the Sample Mean
When we use the sample mean to estimate the population mean, there are some errors.
The standard error of the sample mean is the standard deviation of the difference between the sample mean and the population mean.
For a sample mean calculated from a sample generated from a population with standard deviation , the standard error of the
sample mean is
when population standard deviation () is known.
nX
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Sampling and Estimation
6.165
Standard Error of the Sample Mean
In practice, the population variance is almost always unknown. The standard error of the sample mean is
estimated by,
Note,
XXs
)1( where,1
22
nXXsnssn
i
iX
Sampling and Estimation
6.166
Central Limit Theorem
The central limit theorem: Given a population described by any probability distribution having
mean and finite variance 2, the sampling distribution of the sample mean computed
from samples of size n from this population will
be approximately normal with mean (the population mean) and variance 2/n (the population variance divided by n) when the
sample size n is large.
X
Sampling and Estimation
6.167
Example: Central Limit Theorem
Electronics Associates Industry has 2500 managers on salaries such that = $31,800 and s = $4000.
What is the probability that a random sample of 30 managers will have a
mean salary that lies within $1000 of the population mean?
Sampling and Estimation
6.168
Confidence Intervals
Any estimate has errors. But we know that the estimated parameter must be around the
estimate with high probability.
A confidence interval is an interval for which we can assert with a given probability 1 , called the degree of confidence, that it will contain the
parameter it is intended to estimate.
Note, here we move from a point estimation to internal estimation.
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Sampling and Estimation
6.169
Confidence Intervals
A (1 )% confidence interval for a parameter has the following structure:
Point estimate Reliability factor Standard error the reliability factor is a number based on the
assumed distribution of the point estimate and the degree of confidence (1 ) for the confidence interval
standard error is the standard error of the sample statistic providing the point estimate.
Sampling and Estimation
6.170
Confidence Intervals for the Population Mean
For normally distributed population with known variance.
For large sample, population variance unknown.
nzX 2/
n
szX 2/
Sampling and Estimation
6.171
Confidence Intervals for the Population Mean
For population variance unknown, we have to use t-distribution
The t-distribution is a symmetrical probability distribution defined by a single
parameter known as degrees of freedom
(df).
n
stX 2/
Sampling and Estimation
6.172
(Students) t-Distribution versus the Standard Normal Distribution
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Sampling and Estimation
6.173
Choose Which Reliability Factor?
Sampling from Small
Sample Size
Large
Sample Size
Normal, Known Var z z
Normal, Unknown Var t t (or z)
Nonnormal, Known Var N/A z
Nonnormal, unknown Var N/A t (or z)
Sampling and Estimation
6.174
Example: Confidence Interval for
Population Mean
Now reconsidering that the population variance of the distribution of Sharpe ratio
is unknown, the analyst decides to
calculate the confidence interval using the
theoretically correct t-statistic.
Compare the result obtained under normal distribution.
Sampling and Estimation
6.175
Example: Confidence Interval for
Population Mean
ABC insurance surveys 36 policyholder in order to obtain an estimate of the average age of all policy holders. If average age of a sample of policy
holders 39.5 years with 1.8 years standard deviation, determine a 90%
confidence interval for the average policy holders age.
.
Sampling and Estimation
6.176
Example: Confidence Interval for
Population Mean
Banana Bank is interested in introducing a new computer based training program for use by its employees. A sample of 15 employees is selected to undergo training on the new system and on completion it is found that the duration of training is 53.87 days and s= 6.82 days. Determine a 95% confidence interval estimate.
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Sampling and Estimation
6.177
Example: Confidence Interval for
Population Mean
City insurance has found that from a sample of size 36, with mean age 39.5 years and standard deviation 1.8 years, we can say with 90% confidence that the sampling error associated with the sample mean is 0.5. What sample size would we need to obtain this sampling error with 95% confidence.
Sampling and Estimation
6.178
Selection of Sample Size
What conclusion can we draw from the previous example?
All else equal, a larger sample size decreases the width of the confidence interval because
and reliability factor (or t-critical value) declines in degree of freedom.
size Sample
deviation standard Sample mean sample theoferror Standard
Sampling and Estimation
6.179 5.179
Thank You!
Topic 7:
Hypothesis Testing
Associate Professor Ishaq Bhatti
La Trobe Business School
E-Mail: [email protected]
Slides have been drafted by the La Trobe University, School of Business based on DeFusco et al (2007)
Statistics for Business and Finance
Chapter 7
FIN5SBF
46
Hypothesis Testing
Hypothesis Testing
Statistical inference as two subdivisions: estimation and hypothesis testing
Estimation addresses the question: what is this parameters value? For example, what is the population mean of annual returns for
company XYZ? The answer is usually a confidence interval built around a point
estimate.
A hypothesis testing question is:" Is the value of that parameter equal to ? For example, is the average annual return of XYZ equal to 10%
The statement Average annual return of XYZ is equal to 10% is called a hypothesis
Hypothesis Testing
1.Stating the hypotheses.
2.Identifying the appropriate test statistic
and its probability distribution.
3.Specifying the significance level.
4.Stating the decision rule.
5.Collecting the data and calculating the
test statistic.
6.Making the statistical decision.
Steps in Hypothesis Testing
Hypothesis Testing
Null vs. Alternative Hypothesis (Step 1)
The null hypothesis is the hypothesis to be tested. Null hypothesis can never be accepted. We can either reject the null
(hence accepting the alternative) or not reject the null (and not
accepting the alternative either!)
The alternative hypothesis is the hypothesis accepted when the null hypothesis is rejected.
Eg: H0: Population average risk premium for Canadian equities is less
than or equal to zero.
HA: Population average risk premium for Canadian equities is greater than zero.
Hypothesis Testing
Formulation of Hypotheses (Step 1)
There three ways to form hypothesis
1. H0: = 0 versus Ha: 0
2. H0: 0 versus Ha: > 0 3. H0: 0 versus Ha: < 0
The first formulation is a two-sided test. The other two are one-sided tests.
Eg. we may want to test
H0: Population average risk premium for Canadian equities is less than or equal to zero.
HA: Population average risk premium for Canadian equities is greater than zero.
0 : 0H
: 0aH
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Hypothesis Testing
Test Statistic (Step 2)
A test statistic is a quantity, calculated based on a sample, whose value is the
basis for deciding whether or not to reject
the null hypothesis.
statistic sample theoferror Standard
Hunder parameter population theof Valuestatistic SamplestatisticTest 0
Hypothesis Testing
Test Statistic (Step 2)
Note, if we test a hypothesis of population mean, the standard error of the sample statistic is calculated by
the same formulas as we used the last topic:
In our example, we are testing whether the average can be 0. The test
statistic for our example is
unknown is population theofdeviation standard theif ,nssX
known is population theofdeviation standard theif ,nX
0 0 or
X X
X X
s
Hypothesis Testing
Two types of Errors (Step 3)
In reaching a statistical decision, we can make two possible errors:
We may reject a true null hypothesis (a Type I error), or
Probability of type I error is denoted by the Greek letter alpha,
We may fail to reject a false null hypothesis (a Type II error).
Probability of type II error is denoted by the Greek letter beta,
Hypothesis Testing
Level of Significance (Step 3)
The level of significance of a test is the probability of a Type I error that we are
preparing to accept in conducting a hypothesis
test, is denoted by .
The standard approach to hypothesis testing involves specifying a level of significance
(probability of Type I error) only.
Conventional significance levels: 0.1 (some evidence), 0.05 (strong evidence), 0.01 (very
strong evidence).
FIN5SBF
48
Hypothesis Testing
Level of Significance (Step 3)
Trade-off: all else equal, if we decrease the probability of a type I error by increasing specifying a smaller significance level, we increase the probability of making a Type II error.
The power of a test is the probability of correctly rejecting the null (rejecting the null when it is false).
It is equal to 1 minus the probability of a Type II error.
Hypothesis Testing
Rejection Points (Step 4)
A rejection point (critical value) for a test statistic is a value with which the computed test statistic is compared to decide whether to reject the null hypothesis or not.
For a one-tailed test, we indicate a rejection point using the symbol for the test statistic with a subscript of significance level (e.g., z, t)
For a two-tailed test, the subscript is a half of the significance level (e.g., z/2, t/2)
Hypothesis Testing
Example: Rejection Points of a One-
Sided z-test
If Canadian equity risk premium is normal and its variance is known, then we can use a z-test to test the
hypotheses at the 0.05 level of significance:
H0: 0 (average risk premium is less than or equal to zero) versus
HA: > 0 (average risk premium is greater than zero)
One rejection point exists: z0.05 = 1.65
Hypothesis Testing
Example: Rejection Points of a One-
Sided z-test
So if our sample data yield
we reject the null hypothesis.
Otherwise, we do not reject the null hypothesis.
We cant accept it either!
This is illustrated by the following slide.
Note, H0: 0 versus Ha: < 0, the rejection point is z0.05 = -1.645 and we reject null hypothesis if z < -1.645
01.645
X
X
FIN5SBF
49
Hypothesis Testing
Rejection Point, 0.05 Significance Level, One-Sided Test of the
Population Mean Using a z-Test
Hypothesis Testing
Example: Rejection Point of a Two-Sided
z-test
On the other hand, if the hypotheses are
H0: = 0 (The average Canadian equity risk premium is zero) versus
Ha: 0 (The average Canadian equity risk premium is not equal to zero)
There exists two rejection points. If significance level is still 0.05, the rejection points are:
z0.025 = 1.96 and -z0.025 = -1.96
from the normal distribution table.
Hypothesis Testing
Rejection Points, 0.05 Significance Level, Two-Sided Test of
the Population Mean Using a z-Test
Hypothesis Testing
Example: Rejection Points of a Two-
Sided z-test
So if sample data yield either
we reject the null hypothesis as illustrated by the next
slide. But we do not reject the null hypothesis if
0 01.96 or 1.96
X X
X X
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Hypothesis Testing
Confidence Interval
The (1 ) confidence interval represents the range of values of the test statistic for
which the null hypothesis will not be rejected
at an significance level.
In our previous examples, the 95% confidence intervals are, respectively
0 1.96 , 0 1.96X X
, 0 1.645X
Hypothesis Testing
Collecting Data and Calculating the Test
Statistic (Step 5)
Data collection issues:
Measurement errors
Sample selection bias and time-period bias
Test statistic calculation has shown in the previous examples.
Hypothesis Testing
Making Statistical Decision (Step 6)
Comparing the calculated test statistic with corresponding critical value to
decide whether reject the null
hypothesis or not.
Although we will meet other tests below, the basic principal of statistical decision
making is the same.
Hypothesis Testing
p-Value
An alternative approach is called p-value approach.
The p-value is the smallest level of significance at which the null hypothesis can be rejected.
The smaller the p-value, the stronger the evidence against the null hypothesis and in favor of the alternative hypothesis.
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Hypothesis Testing
Hypothesis Tests Concerning the Mean:
t-tests
Can test that the mean of a population is equal to or differs from some hypothesized value.
Can test to see if the sample means from two different populations differ.
Hypothesis Testing
Tests Concerning a Single Mean: t-test
A t-test is used to test a hypothesis concerning the value of a population
mean, if the variance is unknown and
the sample is large, or
the sample is small but the population is normally distributed, or approximately
normally distributed.
Hypothesis Testing
Tests Concerning a Single Mean
deviation standard sample
mean population theof valueedhypothesiz the
mean sample
freedom of degrees 1n with statistic
where,
/
1
1
s
X
tt
ns
Xt
n
n
Hypothesis Testing
Example: Testing Rio Tinto Mean Return
Sender Equity Fund has achieved a mean monthly return of 1.50% with a sample st. dev. of 3.60% during a 24 months period. Given its level of systematic risk, the share is expected to have earned a 1.10% mean monthly return. Assuming return is normally distributed, is the actual result consistent with an underlying or population mean monthly return of 1.10% with 10% level of significance?
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Hypothesis Testing
Example: Testing Rio Tinto Mean Return
cont.
The hypothesis statement is: H0: the average return is equal to 1.10% The alternative hypothesis is Ha: mean monthly return 0 We use a 2-sided t-test to test for 24 months period (n=24)
Step 1: H0: =0 Ha: 0
Step 2: test statistic =
Step 3: a two tailed test has two rejection points: t/2,df=t0.05,23=1.714 and -t/2,df=-t0.05,23=-1.714
Step 4: reject if 0.544 < -1.714 or 0.544 > 1.714
Can we reject?
1
1.50 1.100.544
/ 3.60 / 24n
Xt
s n
Hypothesis Testing
Example: Testing average Days of
Receivables
FashionDesigns is concerned about a possible slowdown in payments from its customers. The rate of payment is measured by the average number of days in receivables. FashionDesigns has generally maintained an average of 45 days in receivables. A recent random sample of 50 accounts shows a mean number of days in receivables of 49 with a standard deviation of 8 days. Determine whether the evidence supports the suspected condition that customer payments have slowed at 5% level of significance.
Hypothesis Testing
Example: Testing average Days of
Receivables
The hypothesis statement is: H0: number of days in receivable is less than or equal 45 The alternative hypothesis is Ha: number of days in receivable is more than 45 We use a 1-sided t-test to test for 50 accounts (n=50)
Step 1: H0: 45 Ha: >45
Step 2: test statistic =
Step 3: a tone tailed test has one rejection point: t,df=t0.05,49=1.677
Step 4: reject if 3.536 > 1.677
Can we reject?
1
49 453.536
/ 8 / 50n
Xt
s n
Hypothesis Testing
The z-Test Alternative
If the population sampled is normally distributed with known variance, then the
test statistic for a hypothesis test
concerning a single population mean, , is
deviation standard populationknown
mean population theof valueedhypothesiz the
where,
/
n
Xz
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Hypothesis Testing
The z-Test Alternative
If the population sampled has unknown variance and the sample is large, in place
of a t-test, an alternative statistic is
deviation standard populationknown
where
/
s
ns
Xz
Hypothesis Testing
Rejection Points for a z-Test For
= 0.10
1.H0: = 0 versus Ha: 0
Reject the null hypothesis if z > 1.645 or if z < -1.645.
2.H0: 0 versus Ha: > 0
Reject the null hypothesis if z > 1.282
3.H0: 0 versus Ha: < 0
Reject the null hypothesis if z < -1.282
Hypothesis Testing
Rejection Points for a z-Test
For = 0.05
1.H0: = 0 versus Ha: 0
Reject the null hypothesis if z > 1.96 or if z < -1.96.
2.H0: 0 versus Ha: > 0
Reject the null hypothesis if z > 1.645
3.H0: 0 versus Ha: < 0
Reject the null hypothesis if z < -1.645
Hypothesis Testing
Rejection Points for a z-Test
For = 0.01
1.H0: = 0 versus Ha: 0
Reject the null hypothesis if z > 2.576 or if z < -2.576
2.H0: 0 versus Ha: > 0
Reject the null hypothesis if z > 2.326.
3.H0: 0 versus Ha: < 0 Reject the null hypothesis if z < -2.326
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Hypothesis Testing
t-test or z-test?
Population normal and variance known z-test, for both small and large sample.
Variance unknown
Large Sample
(n30) Small Sample
(n 0
3.H0: 1 - 2 0 versus HA: 1 - 2 < 0
Hypothesis Testing
Test Statistics for Difference between Two Population
Means: equal variances
For normally distributed populations with unknown variances, but if the variances
can be assumed to be equal, the t-statistic
is
and degrees of freedom is n1 + n2 - 2
2/1
2
2
1
2
2121
n
s
n
s
XXt
pp
2
)1()1( where
21
2
22
2
112
nn
snsnsp
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Hypothesis Testing
Test Statistics for Difference between Two Population
Means: unequal variances
For normally distributed populations, if population variances are unequal and unknown, the t-statistic is
and degrees of freedom is given by
2/1
2
2
2
1
2
1
2121
n
s
n
s
XXt
2
2
1
2
2
1
2
1
2
1
2
2
2
2
1
2
1
//
n
ns
n
ns
n
s
n
s
df
Hypothesis Testing
Example: Mean Return on S&P 500
The realized mean monthly return on the S&P 500 in the 1980s appears to have been substantially different from that in the 1970s. Was the difference statistically significant?
The data, shown on the next slide, indicate that assuming equal population variances for returns in the two decades is not unreasonable. But if you assumed unequal variances, would you reach a different conclusion?
Assume 5% level of significance
Hypothesis Testing
S&P 500 Monthly Return and Standard
Deviation
Decade No. of months Mean return St. Dev.
1970s 120 0.580 4.598
1980s 120 1.470 4.738
Hypothesis Testing
Example: Mean Return on S&P 500 cont.
We assume equal variance to answer the question: Was the difference statistically significant?
This means, is the difference between the average in 80s and 70s significantly different
from zero?
In other words,80s 70s=0 or 80s 70s0
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Hypothesis Testing
Example: Mean Return on S&P 500 cont.
Step 1: H0: 80s 70s =0 Ha: 80s 70s =0
Step 2: test statistic =
Note that you must find Sp first and then calculate test statistic using the Sp
Step 3: a two tailed test has two rejection points: t/2,df=t0.025,238=1.97 and -t/2,df=-t0.025,238=-1.97
Step 4: reject if 1.477 < -1.96 or 1.477 > 1.96
Can we reject?
1 2 1 21/2
2 2
1 2
1.4767
p p
X Xt
s s
n n
Hypothesis Testing
Mean Differences Samples Not Independent
Reminder: In the previous two t-tests, samples are assumed to be independent
If the samples are not independent, a test of mean difference is done using paired observations.
1. H0: d = d0 versus HA: d d0
2. H0: d d0 versus HA: d > d0 3. H0: d d0 versus HA: d < d0
where d stands for the population mean difference and d0 is a hypothesis value for the population mean difference
Hypothesis Testing
t-statistic for Mean Differences Samples Not Independent
To calculate the t-statistic, we first need to find the sample mean difference:
where di is the difference between two paired
observations (the ith pair)
The sample variance is
n
i
idn
d1
1
)1(2
1
2
nddsn
i
id
Hypothesis Testing
t-statistic for Mean Differences Samples Not Independent
The standard error of the mean difference is
The test statistic, with n 1 df, is,
n
ss d
d
d
d
s
dt 0
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Hypothesis Testing
Hypothesis Tests Concerning Variance
We examine two types:
tests concerning the value of a single population variance and
tests concerning the differences between two population variances.
Hypothesis Testing
Tests Concerning a Single Population
Variance
We can formulate hypotheses as follows:
2
0
2
a
2
0
2
0
2
0
2
a
2
0
2
0
2
0
2
a
2
0
2
0
:H versus:H .3
:H versus:H .2
:H versus:H .1
Hypothesis Testing
Test-Statistic for Tests Concerning a Single
Population Variance
If we have n independent observations from a normally distributed population,
the appreciate test statistic is chi-squared
statistic
where s2 is sample variance,
freedom of degrees 1n with ,)1(
2
0
22
sn
)1(1
22
nXXsn
i
i
Hypothesis Testing
Rejection Points for Tests Concerning a
Single Population Variance
1. Equal to H0: Reject the null if the statistic is greater than or smaller
than
2. Not greater than H0: Reject the null if the statistic is greater than
3. Not less than H0: Reject the null if the statistic is less than
2
2/2
2/1
2
2
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Hypothesis Testing
Tests Concerning the Equality (Inequality) of
Two Variances
Suppose we want to know the relative values of the variances of two populations,
we can formulate one of the following
hypotheses:
2
2
2
1a
2
2
2
10
2
2
2
1a
2
2
2
10
2
2
2
1a
2
2
2
10
:H versus:H .3
:H versus:H .2
:H versus:H .1
Hypothesis Testing
Test Statistic for Tests Concerning the
Equality (Inequality) of Two Variances
Suppose we have two samples, the first has n1 observations with sample
variance and the second has n2
observations with sample variance . If
both populations are normal, the test-
statistic is
freedom of degrees )1(n and )1( with , 212
2
2
1 ns
sF
2
1s2
2s
Hypothesis Testing
Rejection Points for Tests Concerning the
Equality (Inequality) of Two Variances
Convention: Let the sample with larger variance be sample 1 and the other sample 2 F-statistic is always greater than or equal to 1.
Thus, decision rule is:
1. Equal to H0: Reject the null if the statistic is greater than F/2 .
2. Not greater than and Not less than H0: Reject the null if the statistic is greater than F .
Hypothesis Testing
5.232
Thank You!
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233 Slide
Correlation and Simple Linear Regression,p-283-300
The Simple Linear Regression Model
The Least Squares Method
The Coefficient of Determination
Model Assumptions
Testing for Significance
Using the Estimated Regression
Equation for Estimation and Prediction
Residual Analysis: Validating Model Assumptions
Residual Analysis: Outliers and Influential
Observations
Chapter 8
234 Slide
The Simple Linear Regression Model
Simple Linear Regression Model
y = 0 + 1x +
Simple Linear Regression Equation
E(y) = 0 + 1x
Estimated Simple Linear Regression Equation
y = b0 + b1x
y = dept var
^
235 Slide
The Least Squares Method
Least Squares Criterion
min S(yi - yi)2
where
yi = observed value of the dependent variable
for the i th observation
yi = estimated value of the dependent variable
for the i th observation
^
^
236 Slide
Slope for the Estimated Regression Equation
y -Intercept for the Estimated Regression Equation
b0 = y - b1x
where
xi = value of independent variable for i th observation
yi = value of dependent variable for i th observation
x = mean value for independent variable
y = mean value for dependent variable
n = total number of observations
_ _
bx y x y n
x x n
i i i i
i i1 2 2
( ) /
( ) /b
x y x y n
x x n
i i i i
i i1 2 2
( ) /
( ) /
_
_
The Least Squares Method
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237 Slide
Example: Reed Auto Sales
Reed Auto periodically has a special week-long sale. As part of the advertising campaign Reed runs one or more television commercials during the weekend preceding the sale. Data from a sample of 5 previous sales showing the number of TV ads run and the number of cars sold in each sale are shown below.
Number of TV Ads Number of Cars Sold
1 14
3 24
2 18
1 17
3 27
238 Slide
Slope for the Estimated Regression Equation
b1 = 220 - (10)(100)/5 = 5
24 - (10)2/5
y -Intercept for the Estimated Regression Equation
b0 = 20 - 5(2) = 10
Estimated Regression Equation
y = 10 + 5x ^
Example: Reed Auto Sales
239 Slide
The Coefficient of Determination
Relationship Among SST, SSR, SSE
SST = SSR + SSE
Coefficient of Determination
r 2 = SSR/SST
where
SST = total sum of squares
SSR = sum of squares due to regression
SSE = sum of squares due to error
( ) ( ) ( )y y y y y yi i i i 2 2 2( ) ( ) ( )y y y y y yi i i i 2 2 2^ ^
240 Slide
Coefficient of Determination
r 2 = SSR/SST = 100/114 = .88
The regression relationship is very strong since
88% of the variation in number of cars sold can be
explained by the linear relationship between the
number of TV ads and the number of cars sold.
Example: Reed Auto Sales
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241 Slide
The Correlation Coefficient & Hypothesis Testing
The sample correlation coefficient is plus or minus the square root of the coefficient of determination.
Sample Correlation coefficient
Testing for Sample Correlation coefficient p297 r rxy
2r rxy 2
242 Slide
Model Assumptions
Assumptions About the Error Term
The error is a random variable with mean of zero.
The variance of , denoted by 2, is the same for all values of the independent variable.
The values of are independent.
The error is a normally distributed random variable.
243 Slide
Testing for Significance: F Test
Hypotheses
H0: 1 = 0
Ha: 1 = 0
Test Statistic
F = MSR/MSE
Rejection Rule
Reject H0 if F > F
where F is based on an F distribution with 1 d.f. in
the numerator and n - 2 d.f. in the denominator.
244 Slide
Testing for Significance: t Test (p.312)
Hypotheses
H0: 1 = 0
Ha: 1 = 0
Test Statistic
Rejection Rule
Reject H0 if t < -tor t > t
where t is based on a t distribution with
n - 2 degrees of freedom.
tb
sb 1
1
tb
sb 1
1
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245 Slide
Using the Estimated Regression Equation for Estimation and Prediction
Confidence Interval Estimate of E (yp)
Prediction Interval Estimate of yp yp + t/2 sind
where the confidence coefficient is 1 - and t/2 is
based on a t distribution with n - 2 d.f.
/ y t sp yp 2 / y t sp yp 2
246 Slide
F Test
Hypotheses H0: 1 = 0
Ha: 1 = 0
Rejection Rule
For = .05 and d.f. = 1, 3: F.05 = 10.13
Reject H0 if F > 10.13.
Test Statistic
F = MSR/MSE = 100/4.667 = 21.43
Conclusion
We can reject H0.
Example: Reed Auto Sales
247 Slide
t Test
Hypotheses H0: 1 = 0
Ha: 1 = 0
Rejection Rule
For = .05 and d.f. = 3, t.025 = 3.182
Reject H0 if t > 3.182
Test Statistics
t = 5/1.08 = 4.63
Conclusions
Reject H0: 1 = 0
Example: Reed Auto Sales
248 Slide
Point Estimation
If 3 TV ads are run prior to a sale, we expect the mean number of cars sold to be:
y = 10 + 5(3) = 25 cars
Confidence Interval for E (yp)
95% confidence interval estimate of the mean number of cars sold when 3 TV ads are run is:
25 + 4.61 = 20.39 to 29.61 cars
Prediction Interval for yp
95% prediction interval estimate of the number of cars sold in one particular week when 3 TV ads are run is: 25 + 8.28 = 16.72 to 33.28 cars
^
Example: Reed Auto Sales
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249 Slide
Residual Analysis
Residual for Observation i
yi - yi
Standardized Residual for Observation i
where
^
y y
si i
y yi i
y y
si i
y yi i
^
^
s s hy y ii i 1s s hy y ii i 1^
250 Slide
Detecting Outliers
An outlier is an observation that is unusual in comparison with the other data.
Minitab classifies an observation as an outlier if its standardized residual value is < -2 or > +2.
This standardized residual rule sometimes fails to identify an unusually large observation as being an outlier.
This rules shortcoming can be circumvented by using studentized deleted residuals.
The |i th studentized deleted residual| will be larger than the |i th standardized residual|.
Residual Analysis
251 Slide
The End of Chapter 8
252 Slide
Multiple Regression & Issues in Regression Analysis p. 325 text
The Multiple Linear Regression Model
The Least Squares Method
The Multiple Coefficient of Determination
Model Assumptions
Testing for Significance
Using the Estimated Regression Equation
for Estimation and Prediction
Qualitative Independent Variables