FEC 512.03

Probability Distributions

Istanbul Bilgi University

FEC 512 Financial Econometrics-I

Asst. Prof. Dr. Orhan Erdem

FEC 512 Probability Distributions Lecture 3-2

Some Common Probability Distributions

Continuous


Binomial

Poisson

Probability

Distributions

Discrete


Normal

Uniform

Lognormal


Discrete Probability Distribution Example

Probability distribution

X Probability

� 0 0.25

� 1 0.50

� 2 0.25

Experiment: toss 2 coins, random variable �� X = total number of

tails in two tosses.

T

T

T T


The Binomial Distribution

Binomial

Poisson

Probability

Distributions

Discrete



The Binomial Distribution

� Characteristics of the Binomial Distribution:

� A trial has only two possible outcomes – “success”or “failure”

� There is a fixed number, n, of identical trials

� The trials of the experiment are independent of each other

� The probability of a success, p, remains constant from trial to trial

� If p represents the probability of a success, then

(1-p) = q is the probability of a failure


Binomial Distribution Settings

� A manufacturing plant labels items as either defective or acceptable

� A firm bidding for a contract will either get the contract or not

� A marketing research firm receives survey responses of “yes I will buy” or “no I will not”

� New job applicants either accept the offer or reject it


Counting Rule for Combinations

� A combination is an outcome of an experiment where x objects are selected from a group of n objects

)!xn(!x

!nCn

x −=

where:n! =n(n - 1)(n - 2) . . . (2)(1)

x! = x(x - 1)(x - 2) . . . (2)(1)

0! = 1 (by definition)


P(x) = probability of x successes in n trials,

with probability of success p on each trial

x = number of ‘successes’ in sample, (x = 0, 1, 2, ..., n)

p = probability of “success” per trial

q = probability of “failure” = (1 – p)

n = number of trials (sample size)

P(x)n

x ! n xp q

x n x!

( )!====

−−−−−

Example: Flip a coin four times, let x = # heads:

n = 4

p = 0.5

q = (1 - .5) = .5

x = 0, 1, 2, 3, 4

Binomial Distribution Formula


n = 5 p = 0.1

n = 5 p = 0.5

Mean

0

.2

.4

.6

0 1 2 3 4 5

X

P(X)

.2

.4

.6

0 1 2 3 4 5

X

P(X)

0

Binomial Distribution

� The shape of the binomial distribution depends on the values of p and n

� Here, n = 5 and p = .1

� Here, n = 5 and p = .5


Binomial Distribution Characteristics

� Mean

� Variance and Standard Deviation

npE(x)µ ==

npqσ2 =

npqσ =Where n = sample size

p = probability of success

q = (1 – p) = probability of failure


n = 5 p = 0.1

n = 5 p = 0.5

Mean

0

.2

.4

.6

0 1 2 3 4 5

X

P(X)

.2

.4

.6

0 1 2 3 4 5

X

P(X)

0

0.5(5)(.1)npµ ===

0.6708

.1)(5)(.1)(1npqσ

=

−==

2.5(5)(.5)npµ ===

1.118

.5)(5)(.5)(1npqσ

=

−==

Binomial CharacteristicsExamples


A binomial tree of asset prices

“Example”� We wish to know the value of an asset after two time periods.

� Each of the time periods the asset may rise (a success) with a probability of 0.5 or it may fall (a failure) with a probability of 0.5.

� Assume asset price movement in one time period is independent of that in theother time period.

S=50

Su

(55)

Sd

(45)

Su2

(60.50)

Sud=Sdu

(49.5)

Sd2

(40.50)T0 T1

T2


A binomial tree of asset prices

“Example”

� If the asset had previously risen by a factor u, it would either rise again by u toSu2 or would fall by d to Sud.

� If the asset had previoulsy fallen to Sd, it could rise by u to Sud or fall furtherto Sd2

� Suppose that u=1.1, d=0.9, and S=50� Hence the expected value can be calculated as:

µ = (60.50 * 0.25) + (49.50 * 0.50) + (40.50 * 0.25) = 50

� The variance is

σ2 = (60.5 – 50)2 * 0.25 + (49.50 – 50)2 * 0.50 + (40.5 – 50)2 * 0.25 = 50.25


The Poisson Distribution

Binomial

Poisson

Probability

Distributions

Discrete



The Poisson Distribution

� To use binomial distribution, we must be ableto count the # successes and failures. Althoughin many situations you may be able to count #successes, you often cannot count # failures.

� Example: An emergency call center couldeasily count the # calls its unit respond to in 1 hour, but how could it determine how manycalls it didnt receive?


Characteristics of the Poisson Distribution

� The outcomes of interest are rare relative to the possible outcomes

� The average number of outcomes of interest per time or space interval is λλλλ

� The number of outcomes of interest are random, and the occurrence of one outcome does not influence the chances of another outcome of interest

� The probability of that an outcome of interest occurs in a given segment is the same for all segments


Poisson Distribution Formula

where:

t = size of the segment of interest

x = number of successes in segment of interest

λ = expected number of successes in a segment of unit size

e = base of the natural logarithm system (2.71828...)

!x

e)t()x(P

tx λ−λ=


Poisson Distribution (continued)

Ex. Page requests arrive at a webserver at an average rate of 5 every second. If thenumber of requests in a second has a Poissondistribution, find the probabilitythat 15 requests will be made

in a given second.

Here is what the distributionfunction for the above examplelooks like�

( )−

= = =5 155

15 0.0001615!

eP X

0,00

0,01

0,02

0,03

0,04

0,05

0,06

0,07

0,08

0,09

0,10

0,11

0,12

0,13

0,14

0,15

0,16

0,17

0,18

p(X=x)

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23

x=number of page requests in a second


Poisson Distribution Characteristics

� Mean

� Variance and Standard Deviation

λtµ =

λtσ2 =

λtσ =

where λ = number of successes in a segment of unit size

t = the size of the segment of interest


Graph of Poisson Probabilities

0.00

0.10

0.20

0.30

0.40

0.50

0.60

0.70

0 1 2 3 4 5 6 7

x

P(x

)0

1

2

3

4

5

6

7

X

0.6065

0.3033

0.0758

0.0126

0.0016

0.0002

0.0000

0.0000

λλλλt =

0.50

P(x = 2) = .0758

Graphically:

λλλλ = .05 and t = 100


Poisson Distribution Shape

� The shape of the Poisson Distribution depends on the parameters λ and t:

0.00

0.05

0.10

0.15

0.20

0.25

1 2 3 4 5 6 7 8 9 10 11 12

x

P(x

)

0.00

0.10

0.20

0.30

0.40

0.50

0.60

0.70

0 1 2 3 4 5 6 7

x

P(x

)

λt = 0.50 λt = 3.0


The Normal Distribution

Continuous


Probability

Distributions

Normal

Uniform

Lognormal



�The distribution whose pdf is given by

�‘Location is determined by the mean, µ

� Spread is determined by the standard deviation, σ

x

f(x)

µ

σ

−−

=2

2

2

)(

2

1)(

σµ

σπ

x

exf



�‘Bell Shaped’

� Symmetrical

The random variable has an infinite theoretical range: + ∞∞∞∞ to −−−− ∞∞∞∞

x

f(x)

µ

σ


By varying the parameters µ and σ, we obtain

different normal distributions

Many Normal Distributions


The Normal Distribution Shape

x

f(x)

µ

σ

Changing µ shifts the distribution left or right.

Changing σ increases or decreases the spread.


Finding Normal Probabilities

Probability is the

area under the

curve!

a b x

f(x) P a x b( )≤ ≤≤≤≤

Probability is measured by the area under the curve


f(x)

xµ

Probability as Area Under the Curve

0.50.5

The total area under the curve is 1.0, and the curve is symmetric, so half is above the mean, half is below

1.0)xP( =∞<<−∞

0.5)xP(µ =∞<<0.5µ)xP( =<<−∞


Empirical Rules

µ ± 1σ encloses about

68% of x’s

f(x)

xµ µ+1+1+1+1σµ−−−−1111σ

What can we say about the distribution of values around the mean? There are some general rules:

σσ

68.26%


The Empirical Rule

� µ ± 2σ covers about 95% of x’s

� µ ± 3σ covers about 99.7% of x’s

xµ

2σ 2σ

xµ

3σ 3σ

95.44% 99.72%

(continued)


The Standard Normal Distribution

� Also known as the “z” distribution

� Mean is defined to be 0

� Standard Deviation is 1

z

f(z)

0

1

Values above the mean have positive z-values, values below the mean have negative z-values


Transformation to the Standard

Normal Distribution

� Any normal distribution (with any mean and standard deviation combination) can be transformed into the standard normaldistribution (z)

σ

µxz

−=


� If x is distributed normally with mean of 100 and standard deviation of 50, the z value for x = 200 is

� This says that x = 200 is two standard deviations (2 increments of 50 units) above the mean of 100.

Example

2.050

100200

σ

µxz =

−=

−=


Comparing x and z units

z

100

2.00

200 x

Note that the distribution is the same, only the

scale has changed. We can express the problem in

original units (x) or in standardized units (z)

µ = 100

σ = 50


The Standard Normal Table

� The Standard Normal table in the textbooks gives the probability from the mean (zero) up to a desired value for z

z0 2.00

.4772

Example:

P(0 < z < 2.00) = .4772


The Standard Normal Table

The value within the table gives the probability from z = 0up to the desired z value

z 0.00 0.01 0.02 …

0.1

0.2

.4772

2.0P(0 < z < 2.00) = .4772

The row shows the value of z to the first decimal point

The column gives the value of z to the second decimal point

2.0

.

.

.

(continued)


Z Table example

� Suppose x is normal with mean 8.0 and standard deviation 5.0. Find P(8 < x < 8.6)

P(8 < x < 8.6)

= P(0 < z < 0.12)

Z0.120

x8.68

05

88

σ

µxz =

−=

−=

0.125

88.6

σ

µxz =

−=

−=

Calculate z-values:


Z Table example

� Suppose x is normal with mean 8.0 and standard deviation 5.0. Find P(8 < x < 8.6)

P(0 < z < 0.12)

z0.120x8.68

P(8 < x < 8.6)

µ = 8

σ = 5

µ = 0

σ = 1

(continued)


Z

0.12

z .00 .01

0.0 .0000 .0040 .0080

.0398 .0438

0.2 .0793 .0832 .0871

0.3 .1179 .1217 .1255

Solution: Finding P(0 < z < 0.12)

.0478.02

0.1 .0478

Standard Normal Probability Table (Portion)

0.00

= P(0 < z < 0.12)

P(8 < x < 8.6)


Lower Tail Probabilities

� Suppose x is normal with mean 8.0 and standard deviation 5.0.

� Now Find P(7.4 < x < 8)

Z

7.48.0


Lower Tail Probabilities

Now Find P(7.4 < x < 8)…

Z

7.48.0

The Normal distribution is symmetric, so we use the same table even if z-values are negative:

P(7.4 < x < 8)

= P(-0.12 < z < 0)

= .0478

(continued)

.0478


Distributions of Portfolio Returns

Example. Assume that the stock index in a country has an annual return distribution

that is normal with µ = 0.15 and σ = 0.30. What is probability that in a given year the

stock index will exceed an annual return of 100%? What is the probability that theindex

will produce a negative return in a given year?

We first need to transform the normal variable into a standard normal.

Looking up 2.83 in the normal table, we find that F(z) is 0.9977. So 1 – F(z) = 0.0023.

For finding the probability of a negative return, the transformation yields

This time we are interested in F(z), which is 0.3085.

0.15 1.00 0.152.83

0.30 0.30

Xz

− −= = =

0.15 0 0.150.50

0.30 0.30

Xz

− −= = = −


Linear Combinations of Two Normal

Random Variables

� Let X~N(µX,σX2 ) and Y~N(µY,σY

2 ) andσXY=cov(X,Y).

If Z=aX+bY where a,b are constants, thenZ~N(µZ,σZ

2 ) where

µZ=a µX +b µY

σZ2=a2 σX

2 +b2 σY2 +2abσXY=a2 σX

2 +b2 σY2

+2abσX σYρ


Kurtosis and Skewness of Normal

Distribution

� The skewness of a normal distribution is 0. Why?

� The kurtosis of a normal distribution is 3. Hence 3 is a benchmark value for tail thickness of a bell-shaped distribution.

� If kurt(X)>3, the dist. has thicker tails than norm. dist.

� If kurt(X)<3, the dist. has thinner tails than norm. dist.


The Uniform Distribution

Continuous


Probability

Distributions

Normal

Uniform

Lognormal


The Uniform Distribution

� The uniform distribution is a

probability distribution that has

equal probabilities for all possible

outcomes of the random variable


The Continuous Uniform Distribution:

otherwise 0

bxaifab

1≤≤

−

where

f(x) = value of the density function at any x value

a = lower limit of the interval

b = upper limit of the interval

The Uniform Distribution(continued)

f(x) =


Uniform Distribution

Example: Uniform Probability DistributionOver the range 2 ≤ x ≤ 6:

2 6

.25

f(x) = = .25 for 2 ≤ x ≤ 66 - 21

x

f(x)


The Lognormal Distribution

Continuous


Probability

Distributions

Normal

Uniform

Lognormal


The Lognormal Distribution

� Let Z ~N(µ,σ2), and X=eZ r.v. X is said to be log-normally distributed with parameters µ and σ2

or lnX~N(µ,σ2)

� In other words, X is lognormal if its “ln” is normallydistributed.

� If X,Y is lognormally distributed, their linearcombination(i.e. Portfolio of two stocks) may not be lognormal.



� The median of X is eµ, and the expected value of X

is . The expectation is larger than the

median because the lognormal distribution is right-

skewed, and the skew. is more extreme with larger

values of σ.

2

2σµ+e


Example

� Let denote the log-return on an asset and assume that rt ~ N(µ,σ2). Letdenote the simple monthly return, since we knowthat and . Since rt is normally distributed ............. is log-normallydistributed.

1ln( / )t t tr P P −=( )1

1

t t

t

t

P PR

P

−

−

−=

ln(1 )t tr R= + 1tr

te R= +1tr

te R= +


Sample Moments

� Above we introduced the four statistical momentsmean,variance, skewness, kurtosis.

� Given a pdf, we are able to calculate these stat. Moments according to the fomulae.

� In practical applications however, we are faced withthe situation that we observe realizations of a pdf(e.g. The daily return of the IMKB-100 index overthe last year), but we do not know the distributionthat generates these returns. So taking expectationis impossible

� But having the observations x1,…,xn, we can try toestimate the “true moments” out of the sample. These estimates are called sample moments.


Mean (Arithmetic Average)

� The Mean is the arithmetic average of data values

� Sample mean n = Sample Size

n

xxx

n

x

x n

n

ii +++

==∑

= L211


� Measures of variation give information on the spread or variability of the data values.

Variation

Same center,

different variation


Variance

� Average of squared deviations of values from the mean

� Sample variance:

� Sample standard deviation:

1- n

)x(x

s

n

1i

2i

2∑

=

−=

1-n

)x(x

s

n

1i

2i∑

=

−=


Calculation Example:Sample Standard Deviation

Sample

Data (Xi) : 10 12 14 15 17 18 18 24

n = 8 Mean = x = 16

4.24267

126

18

16)(2416)(1416)(1216)(10

1n

)x(24)x(14)x(12)x(10s

2222

2222

==

−−++−+−+−

=

−−++−+−+−

=

L

L


Comparing Standard Deviations

Mean = 15.5

s = 3.33811 12 13 14 15 16 17 18 19 20 21

11 12 13 14 15 16 17 18 19 20 21

Data B

Data A

Mean = 15.5

s = .9258

11 12 13 14 15 16 17 18 19 20 21

Mean = 15.5

s = 4.57

Data C


Coefficient of Variation

� Measures relative variation

� Always in percentage (%)

� Shows variation relative to mean

� Is used to compare two or more sets of data

measured in different units

100%x

sCV ⋅

=

Sample C.V.


Comparing Coefficient of Variation

� Stock A:

� Average price last year = $50

� Standard deviation = $5

� Stock B:

� Average price last year = $100

� Standard deviation = $5

Both stocks have the same standard deviation, but stock B is less variable relative to its price

10%100%$50

$5100%

x

sCVA =⋅=⋅

=

5%100%$100

$5100%

x

sCVB =⋅=⋅

=


Sharpe Ratio

� Let Dt=Rt-Rf where Rf is the riskfree rate of return.

� Sharpe Ratio is

� Reading: “The Sharpe Ratio”, William F. Sharpe

( )

D

DS

σ=

2

1

( )

1

T

t

tD

D D

Tσ =

−=

−

∑1

1 T

t

t

D DT =

= ∑


Skewness

� The moment coefficient of skewness is derived bycalculating the third moment about the mean anddividing by the cube of standard deviation :

( )

( )

− −

−

−

∑

∑

3

32

1

1

X X

n

X X

n


Shape of a Distribution

� Describes how data is distributed

� Symmetric or skewed

Mean = Median Mean < Median Median < Mean

Right-SkewedLeft-Skewed Symmetric

(Longer tail extends to left) (Longer tail extends to right)


Kurtosis

� Skewness indicates the degree of symmetryin the frequency distribution

� Kurtosis indicates the peakedness of thatdistribution


Kurtosis (continued)

( )

( )

4

42

1

1

X X

n

X X

n

−

−

− −

∑

∑


About the Probability Distribution of

Returns� The assumption that period returns(e.g. Daily, monthly,

annually) are normally distributed is inconsistent with thelimited liability feature of most financial instruments, R≥-1.

� There are a number of empirical facts about return distributions

1. While normal distribution is perfectly symmetric about its mean, dailt stock returns are frequently skewed to the right. And few of them are skewed to the left.

2. The sample daily return distributions for many individuals stocksexhibit “excess kurtosis” or “fat tails”. i.e. There is moreprobability in the tails than would be justified by normal distribution. The extent of this excess kurtosis diminishessubstantially, however , when monthly data is used.


Emprical Return Distributions: IMKB-100

Daily

0

100

200

300

400

500

-0.1 0.0 0.1 0.2

Series: XU100RETURNS

Sample 1/04/1993 12/29/2004

Observations 2968

Mean 0.002654

Median 0.001996

Maximum 0.194510

Minimum -0.181093

Std. Dev. 0.031281

Skewness 0.139399

Kurtosis 6.290761

Jarque-Bera 1348.812

Probability 0.000000


Emprical Return Distributions: IMKB-100

Monthly

� Pay attention to Skewness and Kurtosis.

0

4

8

12

16

20

24

-40 -20 0 20 40 60 80

Series: XU100RETURNS

Sample 1993M01 2004M12

Observations 144

Mean 5.899510

Median 5.043156

Maximum 79.78386

Minimum -39.03413

Std. Dev. 17.21817

Skewness 0.903614

Kurtosis 5.585146

Jarque-Bera 59.69430

Probability 0.000000


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FEC 512.03