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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
Probability Distributions
Binomial
Poisson
Probability
Distributions
Discrete
Probability Distributions
Normal
Uniform
Lognormal
FEC 512 Probability Distributions Lecture 3-3
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
FEC 512 Probability Distributions Lecture 3-4
The Binomial Distribution
Binomial
Poisson
Probability
Distributions
Discrete
Probability Distributions
FEC 512 Probability Distributions Lecture 3-5
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
FEC 512 Probability Distributions Lecture 3-6
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
FEC 512 Probability Distributions Lecture 3-7
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)
FEC 512 Probability Distributions Lecture 3-8
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
FEC 512 Probability Distributions Lecture 3-9
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
FEC 512 Probability Distributions Lecture 3-10
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
FEC 512 Probability Distributions Lecture 3-11
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
FEC 512 Probability Distributions Lecture 3-12
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
FEC 512 Probability Distributions Lecture 3-13
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
FEC 512 Probability Distributions Lecture 3-14
The Poisson Distribution
Binomial
Poisson
Probability
Distributions
Discrete
Probability Distributions
FEC 512 Probability Distributions Lecture 3-15
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?
FEC 512 Probability Distributions Lecture 3-16
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
FEC 512 Probability Distributions Lecture 3-17
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=
FEC 512 Probability Distributions Lecture 3-18
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
FEC 512 Probability Distributions Lecture 3-19
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
FEC 512 Probability Distributions Lecture 3-20
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
FEC 512 Probability Distributions Lecture 3-21
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
FEC 512 Probability Distributions Lecture 3-22
The Normal Distribution
Continuous
Probability Distributions
Probability
Distributions
Normal
Uniform
Lognormal
FEC 512 Probability Distributions Lecture 3-23
The Normal Distribution
�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
FEC 512 Probability Distributions Lecture 3-24
The Normal Distribution
�‘Bell Shaped’
� Symmetrical
The random variable has an infinite theoretical range: + ∞∞∞∞ to −−−− ∞∞∞∞
x
f(x)
µ
σ
FEC 512 Probability Distributions Lecture 3-25
By varying the parameters µ and σ, we obtain
different normal distributions
Many Normal Distributions
FEC 512 Probability Distributions Lecture 3-26
The Normal Distribution Shape
x
f(x)
µ
σ
Changing µ shifts the distribution left or right.
Changing σ increases or decreases the spread.
FEC 512 Probability Distributions Lecture 3-27
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
FEC 512 Probability Distributions Lecture 3-28
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( =<<−∞
FEC 512 Probability Distributions Lecture 3-29
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%
FEC 512 Probability Distributions Lecture 3-30
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)
FEC 512 Probability Distributions Lecture 3-31
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
FEC 512 Probability Distributions Lecture 3-32
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
−=
FEC 512 Probability Distributions Lecture 3-33
� 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 =
−=
−=
FEC 512 Probability Distributions Lecture 3-34
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
FEC 512 Probability Distributions Lecture 3-35
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
FEC 512 Probability Distributions Lecture 3-36
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)
FEC 512 Probability Distributions Lecture 3-37
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:
FEC 512 Probability Distributions Lecture 3-38
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)
FEC 512 Probability Distributions Lecture 3-39
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)
FEC 512 Probability Distributions Lecture 3-40
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
FEC 512 Probability Distributions Lecture 3-41
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
FEC 512 Probability Distributions Lecture 3-42
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
− −= = = −
FEC 512 Probability Distributions Lecture 3-43
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
FEC 512 Probability Distributions Lecture 3-44
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.
FEC 512 Probability Distributions Lecture 3-45
The Uniform Distribution
Continuous
Probability Distributions
Probability
Distributions
Normal
Uniform
Lognormal
FEC 512 Probability Distributions Lecture 3-46
The Uniform Distribution
� The uniform distribution is a
probability distribution that has
equal probabilities for all possible
outcomes of the random variable
FEC 512 Probability Distributions Lecture 3-47
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) =
FEC 512 Probability Distributions Lecture 3-48
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)
FEC 512 Probability Distributions Lecture 3-49
The Lognormal Distribution
Continuous
Probability Distributions
Probability
Distributions
Normal
Uniform
Lognormal
FEC 512 Probability Distributions Lecture 3-50
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.
FEC 512 Probability Distributions Lecture 3-51
FEC 512 Probability Distributions Lecture 3-52
� 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
FEC 512 Probability Distributions Lecture 3-53
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= +
FEC 512 Probability Distributions Lecture 3-54
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.
FEC 512 Probability Distributions Lecture 3-55
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
FEC 512 Probability Distributions Lecture 3-56
� Measures of variation give information on the spread or variability of the data values.
Variation
Same center,
different variation
FEC 512 Probability Distributions Lecture 3-57
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∑
=
−=
FEC 512 Probability Distributions Lecture 3-58
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
FEC 512 Probability Distributions Lecture 3-59
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
FEC 512 Probability Distributions Lecture 3-60
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.
FEC 512 Probability Distributions Lecture 3-61
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 =⋅=⋅
=
FEC 512 Probability Distributions Lecture 3-62
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 =
= ∑
FEC 512 Probability Distributions Lecture 3-63
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
FEC 512 Probability Distributions Lecture 3-64
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)
FEC 512 Probability Distributions Lecture 3-65
Kurtosis
� Skewness indicates the degree of symmetryin the frequency distribution
� Kurtosis indicates the peakedness of thatdistribution
FEC 512 Probability Distributions Lecture 3-66
Kurtosis (continued)
( )
( )
4
42
1
1
X X
n
X X
n
−
−
− −
∑
∑
FEC 512 Probability Distributions Lecture 3-67
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.
FEC 512 Probability Distributions Lecture 3-68
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
FEC 512 Probability Distributions Lecture 3-69
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
FEC 512 Probability Distributions Lecture 3-70
Value At Risk