This document provides an overview of common probability distributions including discrete and continuous distributions. It focuses on explaining the binomial, Poisson, and normal distributions in detail. Key aspects of each distribution such as their characteristics, formulas, shapes, and how to calculate probabilities are discussed. Examples are provided to illustrate how to apply each distribution. The document is a lecture on probability distributions presented to students.
1. Probability Distributions
Istanbul Bilgi University
FEC 512 Financial Econometrics-I
Asst. Prof. Dr. Orhan Erdem
2. Some Common Probability
Distributions
Probability
Distributions
Discrete Continuous
Probability Probability
Distributions Distributions
Normal
Binomial
Uniform
Poisson
Lognormal
Lecture 3-2
FEC 512 Probability Distributions
3. Discrete Probability Distribution
Example random variable X = total number of
Experiment: toss 2 coins,
tails in two tosses.
Probability distribution
X Probability
T 0 0.25
1 0.50
T
2 0.25
T T
Lecture 3-3
FEC 512 Probability Distributions
4. The Binomial Distribution
Probability
Distributions
Discrete
Probability
Distributions
Binomial
Poisson
Lecture 3-4
FEC 512 Probability Distributions
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
Lecture 3-5
FEC 512 Probability Distributions
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
Lecture 3-6
FEC 512 Probability Distributions
7. Counting Rule for Combinations
A combination is an outcome of an experiment
where x objects are selected from a group of n
objects
n!
C= n
x
x! (n − x )!
where:
n! =n(n - 1)(n - 2) . . . (2)(1)
x! = x(x - 1)(x - 2) . . . (2)(1)
0! = 1 (by definition)
Lecture 3-7
FEC 512 Probability Distributions
8. Binomial Distribution Formula
n! x n−x
P(x) = pq
x ! (n − x )!
P(x) = probability of x successes in n trials,
with probability of success p on each trial Example: Flip a coin four
times, let x = # heads:
x = number of ‘successes’ in sample,
n=4
(x = 0, 1, 2, ..., n)
p = 0.5
p = probability of “success” per trial
q = (1 - .5) = .5
q = probability of “failure” = (1 – p)
n = number of trials (sample size) x = 0, 1, 2, 3, 4
Lecture 3-8
FEC 512 Probability Distributions
9. Binomial Distribution
The shape of the binomial distribution depends on
the values of p and n P(X) n = 5 p = 0.1
Mean .6
.4
.2
Here, n = 5 and p = .1
0 X
0 1 2 3 4 5
n = 5 p = 0.5
P(X)
.6
.4
Here, n = 5 and p = .5 .2
X
0
0 1 2 3 4 5
Lecture 3-9
FEC 512 Probability Distributions
10. Binomial Distribution
Characteristics
Mean
µ = E(x) = np
Variance and Standard
Deviation
σ = npq
2
σ = npq
Where n = sample size
p = probability of success
q = (1 – p) = probability of failure
Lecture 3-10
FEC 512 Probability Distributions
11. Binomial Characteristics
Examples
Mean = (5)(.1) = 0.5
µ = np n = 5 p = 0.1
P(X)
.6
.4
σ = npq = (5)(.1)(1 − .1) .2
= 0.6708 0 X
0 1 2 3 4 5
µ = np = (5)(.5) = 2.5 n = 5 p = 0.5
P(X)
.6
.4
σ = npq = (5)(.5)(1 − .5) .2
= 1.118 X
0
0 1 2 3 4 5
Lecture 3-11
FEC 512 Probability Distributions
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 the
other time period.
Su2
Su (60.50)
(55)
Sud=Sdu
(49.5)
S=50
Sd
Sd2
(45)
(40.50) T2
T0 T1
Lecture 3-12
FEC 512 Probability Distributions
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 to
Su2 or would fall by d to Sud.
If the asset had previoulsy fallen to Sd, it could rise by u to Sud or fall further
to 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
Lecture 3-13
FEC 512 Probability Distributions
14. The Poisson Distribution
Probability
Distributions
Discrete
Probability
Distributions
Binomial
Poisson
Lecture 3-14
FEC 512 Probability Distributions
15. The Poisson Distribution
To use binomial distribution, we must be able
to count the # successes and failures. Although
in many situations you may be able to count #
successes, you often cannot count # failures.
Example: An emergency call center could
easily count the # calls its unit respond to in 1
hour, but how could it determine how many
calls it didnt receive?
Lecture 3-15
FEC 512 Probability Distributions
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 λ
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
Lecture 3-16
FEC 512 Probability Distributions
17. Poisson Distribution Formula
− λt
( λt ) e x
P( x ) =
x!
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...)
Lecture 3-17
FEC 512 Probability Distributions
18. Poisson Distribution (continued)
Ex. Page requests arrive at a 0,18
0,17
webserver at an average rate 0,16
of 5 every second. If the 0,15
number of requests in a 0,14
0,13
second has a Poisson 0,12
distribution, find the probability 0,11
0,10
that 15 requests will be made p(X=x) 0,09
0,08
in a given second. 0,07
e −5 515 0,06
P ( X = 15 ) = = 0.00016 0,05
0,04
15! 0,03
0,02
0,01
0,00
Here is what the distribution 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23
function for the above example x=number of page requests in a second
looks like
Lecture 3-18
FEC 512 Probability Distributions
19. Poisson Distribution Characteristics
Mean
µ = λt
Variance and Standard
Deviation
σ = λt
2
σ = λt
λ = number of successes in a segment of unit size
where
t = the size of the segment of interest
Lecture 3-19
FEC 512 Probability Distributions
21. Poisson Distribution Shape
The shape of the Poisson Distribution
depends on the parameters λ and t:
λt = 0.50 λt = 3.0
0.70 0.25
0.60
0.20
0.50
0.15
0.40
P(x)
P(x)
0.30 0.10
0.20
0.05
0.10
0.00
0.00
1 2 3 4 5 6 7 8 9 10 11 12
0 1 2 3 4 5 6 7
x
x
Lecture 3-21
FEC 512 Probability Distributions
22. The Normal Distribution
Probability
Distributions
Continuous
Probability
Distributions
Normal
Uniform
Lognormal
Lecture 3-22
FEC 512 Probability Distributions
23. The Normal Distribution
The distribution whose pdf is given by
( x−µ )2
− f(x)
1
2σ 2
f ( x) =
e
2π σ
σ
‘Location is determined by x
the mean, µ µ
Spread is determined by the
standard deviation, σ
Lecture 3-23
FEC 512 Probability Distributions
24. The Normal Distribution
f(x)
‘Bell Shaped’
Symmetrical
σ
The random variable has x
an infinite theoretical µ
range: + ∞ to − ∞
Lecture 3-24
FEC 512 Probability Distributions
25. Many Normal Distributions
By varying the parameters µ and σ, we obtain
different normal distributions
Lecture 3-25
FEC 512 Probability Distributions
26. The Normal Distribution Shape
f(x) Changing µ shifts the
distribution left or right.
Changing σ increases
or decreases the
σ spread.
µ x
Lecture 3-26
FEC 512 Probability Distributions
27. Finding Normal Probabilities
Probability is the
Probability is measured by the area
area under the
curve! under the curve
f(x)
P (a ≤ x ≤ b)
a b x
Lecture 3-27
FEC 512 Probability Distributions
28. Probability as
Area Under the Curve
The total area under the curve is 1.0, and the curve is
symmetric, so half is above the mean, half is below
f(x) P( −∞ < x < µ) = 0.5
P(µ < x < ∞ ) = 0.5
0.5 0.5
x
µ
P(−∞ < x < ∞) = 1.0
Lecture 3-28
FEC 512 Probability Distributions
29. Empirical Rules
What can we say about the distribution of values
around the mean? There are some general rules:
f(x)
µ ± 1σ encloses about
68% of x’s
σ σ
x
µ−1σ µ µ+1
− +1σ
+1
68.26%
Lecture 3-29
FEC 512 Probability Distributions
30. The Empirical Rule
(continued)
µ ± 2σ covers about 95% of x’s
µ ± 3σ covers about 99.7% of x’s
3σ 3σ
2σ 2σ
µ x µ x
95.44% 99.72%
Lecture 3-30
FEC 512 Probability Distributions
31. The Standard Normal Distribution
Also known as the “z” distribution
Mean is defined to be 0
Standard Deviation is 1
f(z)
1
z
0
Values above the mean have positive z-values,
values below the mean have negative z-values
Lecture 3-31
FEC 512 Probability Distributions
32. Transformation to the Standard
Normal Distribution
Any normal distribution (with any mean and
standard deviation combination) can be
transformed into the standard normal
distribution (z)
x −µ
z=
σ
Lecture 3-32
FEC 512 Probability Distributions
33. Example
If x is distributed normally with mean of
100 and standard deviation of 50, the z
value for x = 200 is
x − µ 200 − 100
z= = = 2.0
σ 50
This says that x = 200 is two standard
deviations (2 increments of 50 units) above
the mean of 100.
Lecture 3-33
FEC 512 Probability Distributions
34. Comparing x and z units
µ = 100
σ = 50
100 200 x
0 2.0 z
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)
Lecture 3-34
FEC 512 Probability Distributions
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
.4772
Example:
P(0 < z < 2.00) = .4772
z
0 2.00
Lecture 3-35
FEC 512 Probability Distributions
36. The Standard Normal Table
(continued)
The column gives the value
of z to the second decimal
point
z 0.00 0.01 0.02 …
0.1
The row
shows the 0.2
.
.
value of z to The value within the
.
the first table gives the
.4772
2.0
decimal point probability from z = 0
up to the desired z
2.0
P(0 < z < 2.00) = .4772
value
Lecture 3-36
FEC 512 Probability Distributions
37. Z Table example
Suppose x is normal with mean 8.0 and
standard deviation 5.0. Find P(8 < x < 8.6)
Calculate z-values:
x −µ 8 −8
z= = =0
σ 5
8 8.6 x
x − µ 8.6 − 8 0 0.12 Z
z= = = 0.12
σ 5 P(8 < x < 8.6)
= P(0 < z < 0.12)
Lecture 3-37
FEC 512 Probability Distributions
38. Z Table example
(continued)
Suppose x is normal with mean 8.0 and
standard deviation 5.0. Find P(8 < x < 8.6)
µ=8 µ=0
σ=5 σ=1
x z
8 8.6 0 0.12
P(8 < x < 8.6) P(0 < z < 0.12)
Lecture 3-38
FEC 512 Probability Distributions
39. Solution: Finding P(0 < z < 0.12)
Standard Normal Probability P(8 < x < 8.6)
Table (Portion) = P(0 < z < 0.12)
z .00 .01 .02 .0478
0.0 .0000 .0040 .0080
0.1 .0398 .0438 .0478
0.2 .0793 .0832 .0871
Z
0.00
0.3 .1179 .1217 .1255
0.12
Lecture 3-39
FEC 512 Probability Distributions
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
8.0
7.4
Lecture 3-40
FEC 512 Probability Distributions
41. Lower Tail Probabilities
(continued)
Now Find P(7.4 < x < 8)…
The Normal distribution is
symmetric, so we use the .0478
same table even if z-values
are negative:
P(7.4 < x < 8)
= P(-0.12 < z < 0)
Z
= .0478 8.0
7.4
Lecture 3-41
FEC 512 Probability Distributions
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 the
index
will produce a negative return in a given year?
We first need to transform the normal variable into a standard normal.
X − 0.15 1.00 − 0.15
z= = = 2.83
0.30 0.30
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
X − 0.15 0 − 0.15
z= = = −0.50
0.30 0.30
This time we are interested in F(z), which is 0.3085.
Lecture 3-42
FEC 512 Probability Distributions
43. Linear Combinations of Two Normal
Random Variables
Let X~N(µX,σX2 ) and Y~N(µY,σY2 ) and
σXY=cov(X,Y).
If Z=aX+bY where a,b are constants, then
Z~N(µZ,σZ2 ) where
µZ=a µX +b µY
σZ2=a2 σX2 +b2 σY2 +2abσXY=a2 σX2 +b2 σY2
+2abσX σYρ
Lecture 3-43
FEC 512 Probability Distributions
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.
Lecture 3-44
FEC 512 Probability Distributions
45. The Uniform Distribution
Probability
Distributions
Continuous
Probability
Distributions
Normal
Uniform
Lognormal
Lecture 3-45
FEC 512 Probability Distributions
46. The Uniform Distribution
The uniform distribution is a
probability distribution that has
equal probabilities for all possible
outcomes of the random variable
Lecture 3-46
FEC 512 Probability Distributions
47. The Uniform Distribution (continued)
The Continuous Uniform Distribution:
1
if a ≤ x ≤ b
b−a
f(x) =
0 otherwise
where
f(x) = value of the density function at any x value
a = lower limit of the interval
b = upper limit of the interval
Lecture 3-47
FEC 512 Probability Distributions
48. Uniform Distribution
Example: Uniform Probability Distribution
Over the range 2 ≤ x ≤ 6:
1
f(x) = 6 - 2 = .25 for 2 ≤ x ≤ 6
f(x)
.25
x
2 6
Lecture 3-48
FEC 512 Probability Distributions
49. The Lognormal Distribution
Probability
Distributions
Continuous
Probability
Distributions
Normal
Uniform
Lognormal
Lecture 3-49
FEC 512 Probability Distributions
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
lnX~N(µ,σ2)
or
In other words, X is lognormal if its “ln” is normally
distributed.
If X,Y is lognormally distributed, their linear
combination(i.e. Portfolio of two stocks) may not be
lognormal.
Lecture 3-50
FEC 512 Probability Distributions
52. The median of X is eµ, and the expected value of X
σ2
µ+
is . The expectation is larger than the
2
e
median because the lognormal distribution is right-
skewed, and the skew. is more extreme with larger
values of σ.
Lecture 3-52
FEC 512 Probability Distributions
53. Example
Let rt = ln( Pt / Pt −1 ) denote the log-return on an
asset and assume that rt ~ N(µ,σ2). Let R = ( P P P ) − t −1
t
t
t −1
denote the simple monthly return, since we know
e rt = 1 + Rt . Since rt is
rt = ln(1 + Rt )
that and
normally distributed e t = 1 + Rt is log-normally
r
.............
distributed.
Lecture 3-53
FEC 512 Probability Distributions
54. Sample Moments
Above we introduced the four statistical moments
mean,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 with
the situation that we observe realizations of a
pdf(e.g. The daily return of the IMKB-100 index over
the last year), but we do not know the distribution
that generates these returns. So taking expectation
is impossible
But having the observations x1,…,xn, we can try to
estimate the “true moments” out of the sample.
These estimates are called sample moments.
Lecture 3-54
FEC 512 Probability Distributions
55. Mean (Arithmetic Average)
The Mean is the arithmetic average of data
values
Sample mean n = Sample Size
n
∑x x1 + x 2 + L + x n
i
x= =
i =1
n n
Lecture 3-55
FEC 512 Probability Distributions
56. Variation
Measures of variation give information
on the spread or variability of the
data values.
Same center,
different variation
Lecture 3-56
FEC 512 Probability Distributions
57. Variance
Average of squared deviations of values from the
mean
Sample variance: n
∑ (x − x) 2
i
s2 = i =1
n -1
Sample standard deviation:
n
∑ (x − x) 2
i
s= i=1
n -1
Lecture 3-57
FEC 512 Probability Distributions
58. Calculation Example:
Sample Standard Deviation
Sample
Data (Xi) : 10 12 14 15 17 18 18 24
n=8 Mean = x = 16
(10 − x ) 2 + (12 − x ) 2 + (14 − x ) 2 + L + (24 − x ) 2
s=
n −1
(10 − 16) + (12 − 16) + (14 − 16) + L + (24 − 16)
2 2 2 2
=
8 −1
126
= = 4.2426
7
Lecture 3-58
FEC 512 Probability Distributions
59. Comparing Standard Deviations
Data A
Mean = 15.5
s = 3.338
11 12 13 14 15 16 17 18 19 20 21
Data B
Mean = 15.5
s = .9258
11 12 13 14 15 16 17 18 19 20 21
Data C
Mean = 15.5
s = 4.57
11 12 13 14 15 16 17 18 19 20 21
Lecture 3-59
FEC 512 Probability Distributions
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
Sample C.V.
s
CV = ⋅ 100%
x
Lecture 3-60
FEC 512 Probability Distributions
61. Comparing Coefficient
of Variation
Stock A:
Average price last year = $50
Standard deviation = $5
s $5
CVA = ⋅ 100% = ⋅ 100% = 10%
x $50
Both stocks
have the same
Stock B: standard
deviation, but
Average price last year = $100 stock B is less
variable relative
Standard deviation = $5 to its price
s $5
CVB = ⋅ 100% = ⋅ 100% = 5%
x $100
Lecture 3-61
FEC 512 Probability Distributions
62. Sharpe Ratio
Let Dt=Rt-Rf where Rf is the riskfree rate of
return. T
T
∑
1
D = ∑ Dt (Dt − D) 2
T t =1
σD = t =1
T −1
Sharpe Ratio is
(D)
S=
σD
Reading: “The Sharpe Ratio”, William F. Sharpe
Lecture 3-62
FEC 512 Probability Distributions
63. Skewness
The moment coefficient of skewness is derived by
calculating the third moment about the mean and
dividing by the cube of standard deviation :
( )
∑ X −X
3
n −1
3
2
∑( )
X −X
n −1
Lecture 3-63
FEC 512 Probability Distributions
64. Shape of a Distribution
Describes how data is distributed
Symmetric or skewed
Right-Skewed
Symmetric
Left-Skewed
Mean = Median Median < Mean
Mean < Median
(Longer tail extends to left) (Longer tail extends to right)
Lecture 3-64
FEC 512 Probability Distributions
65. Kurtosis
Skewness indicates the degree of symmetry
in the frequency distribution
Kurtosis indicates the peakedness of that
distribution
Lecture 3-65
FEC 512 Probability Distributions
66. Kurtosis (continued)
∑(X − X )
4
n −1
4
∑(X − X )
2
n −1
Lecture 3-66
FEC 512 Probability Distributions
67. About the Probability Distribution of
Returns
The assumption that period returns(e.g. Daily, monthly,
annually) are normally distributed is inconsistent with the
limited liability feature of most financial instruments, R≥-1.
There are a number of empirical facts about return distributions
While normal distribution is perfectly symmetric about its mean,
1.
dailt stock returns are frequently skewed to the right. And few of
them are skewed to the left.
The sample daily return distributions for many individuals stocks
2.
exhibit “excess kurtosis” or “fat tails”. i.e. There is more
probability in the tails than would be justified by normal
distribution. The extent of this excess kurtosis diminishes
substantially, however , when monthly data is used.
Lecture 3-67
FEC 512 Probability Distributions
68. Emprical Return Distributions: IMKB-100
Daily
500
Series: XU100RETURNS
Sample 1/04/1993 12/29/2004
400 Observations 2968
Mean 0.002654
300 Median 0.001996
Maximum 0.194510
Minimum -0.181093
200 Std. Dev. 0.031281
Skewness 0.139399
Kurtosis 6.290761
100
Jarque-Bera 1348.812
Probability 0.000000
0
-0.1 0.0 0.1 0.2
Lecture 3-68
FEC 512 Probability Distributions
69. Emprical Return Distributions: IMKB-100
Monthly
24
Series: XU100RETURNS
Sample 1993M01 2004M12
20
Observations 144
16 Mean 5.899510
Median 5.043156
Maximum 79.78386
12
Minimum -39.03413
Std. Dev. 17.21817
8 Skewness 0.903614
Kurtosis 5.585146
4
Jarque-Bera 59.69430
Probability 0.000000
0
-40 -20 0 20 40 60 80
Pay attention to Skewness and Kurtosis.
Lecture 3-69
FEC 512 Probability Distributions
70. Value At Risk
Lecture 3-70
FEC 512 Probability Distributions