Fec512.02

Random Variables and
Summary Measures

Istanbul Bilgi University
FEC 512 Financial Econometrics-I
Asst. Prof. Dr. Orhan Erdem

Introduction to Probability
Distributions
Random Variable
Represents a numerical value from a
random event
Random
Variables

Discrete Continuous
Random Variable Random Variable

Lecture 2-2
FEC 512 Probability Distributions

Definitions

The r.v. is discrete if it takes countable
number of values. The discrete r.v. X has
probability density function (pdf) f:R→[0,1]
fiven by f(x)=P(X=x)
The r.v. is continuous if its takes
uncountable number of values.

Lecture 2-3

Examples

Stock prices are discrete random variables,
because they can only take on certain values,
such as 10.00TL, 10.01TL and 10.02TL and
not 10.005TL, since stocks have a minimum
tick size of 0.01TL.
By way of contrast, stock returns are
continuous not discrete random variables,
since a stock's return could be any number.

Lecture 2-4

Dicrete Random Variables: Examples
Roll a die twice: Let x be the number of
times 4 comes up (then x could be 0, 1, or 2
times)

Toss a coin 5 times.
Let x be the number of heads
(then x = 0, 1, 2, 3, 4, or 5)

Lecture 2-5

Discrete Probability Distribution

Experiment: Toss 2 Coins. Let x = # heads.
4 possible outcomes
Probability Distribution
T T x Value Probability
0 1/4 = .25
T H 1 2/4 = .50
2 1/4 = .25
H T Probability
.50

H .25
H
0 1 2 x
Lecture 2-6

Discrete Probability Distribution
Function (P.d.f.)
0 ≤ P(xi) ≤ 1 for each xi
Σ P(xi) = 1

Lecture 2-7

Cumulative Distribution Function(c.d.f.)

Cumulative distribution function of X is
FX(x)=P(X≤x)
If X has a pdf then
u=x

∑f
FX ( x) = u
u = −∞

Example: Draw the c.d.f of the prev. example

Lecture 2-8

Summary Measures: Location

Expected Value of a discrete distribution
(Weighted Average)

E(x) = Σxi P(xi)
x P(x)
Example: Toss 2 coins, 0 .25
x = # of heads, 1 .50
compute expected value of x: 2 .25

E(x) = (0 x .25) + (1 x .50) + (2 x .25)=1.0

Lecture 2-9

The Allais Example-1

1 Lottery
Probability 1
Outcome 500,000
2.Lottery
Probability 0.10 0.89 0.01
Outcome 2,500,000 500,000 0

Which one do you prefer?
It is common for ind. to express 1.Lottery is better than
2.Lottery

Lecture 2-10

The Allais Example-2

1 Lottery
Probability 0 0,11 0.89
Outcome 2,500,00 500,000 0
2.Lottery

Probability 0,10 0 0.90

Outcome 2,500,00 500,000 0

Which one do you prefer? It is common for ind. to express
2..Lottery is better than 1.Lottery

Lecture 2-11

Summary Measures: Dispersion

Standard Deviation of a discrete distribution
n

∑{x
σx = − E(x)} P(x i ) 2
i
i=1

where:
E(x) = Expected value of the random variable
P(x) = Probability of the random variable having
the value of x
Lecture 2-12

Summary Measures: Dispersion

Example: Toss 2 coins, x = # heads,
compute standard deviation (recall E(x) = 1)
n

∑{x
σx = − E(x)} P(x i ) 2
i
i=1

σ x = (0 − 1)2 (.25) + (1 − 1)2 (.50) + (2 − 1)2 (.25) = .50 = .707

Possible number of heads
= 0, 1, or 2

Lecture 2-13

Example: Random Walk

Assume that at each time step the price can
either increase or decrease by a fixed
amount ∆>0. Suppose that
P1: is the probability of an increase (0< P1 <1)
P2 : is the probability of a decrease. (0< P2 <1)
Random Variable:
If X is the change in a single step, then the set of
possible values of X is {x1= ∆, x2=- ∆} and their
probabilities are {P1, P2}
What is the exp. value of a random walk if P1=P2 =0.5?

Lecture 2-14

Chebyshev’s Inequality

Let X be a r.v. with expected value µ and finite
variance σ2. Then for any real number m> 0,
1
P( X − µ ≥ mσ ) ≤ 2
m
or
1
P( X − µ ≤ mσ ) ≤ 1 − 2
m

Lecture 2-15

Two Discrete Random Variables

Expected value of the sum of two discrete
random variables:

E(x + y) = E(x) + E(y)
= Σ x P(x) + Σ y P(y)

Lecture 2-16

Conditional Expectation

The conditional pdf of Y given X=x written
fYlX(y l x)=P(Y=ylX=x).
E(Y l X=x) is called conditional expectation of
Y given X, defined as
E(Y l X)=ΣyfYlX(y l x)
Although conditional expectation sounds like
a number it is actually a r.v.

Lecture 2-17

Bivariate Distributions

Situations where we are interested at the
same time in a pair of r.v. Defined over a joint
sample space.
If X and Y are disrete r.v., we write the prob
that X will take on the value x and Y will take
on the value y as P(X=x,Y=y), the joint pdf.
If X and Y are cont r.v. the joint pdf of X,Y is
the function fX,Y(x,y) which display the joint
distribution of X,Y.
Lecture 2-18

Example

Determine the value of k for which the
function given by f(x,y)=kxy for x=1,2,3;
y=1,2,3 can serve as a joint pdf.
Solution: Substituting values of x,y we get
f(1,1)=k; f(1,2)=2k; …f(3,3)=9k
k+2k+3k+2k+4k+6k+3k+6k+9k=1
36k=1 and k=1/36

Lecture 2-19

Example: Conditional Pdf
X
0 1 2
0 1/6 1/3 1/12 7/12
Y 1 2/9 1/6 - 7/18
2 1/36 - - 1/36
5/12 ½ 1/12 1
1/ 6 6
P( Y = 0 X = 0) = =
5 / 12 15
2/9 8
P( Y = 1 X = 0) = =
5 / 12 15
1 / 36 1
P( Y = 0 X = 0) = =
5 / 12 15
Lecture 2-20

Continuous Random Variables

has a probability density function (pdf) fX such
that

(a ) f ( x) ≥ 0 fo r a ll x ,
+∞

∫ f ( x ) d x = 1.
(b )
-∞

( c ) F o r a n y a , b , w ith - ∞ < a < b < + ∞ ,
b

∫
w e have P (a ≤ X ≤ b ) = f ( x)dx
a

Examples: Changes in stock prices

Lecture 2-21

Cumulative Distribution Function

* Cumulative distribution function (CDF) of X is
FX ( x ) = P ( X ≤ x )

If X has a pdf then
x

∫f
FX ( x ) = (u )du
X
−∞

Lecture 2-22

Moments

Lecture 2-23

Example: Continuous Probability
Distributions
Ex. Suppose that X is a continuous random variable with pdf
f ( x) = 2 x, 0 < x < 1,
= 0, elsewhere.

Hence the cdf is given by

1,2

F ( x) = 0, x ≤ 0,
if 1

x 0,8

∫ 2s ds
= =x, if 0 < x ≤ 1,
2

F(x)
0,6

0 0,4

= 1, if x > 1. 0,2

0
The graph of F(x) 0 0,2 0,4 0,6 0,8 1 1,2
x

Lecture 2-24

Marginal Distributions
X
0 1 2
0 1/6 1/3 1/12 7/12
Y 1 2/9 1/6 - 7/18
2 1/36 - - 1/36
5/12 ½ 1/12 1
Marginal Distribution of Y :
7 7 1
P(Y = 0) = ; P(Y = 1) = ; P(Y = 2) =
12 18 36

Lecture 2-25

Conditional Distribution and Expectation
i. Discrete Case

P( A ∩ B)
Before we have seen P( A B) = , Similarly
P( B)
P( X = x Y = y )
P( X = x Y = y ) = is the conditional pdf of X, Y.
P (Y = y )
Remember that Y is fixed here.

Lecture 2-26

Conditional Distribution and Expectation
ii. Continuous Case
(y x ) is given by
The conditional pdf, written as f Y X

fY X (y x) =
f ( x, y )
.
f X ( x)
∞
E (Y X ) = ∫ yf ( y x)dy
YX
−∞

Lecture 2-27

Martingale
Given an information set available at time t, I t ,
a sequence of r.v. Pt is called a martingale w.r.t info set I t if
E[ Pt +1 I t ] = Pt

Lecture 2-28

Skewness

The skewness of a r.v. measures the symmetry of a
dist. About its mean value.

{ }
E [ X − E ( X )]3
Skew( X ) =
σ x3
n

∑(X − E ( X ))3 P( X i )
i
i =1
= if X is discrete.
σx 3

∞

∫ ( X − E ( X ))3 f x
= −∞
if X is continuous
σx 3

Lecture 2-30

Kurtosis

The kurtosis of a r.v. measures the thickness in the
tails of a distribution.
{ }
E [ X − E ( X )] 4
Kurt ( X ) =
σ x4
n

∑ (X − E ( X )) 4 P ( X i )
i
i =1
= if X is discrete.
σ x4
∞

∫ (X − E ( X )) 4 f x
= −∞
if X is continuous
σx 4

Lecture 2-31

Example

x 0 1 2
P(x) 0.25 0.5 0.25

We know that E(X)=µ=1, σ=0.707 from previous
example.

Skew(X)=[(0-1)30.25+(1-1)3 0.5+(2-1)30.25] /(0.707)3=0.
So it is symmetric.

H.W. Find its kurtosis.
Lecture 2-32

Covariance

Covariance between two r.v.

σ XY = E [{X − E ( X )}{Y − E (Y )}]

Lecture 2-33

Covariance (cont.)

If X,Y are discrete r.v:
σxy = Σ [xi – E(x)][yj – E(y)]P(xiyj)
where:
P(xi ,yj) = joint probability of xi and yj.

Lecture 2-34

Useful Formulas

Lecture 2-35

Interpreting Covariance

Covariance between two discrete random
variables:
σxy > 0 x and y tend to move in the same
direction
σxy < 0 x and y tend to move in opposite
directions
σxy = 0 x and y do not move closely together

Lecture 2-36

Correlation Coefficient

The Correlation Coefficient shows the
strength of the linear association between
two variables
σxy
ρ=
σx σy
where:
ρ = correlation coefficient (“rho”)
σxy = covariance between x and y
σx = standard deviation of variable x
σy = standard deviation of variable y
Lecture 2-37

Interpreting the Correlation Coefficient

The Correlation Coefficient always falls between -1
and +1
ρ=0 x and y are not linearly related.

The farther ρ is from 0, the stronger the linear relationship:

ρ = +1 x and y have a perfect positive linear relationship
ρ = -1 x and y have a perfect negative linear relationship
* A strong nonlinear relationship may or or may not imply a
high correlation

Lecture 2-38

Lecture 2-39

Independence

A r.v. X is independent of Y if knowledge
about Y does not influence the likelihood that
X=x for all possible values of x. and y.
(Similarly for Y)
Holds for both type of r.v.

Lecture 2-40

Independence

The r.v. X and Y are independent if and only if
f X,Y (x, y) = f X (x)f Y (y) for cont r.v.
or P(X = x andY = y) = P(X = x)P(Y = y) for disc.r.v.
for all x, y ∈ R.
If X and Y are independent then E(XY) = E(X)E(Y)
If E(XY) = E(X)E(Y), then X and Y are uncorrelated
Thus independence ⇒ E(XY) = E(X)E(Y) ⇒ uncorrelatedness.
The converse is not true.

Lecture 2-41

Linear Functions of a Random Variable

Let X be a r.v. Either discrete or cont. E(X)=µ,
Var(X)=σ2.Define a new r.v. Y as
Y=aX+b.
Then E(Y)=aE(X)+b=a µ+b
Var(Y)=a2 σ2

Lecture 2-42

Linear Combinations of Two Random
Variables
Let X~(µX,σX2 ) and Y~(µY,σY2 ) and
σXY=cov(X,Y).
If Z=aX+bY where a,b are constants, then
Z~(µZ,σZ2 ) where
µZ=a µX +b µY
σZ2=a2 σX2 +b2 σY2 +2abσXY=a2 σX2 +b2 σY2
+2abσX σYρ

Lecture 2-43

Linear Combinations of N Random
Variables

Lecture 2-44

Diversification
As long as security returns are not positively
correlated, diversification benefits are possible. The
smaller the correlation between security returns, the
greater the cost of not diversifying.
σZ2=a2 σX2 +b2 σY2 +2abσX σYρ
Sigma(Z)

3

2.5

2

1.5

1

0.5

0
-1 -0.5 0 0.5 1

Correlation

Lecture 2-45

Fec512.02

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