Lecturer: Nguyen Duong Nguyen, Mathematics Department, Faculty of Basic Science, FTU
Chapter 4. Joint distributions (Two-way Random Variable)
4.1 Definition
Let X and Y be random variables. The pair (X, Y) is called a (two-dimensional) random
vector.
If X and Y are discrete random variables, (X, Y) is called a discrete random vector.
If X and Y are continuous random variables, (X, Y) is called a continuous random vector.
4.2 The probability distribution of a (two-dimensional) random vector
4.2.1 Probability distribution table of a discrete (two-dimensional) random vector
Let (X, Y) is a discrete two-dimensional random vector. Suppose that the set of
values that X can take is {x1, x2, …, xn} and the set of values that Y can take are {y1, y2,
…, ym}.
Denote
P(xi, yj) = P(X=xi, Y=yj), i  1, n; j  1, m
We have the probability distribution table of a discrete two-dimensional random vector
(X, Y) as follows:
X
x1
x2
…
xi
…
xn
y1
P(x1, y1)
P(x2, y1)
…
P(xi, y1)
…
P(xn, y1)
y2
P(x1, y2)
P(x2, y2)
…
P(xi, y2)
…
P(xn, y2)
…
…
…
…
…
…
…
yj
P(x1, yj)
P(x2, yj)
…
P(xi, yj)
…
P(xn, yj)
…
…
…
…
…
…
…
ym
P(x1, ym)
P(x2, ym)
Y
Note.
1
P(xi, ym)
P(xn, ym)
Lecturer: Nguyen Duong Nguyen, Mathematics Department, Faculty of Basic Science, FTU
0 ≤ P(xi, yj) ≤ 1, i = 1, n , j = 1, m ;
n
m
 P(x , y )  1
i
i 1 j1
j
Then, the marginal probability distribution table of the component X is:
X
x1
x2
...
xi
...
xn
P
P(x1)
P(x2)
...
P(xi)
...
P(xn)
where
m
 P(x , y ) , i = 1, n
P(xi) =
i
j1
j
n
Note.
 P(x )  1
i
i 1
The marginal probability distribution table of the component Y is:
Y
y1
y2
...
yi
...
ym
P
P(y1)
P(y2)
...
P(yi)
...
P(ym)
where
n
P(yj) =
 P(x , y ) , j = 1, m .
i
i 1
j
m
Note.
 P(y )  1
j1
j
Example. Let the probability distribution table of a two-dimensional random vector (X,
Y) as follows:
100
150
200
0
0.1
0.05
0.05
1
0.05
0.2
0.15
2
0
0.1
0.3
Y
X
Find the marginal distribution of each component.
2
Lecturer: Nguyen Duong Nguyen, Mathematics Department, Faculty of Basic Science, FTU
Solution.
We have
P(X = 100) = 0.1 + 0.05 + 0 = 0.15
P(X = 150) = 0.05 + 0.2 + 0.1 = 0.35
P(X = 200 ) = 0.05 + 0.15 + 0.3 = 0.5
We have the following probability distribution table of X:
X
100
150
200
P
0.15
0.35
0.5
P(Y=0) = 0.1 + 0.05 + 0.05 = 0.2
P(Y=1)= 0.05 + 0.2 + 0.15 = 0.4
P(Y=2)= 0 + 0.1 + 0.3 = 0.4
We have the following probability distribution table of Y:
Y
0
1
2
P
0.2
0.4
0.4
4.2.2 The joint distribution function
1) Definition. The joint distribution function (joint cdf) of (X, Y), denoted F(x, y), is
defined as
F(x, y) = P(X < x, Y < y),
for x, y  R .
2) Properties
Property 1. 0 ≤ F(x, y) ≤ 1
Property 2. F(x, y) is a function that does not decrease with each variable or with both
variables, which means
F(x1, y) ≤ F(x2, y) when x1 < x2,
F(x, y1) ≤ F(x, y2) when y1 < y2,
F (x1, y1) ≤ F(x2, y2) when x1 < x2, y1 < y2.
Property 3. If at least one variable approaches  , F(x, y) approaches zero, which
means
3
Lecturer: Nguyen Duong Nguyen, Mathematics Department, Faculty of Basic Science, FTU
F(, y) = lim F(x, y) = 0,
x 
F(x, ) = lim F(x, y) = 0,
y 
F(, ) = lim F(x, y) = 0.
x 
y 
Property 4. If one variable approaches  , F(x, y) approaches the distribution function
of the other variable, which means
F(, y)  lim F(x, y)  FY (y)
x 
and
F(x, )  lim F(x, y)  FX (x) .
y
Property 5. If both variables approaches  , F(x, y) approaches 1, which means
F(, )  lim F(x, y)  1 .
x 
y
Note. From the above properties, we can deduce:
- The probability that the random vector (X, Y) takes the value in the rectangle
( x1 < X < x2, y1 < Y < y2) is defined as follows
P (x1 < X < x2, y1 < Y < y2) = F(x1, y1) + F(x2, y2) – F(x1, y2) – F(x2, y1)
- The probability that the random vector (X, Y) takes the value in the rectangle
( x1 < X < x2, Y < y) is defined as follows
P(x1 < X < x2, Y < y) = F(x2, y) – F(x1, y)
- Probability that a random vector (X, Y) takes the value in the rectangle
( X < x, y1 < Y < y2) is defined as follows
P ( X < x, y1 < Y < y2) = F(x, y2) – F(x, y1)
Example. Let the joint distribution function of a two-dimensional random vector (X, Y)
as follows:
4
Lecturer: Nguyen Duong Nguyen, Mathematics Department, Faculty of Basic Science, FTU

   
s inxsiny,if (x,y)  D  0,   0, 
F(x, y)  
 2  2
0,
if (x, y)  D

Find the probability that the random variable (X, Y) takes the value in the
rectangle bounded by the lines x = 0, x 



, y  và y  .
6
3
4
Solution. We have
P( 0  X 
 

 
 
  
 
;  Y  ) = F  ,   F  0,   F  ,   F  0, 
4 6
3
4 3
 3
4 6
 6


=  sin
=
 

 
 




 sin    sin 0   sin    sin  sin    sin 0   sin 
4 
3
3 
4 
6
6


2 3 1
 =

2  2 2
6 2
.
4
4.2.3 The joint probability density function (joint pdf) of a continuous (twodimensional) random vector
1) Definition. The joint probability density function of a continuous two-dimensional
random vector (X, Y), denoted f(x, y), is the second-order mixed partial derivative of the
joint distribution function:
f  x, y   Fxy (x, y) ,
for x, y  R .
Example. Find the joint probability density function of a two-dimensional random vector
 X, Y  if its joint distribution function is:


F(x, y) = sinx.siny (0  x  ,0  y  ) .
2
2
Solution. We have
Fx  cosxsiny
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Lecturer: Nguyen Duong Nguyen, Mathematics Department, Faculty of Basic Science, FTU
So, the joint probability density function of a continuous two-dimensional random vector
 X, Y  is:


f  x, y   Fxy  cosx.cosy , (0  x  ,0  y  )
2
2
2) Properties
Property 1. f(x, y) ≥ 0.
 
Property 2.
  f (x, y)dxdy  1 .
 
Example. The random vector (X, Y) has joint pdf as follows:
C(2x  y), if 2 < x < 6, 0 < y < 5
f (x, y)  
0, with other values of x, y
Find the constant C.
Solution. We have
6
5

25
1    f (x, y)dxdy    C(2x  y)dxdy  C    (2x  y)dy dx  C  (10x  )dx  210C
2
 
2 0
20
2

 
6 5
6
Thus,
C
1
.
210
x y
Property 3. F(x, y) 
  f (u, v)dudv , where
F(x, y) is the joint distribution function of
 - 
(X, Y) .
Example. Find the joint distribution function of the random vector (X, Y) according to
the following joint probability density function:
Solution. We have
y
x
dudv
1 
du  
dv 
F(x, y)    f (u, v)dudv    2




 1  u 2    1  v2 
 (1  u 2 )(1  v 2 ) 2  
 - 
 - 


x y
x y
6
Lecturer: Nguyen Duong Nguyen, Mathematics Department, Faculty of Basic Science, FTU
1 1
1
1
  arctanx+  .  arctany+ 
2 
2

Property 4. Let f1, f2 be the marginal probability density function of components X and
Y respectively. Then
;
Example. The joint probability density function of a two-dimensional random vector
(X, Y) has the form:
Find the marginal probability density function of the components.
Solution. We have

f1 ( x) 


f ( x, y )dy 


3 3


e
4 x 2  6 xy 9 y 2
3e 3 x
dy 
2

e

.
 (3 y  x ) 2
d (3 y  x) 


(Here we apply the integral
e
u2
3e 3 x

2
.  
3

e 3 x
2
du   )


f2 ( y) 


f ( x, y )dx 




3 3


e
4 x 2  6 xy 9 y 2
27
3 3e 4
dx 
2
y 2 
e
3
 (2 x  y )2
2


27
3
3 3e 4
d (2 x  y ) 
2
2
y2
. 
3 3  274 y 2
e
2 
Property 5. P[(X, Y)  D] =
 f (x, y)dxdy .
D
Example 1. Find the probability that (X, Y) takes the value in a rectangle with vertices
K(1; 1), L( 3 ; 1), N( 3 ; 0), M(1; 0) if the random vector (X, Y) has the following joint
probability density function:
f(x, y) =
1
.
 (1  x 2 )(1  y 2 )
2
Solution. We have
7
Lecturer: Nguyen Duong Nguyen, Mathematics Department, Faculty of Basic Science, FTU
P[(X, Y)  D] =
1
1
D 2 (1  x 2 )(1  y2 ) dxdy = 2
3
1
dx
dy
1
1 1  x 2 0 1  y2 = 48 .
Example 2. Let the joint probability density function of a two-dimensional random
vector
(X, Y) as follows:
 1
(2x  y), if 2 < x < 6, 0 < y < 5

f (x, y)   210
0, with other values of x, y
Find the probability P(3 < X < 4, Y > 2).
Solution. We have
Example 3. Let the joint probability density function of a two-dimensional random
vector
(X, Y) as follows:
8xy, nÕu 0 < x < 1, 0 < y < x
f (x, y)  
0, with other values of x, y
Find the probability P(0 < X <
1
, 0 < Y < x).
2
Solution. We have
1
P(0 < X < , 0 < Y < x) =
2
1
2 x
1
2 x
1
2
  f (x, y)dxdy    8xydxdy  4 x dx  0.0625 .
3
0 0
0 0
0
4.3 Conditional distribution and Independence
4.3.1 Conditional Probability Distribution Table
Consider a discrete two-dimensional random vector (X, Y), where the set of values
that X can take is {x1, x2, …, xn} and the set of the values that Y can take is {y1, y2, …,
ym}.
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Lecturer: Nguyen Duong Nguyen, Mathematics Department, Faculty of Basic Science, FTU
The conditional probability distribution table of the component X given Y = yj has
the form
X/Y=yj
x1
x2
...
xi
...
xn
P
P(x1/yj)
P(x2/yj)
…
P(xi/yj)
…
P(xn/yj)
where P(xi/yj) = P(X = xi/Y = yj) (i  1, n; j  1, m) is the conditional probability of the
component X having a value of xi provided that the component Y has a value of yj
calculated by the formula:
P(xi/yj) =
n
Note.
 P(x
i 1
i
P(x i , y j )
P(Y  y j )
,i  1, n
/ yj)  1
Similarly, the conditional probability distribution table of the component Y given
X = xi has the form
Y/X=xi
y1
y2
…
yj
…
ym
P
P(y1/xi)
P (y2/xi)
…
P(yj/xi)
…
P(ym/xi)
where the conditional probabilities P(yj/xi) are calculated using the formula
P(yj/xi) =
P(x i , y j )
P(X  x i )
, j  1, m
m
Note.
 P(y / x )  1
j1
j
i
Example. Let the probability distribution table of a two-way random vector (X, Y),
where
X = "Revenue", Y = "Advertising cost" as follows: (Unit: Million VND)
Y
X
100
150
200
0
0.1
0.05
0.05
1
0.05
0.2
0.15
9
Lecturer: Nguyen Duong Nguyen, Mathematics Department, Faculty of Basic Science, FTU
2
0
0.1
0.3
Find the probability distribution of the revenue without advertising and when
advertising costs is 2 million dong.
Solution.
+) We have
P(Y = 0) = 0.1 + 0.05 + 0.05 = 0.2
0.1
P(X  100, Y  0)
=
= 0.5
0.2
P(Y  0)
P(X=100/Y=0) =
P(X=150/Y=0) =
0.05
P(X  150, Y  0)
=
= 0.25
P(Y  0)
0.2
P(X=200/Y=0) =
0.05
P(X  200, Y  0)
=
= 0.25
P(Y  0)
0.2
So, the probability distribution of revenue without advertising is
X/Y =0
100
150
200
P
0.5
0.25
0.25
+) We have
P(X=100/Y=2) =
0
P(X  100, Y  2)
=
=0
0.4
P(Y  2)
P(X=150/Y=2) =
0.1
P(X  150, Y  2)
=
= 0.25
0.4
P(Y  2)
P(X=200/Y=2) =
0.3
P(X  200, Y  2)
=
= 0.75
0.4
P(Y  2)
So, the probability distribution of the revenue when the advertising cost is 2 million is
10
Lecturer: Nguyen Duong Nguyen, Mathematics Department, Faculty of Basic Science, FTU
X/Y =2
100
150
200
P
0
0.25
0.75
4.3.2 The conditional probability density function (The conditional pdf)
Assume (X, Y) is a continuous two-dimensional random vector with joint pdf
f (x, y) . The conditional probability density function of the component X given Y = y,
denoted f(x/y), is defined as:
f(x/y) =
f (x, y)

f 2 (y)
f (x, y)

.
 f (x, y)dx

Similarly, the conditional probability density function of the component Y given X
= x, denoted by f(y/x), is defined as:
f(y/x) =
f (x, y)

f1 (x)
f (x, y)

.
 f (x, y)dy

Note: f(x/y)  0 and

 f (x / y)dx  1

f(y/x)  0 and

 f (y / x)dy  1 .

Example. The joint probability density function of a two-dimensional random variable
has the form:
Find the conditional probability density function of the components.
Solution. The marginal pdf of X is
f1 ( x) 
3

The marginal pdf of Y is
11
e3 x .
2
Lecturer: Nguyen Duong Nguyen, Mathematics Department, Faculty of Basic Science, FTU
f2 ( y) 
3 3  274 y 2
.
e
2 
We have the conditional probability density function of X given Y  y is
and the conditional probability density function of Y given X  x is
● On the basis of conditional probability distributions, we have the following
formulas:
+) If (X, Y) is a discrete two-dimensional random vector, where the set of values that X
can take is {x1, x2, …, xn} and the set of values that Y can take is {y1, y2, …, ym} , we
have
P(x i , y j )  P(X  x i )P(y j / x i )= P(Y=y j )P(x i / y j ) , for i  1,n; j  1,m .
Especially: X and Y are independent if and only if
P(xi, yj) = P(X = xi).P(Y = yj), for i  1,n; j  1,m .
+) If (X, Y) is a continuous two-dimensional random vector, we have
f(x, y) = f1(x)f(y/x) = f2(y)f(x/y).
Especially: X and Y are independent if and only if
f  x, y   f1  x  f 2  y  ,
for all x, y e R.
4.4 Characteristic parameter of two-dimensional random vector
4.4.1 The expected value and the variance of the component random variable
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Lecturer: Nguyen Duong Nguyen, Mathematics Department, Faculty of Basic Science, FTU
1) If (X, Y) is a discrete two-dimensional random vector, where the set of values that X
can take is {x1, x2, …, xn} and the set of values that Y can take is {y1, y2, …, ym}, we
have
Example. Let (X, Y) be a two-dimensional random vector with the following probability
distribution table:
100
150
200
0
0.1
0.05
0.05
1
0.05
0.2
0.15
2
0
0.1
0.3
Y
X
Find E(X), V(X), E(Y), V(Y).
Solution.
E(X) = 100.P(X = 100) + 150.P(X = 150) + 200.P(X = 200)
= 100(0.15) + 150(0.35) + 200(0.5) = 167.5
V(X) = (1002.P(X = 100) + 1502.P(X = 150) + 2002.P(X = 200)) – [E(X)]2
= (1002(0.15) + 1502(0.35) + 2002(0.5)) – (167.5)2 = 29375 – 28056.25 = 1318.75
E(Y) = 0.P(X = 0) + 1.P(X = 1) + 2.P(X = 2) = 0(0.2) + 1(0.4) + 2(0.4) = 1.2
V(Y) = (02.P(Y = 0) + 12.P(Y = 1) + 22.P(Y = 2)) – [E(Y)]2
= (02.(0.2) + 12.(0.4) + 22.(0.4)) – (1,2)2 = 2 – 1.44 = 0.56
b) If (X, Y) is a continuous two-dimensional random vector , we have
13
Lecturer: Nguyen Duong Nguyen, Mathematics Department, Faculty of Basic Science, FTU

 
 xf (x)dx    xf (x, y)dxdy ,
E(X) =
1

 
 

V(X) =

  x f (x, y)dxdy - [E(X)] ,
2
x 2f1 (x)dx - [E(X)]2 =
 


E(Y) =

 
yf 2 (y)dy 

  yf (x, y)dxdy ,
 
 

V(Y) =

2
y 2f 2 (y)dy - [E(Y)]2 =
  y f (x, y)dxdy -[E(Y)] .
2
2
 

Example. Let the probability distribution density function of a two-dimensional random
vector (X, Y) as follows:
 1
(2x  y), if 2 < x < 6, 0 < y < 5

f (x, y)   210

with other values of x, y
0,
Find E(X), V(X), E(Y), V(Y).
Solution. We have
 
6 5
6
1
1
2144 268
x(2x  y) dxdy =
x(4x  5)dx 

E(X) =   xf (x, y)dxdy =  

210
84 2
504
63
2 0
 
 
1 2
 268 
x (2x  y) dxdy - 
V(X) =   x f (x, y)dxdy  [E(X)] =  

210
 63 
2 0
 
6 5
2
2
2
2
2
5036
1
 268  1220  268 
x 2 (4x  5)dx - 
=
-
.
 =
 =

84 2
3969
63  63 
 63 
6
 
E(Y) =

1 5
170
1
y(2x

y)
dxdy
y(2y  16)dy =
=

2 0 210
105 0
63
6 5
yf (x, y)dxdy =
 
 
V(Y) =

 
6 5
y 2 f (x, y)dxdy - [E(Y)]2 =
1
  210 y (2x  y)dxdy - [E(Y)]
2 0
14
2
2
Lecturer: Nguyen Duong Nguyen, Mathematics Department, Faculty of Basic Science, FTU
2
2
1 5 2
1175  170 
16225
 170 
y (2y  16)dy - 

=
 =
 =

105 0
7938
126  63 
 63 
4.4.2 Covariance and correlation coefficient
Covariance and correlation coefficient are numbers that characterize the degree of
dependence between random variables.
1) Covariance
a) Definition. The covariance of the random variables X and Y, denoted Cov(X, Y), is
defined as:
Cov(X, Y) = E{[X – E(X)][Y – E(Y)]} = E(XY) – E(X)E(Y)
Especially:
i) If (X, Y) is a discrete random vector then
n
Cov(X, Y) =
m
 x y P(x , y )  E(X)E(Y)
i 1 j1
i
j
i
j
ii) If (X, Y) is a continuous random vector then
 
Cov(X, Y) =
  xyf(x,y)dxdy  E(X)E(Y)
 
b) Properties
Property 1. If X, Y are independent, Cov(X, Y) = 0.
The converse is not true, which means if Cov(X, Y) = 0, X and Y can be independent or
dependent.
Example 1. Let X and Y be two random variables with a probability distribution table as
follows:
Y
-1
0
1
-1
4/15
1/15
4/15
0
1/15
2/15
1/15
1
0
2/15
0
X
Cov(X, Y) = 0 but X and Y are not independent because
15
Lecturer: Nguyen Duong Nguyen, Mathematics Department, Faculty of Basic Science, FTU
1
9 5
4
 P(X = -1).P(Y = -1) =
. = .
15
15 15
5
P(X = -1, Y = -1) =
Property 2. Let X, Y be two dependent random variables and a, b be real numbers. Then
V (aX  bY )  a 2V ( X )  b 2V (Y )  2abCov( X , Y )
Especially: V ( X  Y )  V ( X )  V (Y )  2Cov( X , Y ) ,
V ( X  Y )  V ( X )  V (Y )  2Cov( X , Y ) .
Example 2. There are two types of stocks A and B sold in the stock market and their
interest rates are two random variables X and Y respectively (unit: %). Suppose (X, Y)
has the following probability distribution table:
Y
-2
0
5
10
0
0
0.05
0.05
0.1
4
0.05
0.1
0.25
0.15
6
0, 1
0.05
0.1
0
X
a) If the goal is to achieve the maximum expected return, what ratio should you invest in
both stocks?
b) In order to minimize interest rate risk, what ratio should you invest in these two
stocks?
Solution.
Y
-2
0
5
10
P(xi)
0
0
0.05
0.05
0.1
0.2
4
0.05
0.1
0.25
0.15
0.55
6
0.1
0.05
0 ,1
0
0.25
P(yj)
0.15
0.2
0.4
0.25
X
E(X) = 0(0.2) + 4(0.55) + 6(0.25) = 3.7%;
V(X) = (02(0.2) + 42(0.55) + 62(0.25)) – 3.72 = 4.11
E(Y) = 4.2%; V(Y) = 17.96
16
Lecturer: Nguyen Duong Nguyen, Mathematics Department, Faculty of Basic Science, FTU
a) If we denote α as the ratio of investment in stock A, then we have a ratio of investment
in stock B of (1 - α). So, we have to find α such that E(αX + (1 - α)Y)
max
We have
E(αX + (1 - α)Y) = αE(X) + (1 - α)E(Y) = α(3.7) + (1 - α)(4.2) = 4.2 – (0.5)α
⇒ E(αX + (1 - α)Y)
max when α = 0. That is, if we want to achieve the maximum
expected interest rate, we must invest in buying all stock B.
b) Determine α such that V(αX + (1 - α)Y)
We have: P(X = 0, Y = -2)
min
P(X = 0) ).P(Y = -2) = (0.2)(0.15) = 0.03 ⇒ X, Y
are two dependent variables.
Therefore,
V(αX + (1 - α)Y) = α2V(X) + (1 - α)2V(Y) + 2α(1 - α)Cov(X, Y)
Cov(X, Y) =
 x y p(x , y )
i
j
i
j
- E(X)E(Y) = (12.4) – (3.7)(4.2) = -3.14
V(αX + (1 - α)Y) = (4.11)α2 +(17.96)(1 - α)2 + 2α(1 - α)(-3.14)
= (28.35)α2 – (42.2)α + 17.96 = f(α)
f      (56.7)  42.2  0 ⇒ α = 0.7443.
f ()  56.7  0
So, V(αX + (1 - α)Y)
min when α = 0.7443.
In conclusion: If investing in stocks A and B at the ratio of 74.43% and 25.57% , we
will have the lowest level of risk.
From the above definition, we can see that the covariance has a unit of measure
equal to the product of the measurement units of the random variables X and Y.
Therefore, the covariance will have different values depending on the unit of measure of
those random variables. Therefore, covariance is usually not significant and does not
accurately reflect the degree of dependence between random variables X and Y. To
overcome this limitation, a parameter called correlation coefficient is introduced.
2) Correlation coefficient
17
Lecturer: Nguyen Duong Nguyen, Mathematics Department, Faculty of Basic Science, FTU
a) Definition. The correlation coefficient of two random variables X and Y, denoted by
XY , is the ratio between the covariance and the product of the standard deviations of
those random variables:
XY 
Cov(X, Y)
Cov(X, Y)
.

X .Y
V(X)V(Y)
b) Properties
Property 1. XY = YX .
Property 2. -1  XY  1.
Property 3. If X and Y are independent, we have XY = 0.
The converse of Property 3 is not true; the correlation coefficient can be 0 even if the
random variables are dependent.
Property 4. XY  1 if and only if Y = aX + b, where a > 0.
Property 5. XY  1 if and only if Y = aX + b, where a < 0.
c) Meaning. The correlation coefficient XY is used to measure the degree of linear
dependence between two random variables X and Y. The closer |  XY | is to 1, the
stronger is the linear dependence between X and Y. The closer
is to 0, the weaker
is the linear dependence between X and Y. In particular, if  XY = 0, then the two
variables X and Y have no linear relationship.
Definition. Two random variables X and Y are said to be correlated if XY  0 . Otherwise
, if XY  0 , we say that X and Y are uncorrelated.
Note.
i) If two random variables are correlated, they are also dependent. But the converse is not
true, which means if two random variables are dependent, they may be correlated but also
may be uncorrelated.
Example 3. Two random variables X and Y in Example 1 are dependent but they are
uncorrelated.
18
Lecturer: Nguyen Duong Nguyen, Mathematics Department, Faculty of Basic Science, FTU
ii) If two random variables X and Y are independent, they are uncorrelated.
Example 4. Let the probability distribution table of a two-dimensional random vector (X,
Y), where X = "Revenue", Y = "Advertising cost" as follows: (Unit: Million VND) )
Y
X
100
150
200
0
0.1
0.05
0.05
1
0.05
0.2
0.15
2
0
0.1
0.3
Are revenue and advertising costs correlated?
Solution. Cov(X, Y) =
 x y P(x , y )  E(X)E(Y)
i
j
i
j
= ((100)(0)(0.1) + (100)(1)(0.05) + (100)(2)(0) + (150)(0)(0.05) +
(150)(1)(0.2) + (150)(2)(0.1) + (200)(0)(0.05) + (200)(1)(0.15) + (200)(2)(0.3)) –
(167.5)(1.2)
= 215 – 201 = 14
 XY 
Cov(X, Y)
14

 0.5152  0.
V(X).V(Y)
(1318.75)(0.56)
So, revenue and advertising costs are correlated.
Example 5. Let the joint probability density function of a two-dimensional random
vector (X, Y) as follows:
 1
(2x  y),if 2 < x < 6, 0 < y < 5

f (x, y)   210
0,
with other values of x, y
Are X and Y correlated?
Solution. We have
 
Cov(X, Y) =

 xyf(x,y)dxdy - E(X)E(Y) =
 
19
6 5
1
  xy 210 (2x  y)dxdy
2 0
- E(X)E(Y)
Lecturer: Nguyen Duong Nguyen, Mathematics Department, Faculty of Basic Science, FTU
=
XY 
80 268 170
200

.

7
63 63
3969
Cov(X, Y)
200 / 3969

 0.03129
V(X).V(Y)
5036 / 3969 16225 / 7938
So, X and Y are correlated.
4.5 Conditional Expectation
1) If (X, Y) is a discrete random vector, where the set of possible values X can take is
 x1 , x 2 ,
, x n  , and the set of values that Y can take is  y1 , y 2 , , y m  , the conditional
expectation of random variable Y given X = xi is defined as follows:
.
Similarly, the conditional expectation of X given Y = yj is defined as follows:
.
2) If (X, Y) is a continuous random vector, the conditional expectation of the random
variable Y given X = x is defined as follows:
E Y / X  x 

 yf(y/x)dy ,

where f(y/x) is the conditional probability density function of Y given X = x.
Similarly, the conditional expectation of X given Y = y is defined as follows:
E X / Y  y 

 xf(x/y)dx ,

where f(x/y) is the conditional probability density function of X given Y = y.
Example 1. Let the probability distribution table of two-dimensional random vector (X,
Y), where X = "Revenue", Y = "Advertising cost" as follows: (Unit: Million VND) )
Y
X
0
100
150
200
0.1
0.05
0.05
20
Lecturer: Nguyen Duong Nguyen, Mathematics Department, Faculty of Basic Science, FTU
1
0.05
0.2
0.15
2
0
0.1
0.3
Determine average revenue by advertising cost.
Solution. Average revenue by advertising cost is the conditional expectation of X over Y.
We have the probability distribution of revenue without advertising is
X/Y =0
100
150
200
P
0.5
0, 25
0.25
⇒ E(X/Y=0) = (100)(0.5) + (150)(0.25) + (200)(0.25) = 137.5
The probability distribution table of revenue when the advertising cost is 1 million
is
X/Y =1
100
150
200
P
0.05
 0.125
0.4
0.2
 0.5
0.4
0.15
 0.375
0.4
⇒ E(X/Y=1) = (100)(0.125) + (150)(0.5) + (200)(0.375) = 162.5
The probability distribution table of revenue when the advertising cost is 2 million
dong is
X/Y =2
100
150
200
P
0
0.25
0.75
⇒ E(X/Y=2) = 100.0 + 150.0.25 + 200.0.75 = 187.5
Example 2. The random vector (X, Y) has the joint probability density function as
follows:
8xy,if 0<x<1;0<y<x
f (x, y)  
0, with other values of x, y
Find the conditional expectation E(X/Y=y), E(Y/X=x).
Solution. The probability density function of X is
x
3
 8xydy,if 0 < x < 1 4x ,
f1 (x)   f (x, y)dy   0

0,

0,
if x  (0;1) 


The probability density function of Y is
21
if 0 < x < 1
if x  (0;1)
Lecturer: Nguyen Duong Nguyen, Mathematics Department, Faculty of Basic Science, FTU
1
2
  8xydx,if 0 < y < 1 4y(1  y ), if 0 < y < 1
f2 (y)   f (x, y)dx   y

if y  (0;1)

0,
0,
if
y

(0;1)


The conditional probability density function of X given Y  y is
 2x
,if y<x<1
f(x,y) 
f(x/y)=
 1  y 2
f2 ( y) 
with other values of x
0,
The conditional probability density function of Y given X  x is
 2y
f(x,y)  2 ,if 0<y<x
f(y/x)=
 x
f1 (x ) 
0, with other values of y
So, the conditional expectation of X given Y = y is
2x
2(1  y3 ) 2(1  y  y 2 )
E(X / Y  y)   xf (x / y)dx   x
dx 

2
3(1  y 2 )
3(1  y)

y 1 y

1
and the conditional expectation of Y given X = x is
E(Y / X  x) 

x

0
 yf (y / x)dy   y
2y
2x
dy 
.
2
x
3
22