STAT 111
Chapter6
Jointly Distributed Random
Variables
1
In many experiments it is necessary to consider the properties of
two or more random variables simultaneously. In the following,
we shall be concerned with the bivariate case, that is, with
situations where we are interested at the same time in a pair of
random variables. Later, we shall extend this discussion to the
multivariate case, covering any finite number of random
variables,
If X and Y are two discrete random variables, the probability
distribution for their simultaneous occurrence can be represented
by a function with values f (x, y) for any pair of values (x, y)
within the range of the random variables X and Y. It is common to
refer to this function as the joint probability distribution
of X and Y, and it is defined as
f (x, y) = P(X = x, Y = y)
2
Definition
x
The function f (x, y) is a joint probability distribution (or joint
probability mass function) of the discrete random variables X and Y
if and only if its values satisfy the conditions;
1. f ( x, y )  0 for all (x, y)
2. ∑ ∑ f (x ,y ) = 1, where the double summation extends over all
x y
Based on this definition , the joint cumulative distribution
function is the joint probably that X ≤ x and Y ≤ y ,given by
F (x , y ) = p (X ≤ x ,Y≤ y)= ∑
xi ≤ x
for - ∞ < x<∞
,
∑ ƒ ( xi , yi)
yi ≤ y
- ∞< y <∞
3
Example
Let X denote the number of heads and Y the number of
heads minus the number of tails when coins are tossed. Find
the joint probability distribution of X and Y.
Possible values (0,-3) ,(1,-1), (2,1), (3,3)
P(X=0, Y= -3) = 1 ,
8
P(X=1 , Y=-1)= 3 , …..
8
These probabilities can most easily be expressed in tabular from as in this
table
Y/X
-3
-1
1
3
Sum
0
1/8
0
0
0
1/8
1
0
3/8
0
0
3/8
2
0
0
3/8
0
3/8
3
0
0
0
1/8
1/8
Sum
1/8
3/8
3/8
1/8
1
4
Example
Suppose that 3 balls are randomly selected from an urn containing 3 red, 4 white
and 5 blue balls. Let X denotes the number of red balls chosen. Let Y denotes the
number of white balls chosen. Find the joint probability function of X and Y,
P(X + Y ≤ 2), P(X = 1, Y = 2), p(X = 0, 0 ≤ Y≤ 2) , and P(X > Y).
X = # R , Y= #W
f (0,0)  P( X  0, Y  0)  C35 / C312
1. X + Y ≤ 2
for the following values of X and Y
X/Y
{(0. 0), (1, 0), (2, 0), (0, 1),(1, 1), (0,2)}therefore,
P(X+Y≤2)= 10 + 40 + 30 + 30 + 60 + 15
220 220 220 220 220 220
2. P (X= 1,Y=2) =f (1,2) = 18/220
3R
5B
4W
0
1
3
2
3
0
10/220
40/220 30/220
4/220
1
30/220
60/220 18/220
0
2
15/220
12/220
0
0
3
1/220
0
0
0
3. P(X = 0, 0 ≤ Y≤ 2) = ƒ (0, 0) + ƒ(0,1) + ƒ(0, 2 ) = 10 + 40 + 30
220 220 220
5
Example
Let the joint probability mass function of X and Y is given in the
following table Y
X
0
1
2
0
1
2
1/16
1/16
1/8
1/8
1/8
1/4
1/8
1/16
1/16
Find
1. P(X= 1, Y≤ 0) = ƒ(1,0) =1/8
2. P(X = 2,Y ≤ 0) =ƒ(2, 0) = 1/8
3. P(X = 2,X + Y = 4) = ƒ(2,2) = 1/16
4. P(l ≤ X<3, Y≥ 1) = ƒ(1, l)+ƒ(l,2) + ƒ(2, l) + ƒ(2,2) =
5.P(X ≤ 2) = 1 - P(X > 2) = 1 - 0 = 1
1 + 1 + 1+ 1
8
4 16 16
6. F(l, 1) = (P(X≤1,Y≤1) = f (1,0)+ f (1, l)+f (0,0)+f (0,1) =
6
1 + 1 + 1 + 1
8
8
16 16
Example
Determine the value of c so that the following functions represent
joint probability distributions of the random variables X and Y;
1. ƒ(x,y) = c x y,
for x = 1,2,3.
y=1,2,3.
∑ ƒ(x+ y) = c [1+2+3+2+4+6+3+6+9] = 1
 c = 1/36
2. ƒ(x , y) = c│ x-y │,
for x= -2 ,0 ,2
y =1,2,3.
∑ ƒ(x , y) = c[3+4+5+1+2+3+1+0+1]=1
 c = 1/20
7
All the preceding definitions concerning two random variables can
be generalized to the multivariate case, where there are n random
variables. The values of the joint probability distribution of the
discrete random variables X1,X2 , …, Xn, defined over the same
sample space S, are given by
f(x1,x2, ...,xn) = P(X1=X 1,X2, ..., Xn = xn)
for all n-tuple (x1, x2, ..., xn) within the range of the random
variables. Also the values of their joint cumulative distribution
function are given by
F ( x1 , x2 ,… , xn ) = P ( X1 ≤ x1 , X2 ≤ x2, …Xn ≤ xn)
8
Example
Considering n flips of a balanced coin, let X 1 be the number of heads
(0 or 1) obtained on the first flip, X2 the number of heads obtained on
the second flip, ..., and Xn the number of heads obtained on the nth flip.
find the joint probability distribution of these n random variables.
f(x1 , x2 , …,xn)= P(X1 = x1 , X2 = x2 ,..,Xn=xn) = 1 x 1 x 1 x…x 1 = 1
2 2 2
2 2n
Where each X: can take on the value 0 or 1
9
Marginal Distributions
If X and Y are discrete random variables and f(x, y) is the value
of their joint probability distribution at (x, y), the function given
by
f x (x) =∑ ƒ(x,y)
y
for each x within the range of X, is called the marginal distribution of
X
Similarly , the function given by
f y(y) =∑ ƒ(x,y)
x
for each y within the range of y , is called the marginal distribution of
y note that , because the probabilities are obtained from the margins ,
we call them marginal distribution
10
Example
Assume that the random variable X and Y have the following joint
prob. Mass function. Find the marginal distribution of X and Y.
X\Y
0
1
2
3
sum
0
10/220
30/220
15/220
1/220
56/220
1
40/220
60/220
12/220
0
112/220
2
30/220
1 8/220
0
0
4 8 /220
3
4/220
0
0
0
4/220
Sum
84/220
1 08/220
27/220
1/220
1
Marginal of x
X
ƒ(x)
0
2
1
84/220 108/220
3
27/220
Sum
1/220
1
Marginal of y
Y
ƒ(y)
0
1
56/220 112/220
2
48/220
3
4/220
11
Example
Assume that the random variables X and Y have the joint
probability mass function given as
f(x ,y)=
λx+y e -2λ
x!y!
x = 0 , 1 , 2 ,..
y=0,1,2,……
Find the marginal distribution of X
ƒ(x)= ∑
λx+y
e-2λ
=
x! y!
=
λxe-2λ
x!
∞
Y
x -2λ λ
λ
λ
∑ = e e
Y=0 Y!
X!
λx e -λ
x!
[Using e λ =
∞
∑ λt ]
t =0
t!
12
Example
Let the joint distribution of X and Y be given as
f(x , y) = x + y
x = 0,l,2,3 y = 0,1,2,
30
Find the marginal distribution function of X and Y.
marginal of X
2
2
y 0
y 0
f ( x)   f ( x, y) 
x y 1
1
x 1
 [( x  0)  ( x  1)  ( x  2)] 
[3x  3] 
30
30
30
10
Similarly, f(y) = (3+2x)/15
Or , since the joint is
y x
0
1
2
0
0 1/30 1/15
1 1/30 1/15 1/10
2 1/15 1/10 2/15
1/10 1/5 3/10
the marginal of x
3
1/10
2/15
1/6
2/5
x
f(x)
0
1
2
3
1/10
1/5 of
3/10
the
marginal
y
2/5
Y
0
1
2
f(y)
1/5
1/3
7/15
13
Conditional Distribution:
Recall that for any two events A and B, the conditional probability of
A given B is defined as;
P( A\ B) =
P (A ∩ B)
P (B)
Provided P(B) ≠ 0. Suppose now that A and B are the events X = x and
Y = y. So that we can write
P(X = x\ Y = y)= P (X=x ,Y=y)
(1)
P (Y = y)
Provided P(Y = y) = fY(y) ± 0, where P(X = x, Y = y) = f(x, y) is the value
of the joint probability distribution of X and Y at (x, y) and fy (y) is the
value of the marginal distribution of Y at y. Denoting the probability in
equation 1 by f(x\y) to indicate that x is a variable and y is fixed, then we
have the following definition.
14
Definition:
If f(x, y) is the value of the joint probability' distribution of the
discrete random variable X and Y at (x, y) and fy (y) is the value of the
marginal distribution of Y at y, the function given by
ƒ(x\y) = P(X = x\ Y = y) = ƒ (x , y)
ƒY (y)
for each x within the range of X, is called the conditional
distribution of X given Y = y. Similarly, if fx(x) is the value of
the marginal distribution of X at x
f(y/x) = P ( Y = y / X = x ) = ƒ (x , y)
ƒx (y)
For each y within the range of Y , is called the conditional
distribution of Y given X = x
15
Example 1
The joint probability mass function of X and Y
is given by Marginal of Y
f(1,1) = 1
f(1,2) = 1
f(2,1)= 1
8
4
8
f(2,2)=
2
1.Compute the conditional mass function of X given Y = i, i =1,2
2.Compute P(XY ≤ 3) = f(1, 1) + f(1, 2) + f(2, 1) = 1/2
3. P(X/Y> l) = f(2, 1)= 1/8
Y
x
1
2
1/8
1/8
1
2
1/4
1/2
sum
3/8
5/8
Marginal of y
y
f(y)
1
2/8
2
6/8
x
f(x\y=1)
The conditional mass f n of X/Y =1
y
The conditional of X/Y=2
f(x\y=2)
1
1/3
2
2/3
Sum
1
1
Sum
2/8
6/8
1
1
1/2
2
1/2
16
Independent Random Variables:
In chapter 3 we stated that two events are independent if
and only if
P(A ∩ B) = P(A) P(B)
that is, their joint probability is equal to the product of their
marginal probabilities. In the following the notion of
independence of random variables is presented.
17
Definition
The random variables X and Y are said to be independent if and
only if
ƒ(x, y) = ƒx(x) fY(y)
for all possible values of X and Y.
In terms of the Joint cumulative distribution function, X and Y
are independent if and only if
F(x, y) = Fx(x) FY(y)
For all possible values of X and Y
Or
P(X є A, Y є B) = P(X Є A) P(Y ЄB) for every A, B.
18
Thus, loosely speaking, X and Y are independent if knowing the
value of one does not change the distribution of the other.
Random variables that are not independent are said to be
dependent.
checking for Independence of discrete random variables requires
a very thorough investigation since it is possible to have the
product of the Marginal distributions equal to the joint
probability distribution for some but not all combinations of (
x,y) . If one can find any point ( x , y ) for which
ƒ(x , y) ≠ ƒx (x) . ƒy(y) ;
then the discrete variable X and Y are not independent
19
If X is independent of Y, then the conditional mass function is the
same as the unconditional ones. This follows because
if X is independent of Y, then
ƒ(x\y) = P(X = x\Y = y)
= P(X = x,Y = y)
P(Y = y)
= P(X = x)P(Y = y)
P( Y = y)
= P(X = x)
20
Definition
Let X1, X2, ..., Xn, be n discrete random variable having
densities ƒ1, f2, ..., fn respectively. These random variables are
said to be independent if their joint density function f is given
by
ƒ(x1,x2,..„xn)=ƒX1(xl)ƒX2(x2)...ƒXn(xn)
for all (x1, x2, ..., xn) within their range.
21
Example
Show that the random variables of Example 1 are not independent,
f (1,1) = 1
8
f(1,2)=
1
4
f (2,1)=
f (2,2)=
1
2
1
8
f x (1) = 3
8
, fx (2)= 5 , fy (1)= 2 ,
8
8
fy (2)=3 
4
f(1,1)= 1 ≠ 6 = fx (1) . fy (1)  X and Y are not independent
8 64
22
Example
The random variables X and Y are specified by
P(X=1)= 2
P(X=0)= 1
3
3
P(Y= 1)= 1
P( Y= -1) = 3
4
4
Construct the joint distribution of X and Y assuming that
X and Y are independent
0
1
-1
1/4
½
1
1/12
2/12
Y
X
23