Statistics 510: Notes 18
Reading: Sections 6.3-6.5
I. Sums of Independent Random Variables (Chapter 6.3)
It is often important to be able to calculate the distribution
of X  Y from the distribution of X and Y when X and
Y are independent.
Sum of Discrete Independent Random Variables:
Suppose that X and Y are discrete independent random
variables. Let Z  X  Y . The probability mass function
of Z is
P ( Z  z )   P ( X  x, Y  z  x )
x
  P ( X  x ) P (Y  z  x )
x
where the second equality uses the independence of X and
Y.
Example 1: Sums of independent Poisson random
variables. Suppose X and Y are independent Poisson
random variables with respective means 1 and 2 . Show
that Z  X  Y has a Poisson distribution with mean
1  2 .
1
Sums of Continuous Independent Random Variables:
Suppose that X and Y are continuous independent random
variables. Let Z  X  Y . The CDF of Z is
FZ (a )  P ( Z  a )
 P( X  Y  a)


f XY ( x, y )dxdy
x  y a
2


f X ( x) fY ( y )dxdy (using X , Y independent)
x ya


a y

 
 F
 

a y
 


X
f X ( x) fY ( y )dxdy
f X ( x)dxfY ( y )dy
(a  y ) fY ( y )dy
By differentiating the cdf, we obtain the pdf of Z  X  Y :
d 
f Z (a) 
FX (a  y ) fY ( y )dy


da
 d

F (a  y ) fY ( y )dy
 da X



f X (a  y ) fY ( y )dy
Example 2: Sum of independent uniform random variables.
If X and Y are two independent random variables, both
uniformly distributed on (0,1), calculate the pdf of
Z  X Y .
3
Sum of independent normal random variables:
Proposition 6.3.2: If X i , i  1, , n, are independent
random variables that are normally distributed with
2
respective means and variances i ,  i , i  1, , n, then
n
X
i 1
n
i
is normally distributed with mean
2
i
.

i 1
i
and variance
n

i 1
The book contains a proof of Proposition 6.3.2 that
calculates the distribution of X 1  X 2 using the formula for
the sum of two independent random variables established
above, and then applying induction. We will prove
Proposition 6.3.2 in Chapter 7.7 using Moment Generating
Functions.
4
Example 3: A club basketball team will play a 44-game
season. Twenty-six of these games are against class A
teams and 18 are against class B teams. Suppose that the
team will win each against a class A team with probability
.4 and will win each game against a class B team with
probability .7. Assume also that the results of the different
games are independent. Approximate the probability that
the team will win 25 games or more.
5
II. Conditional Distributions (Chapters 6.4-6.5)
(1) The Discrete Case:
Suppose X and Y are discrete random variables. The
conditional probability mass function of X given Y  y j is
the conditional probability distribution of X given Y  y j .
The conditional probability mass function of X|Y is
p X |Y ( xi | y j )  P( X  xi | Y  y j )


P( X  xi , Y  y j )
P(Y  y j )
p X ,Y ( xi , y j )
pY ( y j )
(this assumes P (Y  y j )  0 ). This is just the conditional
probability of the event X  xi given that Y  y j .
If X and Y are independent random variables, then the
conditional probability mass function is the same as the
unconditional one. This follows because if X is
independent of Y, then
p X |Y ( 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)

6
Example 3: In Notes 17, we considered the situation that a
fair coin is tossed three times independently. Let X denote
the number of heads on the first toss and Y denote the total
number of heads.
The joint pmf is given in the following table:
Y
X
0
1
2
3
0
1/8
2/8
1/8
0
1
0
1/8
2/8
1/8
What is the conditional probability mass function of X
given Y? Are X and Y independent?
(2) Continuous Case
If X and Y have a joint probability density function
f ( x, y ) , then the conditional pdf of X, given that Y=y , is
defined for all values of y such that fY ( y)  0 , by
7
f X |Y ( x | y) 
f X ,Y ( x, y)
fY ( y ) .
To motivate this definition, multiply the left-hand side by
dx and the right hand side by (dxdy ) / dy to obtain
f X ,Y ( x, y )dxdy
f X |Y ( x | y )dx 
fY ( y )dy
P{x  X  x  dx, y  Y  y  dy}
P{ y  Y  y  dy}
 P{x  X  x  dx | y  Y  y  dy}

In other words, for small values of dx and dy ,
f X |Y ( x | y ) represents the conditional probability that X is
between x and x  dx given that Y is between y and y  dy .
The use of conditional densities allows us to define
conditional probabilities of events associated with one
random variable when we are given the value of a second
random variable. That is, if X and Y are jointly continuous,
then for any set A,
P{X  A | Y  y}   f X |Y ( x | y)dx .
A
In particular, by letting A  (, a] , we can define the
conditional cdf of X given that Y  y by
a
FX |Y (a | y)  P( X  a | Y  y)   f X |Y ( x | y)dx .

8
Example 4: Suppose X and Y are two independent random
variables, both uniformly distributed on (0,1). Let
T1  min{ X , Y }, T2  max{ X , Y } . What is the conditional
distribution of T2 given that T1  t ? Are T1 and
T2 independent?
9
10