Page 183 - Fundamentals of Probability and Statistics for Engineers

166 Fundamentals of Probability and Statistics for Engineers

it be given that, on average, a telephone user is on the phone 5 minutes per
hour; an estimate of p is

5 1
p ˆ ˆ :
60 12
The solution to this problem is given by

2
3 1 11 3 1
3
p 2†‡ p 3†ˆ ‡
X X
2 12 12 3 12
11
ˆ ˆ 0:0197:
864

Example 6.3. Problem: let X 1 and X 2 be two independent random variables,
both having binomial distributions with parameters (n 1 , p) and (n 2 , p), respect-
ively, and let Y ˆ X 1 ‡ X 2 . Determine the distribution of random variable Y .
Answer: the characteristic functions of X 1 and X 2 are, according to the first
of Equations (6.8),
jt n 1 jt n 2
X 1 t†ˆ pe ‡ q† ; X 2 t†ˆ pe ‡ q† :
In view of Equation (4.71), the characteristic function of Y is simply the
(t) (t). Thus,
product of X 1 and X 2
t†
Y t†ˆ X 1 t† X 2
jt
ˆ pe ‡ q† n 1 ‡n 2 :
By inspection, it is the characteristic function corresponding to a binomial
distribution with parameters (n 1 ‡ n 2 , p). Hence, we have

n 1 ‡ n 2 k n 1 ‡n 2 k
p k†ˆ p q ; k ˆ 0; 1; ... ; n 1 ‡ n 2 :
Y
k
Generalizing the answer to Example 6.3, we have the following important
result as stated in Theorem 6.1.
Theorem 6.1: The binomial distribution generates itself under addition of
independent random variables with the same p.

Example 6.4. Problem: if random variables X and Y are independent binomial

distributed random variables with parameters (n 1 , p) and (n 2 , p), determine the
conditional probability mass function of X given that
X ‡ Y ˆ m; 0 m n 1 ‡ n 2 :

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