SUM OF A RANDOM NUMBER OF INDEPENDENT RANDOM VARIABLES
Properties of the Sum of a Random Number of Independent Ran dom Variables
Let Xl , X2, . . . be identically distributed random variables with mean E[X] and variance var(X). Let N be a random variable that takes nonnegative in
teger values. We assume that all of these random variables are independent, and we consider the sum
Y = X1 + · · · + XN·
Then:
E[Y] = E[N] E[X] .
var(Y) = E[N] var(X) + (E[X] ) var(N).
We have
lvly (s) = l\,fN ( log Mx (s)) .
Equivalently, the transform lvly (s) is found by starting with the trans form MN(S) and replacing each occurrence of e8 with lvIx (s).
Example 4.34. A remote village has three gas stations. Each gas station is open on any given day with probability 1/2. independent of the others. The amount of gas available in each gas station is unknown and is uniformly distributed between 0 and 1000 gallons. We wish to characterize the probability law of the total amount of gas available at the gas stations that are open.
The number N of open gas stations is a binomial random variable with p= 1/2 and the corresponding transform is
The transform !Ylx {s) associated with the amount of gas available in an open gas station is
elOOOs _ 1
l\/x { s) =
lOs 0 0 .
Sec. 4.5 Sum of a Random Number of Independent Random Variables 243
The transform associated with the total amount Y available is the same as MN (S), except that each occurrence of eS is replaced with Mx (s), i.e.,
My (s) =8 ( ( 1 + ))
1 elOOOS _ 1
1000s
Example 4.35. Sum of a Geometric Number of Independent Exponen
tial Random Variables. Jane visits a number of bookstores, looking for Great Expectations. Any given bookstore carries the book with probability p, indepen dent of the others. In a typical bookstore visited, Jane spends a random amount of time, exponentially distributed with parameter A, until she either finds the book or she determines that the bookstore does not carry it. We assume that Jane will keep visiting bookstores until she buys the book and that the time spent in each is independent of everything else. We wish to find the mean, variance, and PDF of the total time spent in bookstores.
The total number N of bookstores visited is geometrically distributed with pa rameter p. Hence, the total time Y spent in bookstores is the sum of a geometrically distributed number N
of independent exponential random variables Xl, X2, We have
E[Y] = E[N] E [ X ] = .
Using the formulas for the variance of geometric and exponential random variables, we also obtain
var(Y) 1 = E[N] var(X)
+ ( )2 E[X] var(N) =p. +
1 I-p
A2 A2 . �= A2p2 ·
In order to find the transform My (s), let us recall that
Mx (s)
Then, My (s) is found by starting with MN (S) and replacing each occurrence of eS with Mx (s). This yields
PA
pMx (s)
My (s)
= A-S
1 - (1 - p)Mx (s) -
1 - ( 1 - p)
A-S
which simplifies to
My (s) =
p -s
We recognize this as the transform associated with an exponentially distributed random variable with parameter PA, and therefore,
fy
(y) -P>'Y , pAe
y � o.
244 Further Topics on Random Variables Ch ap. 4 This result can be surprising because the sum of a fixed number n of indepen
dent exponential random variables is not exponentially distributed. For example, if n=
2, the transform associated with the sum is (>'/(>' - s ) ) 2, which does not
correspond to an exponential distribution. Example 4.36. Sum of a Geometric Number of Independent Geometric
Random Variables. This example is a discrete counterpart of the preceding one. We let N be geometrically distributed with parameter p. We also let each random variable Xi be geometrically distributed with parameter q. We assume that all of these random variables are independent. Let Y = Xl +... + XN . We have
Mx( s ) =
To determine My (s), we start with the formula for MN (S) and replace each occur rence of eS with M x (s) . This yields
pMx (s) lY!y (s) =
1 _ (1 - p)Mx (s) '
and, after some algebra,
My ( s) =
1- -
We conclude that Y is geometrically distributed, with parameter pq.