Page 441 - A First Course In Stochastic Models

436 APPENDICES

Wald’s equation
The result (A.9) remains valid when the assumption that the random variable N is
independent of the sequence X 1 , X 2 , . . . is somewhat weakened. Suppose that the
following conditions are satisfied:

(i) X 1 , X 2 , . . . is a sequence of independent and identically distributed random
variables with finite mean,
(ii) N is a non-negative, integer-valued random variable with E(N) < ∞,
(iii) the event {N = n} is independent of X n+1 , X n+2 , . . . for each n ≥ 1.

Then it holds that
N

E X k = E(X 1 )E(N). (A.12)
k=1
This equation is known as Wald’s equation. It is a very useful result in applied
probability. To prove (A.12), let us first assume that the X i are non-negative. The
following trick is used. For n = 1, 2, . . . , define the random variable I k by

1 if N ≥ k,
I k =
0 if N < k.

N
∞
Then k=1 X k = k=1 X k I k and so
N ∞ ∞

E X k = E X k I k = E(X k I k ),
k=1 k=1 k=1
where the interchange of the order of expectation and summation is justified by the
non-negativity of the random variables involved. The random variable I k can take
on only the two values 0 and 1. The outcome of I k is completely determined by the
event {N ≤ k−1}. This event depends on X 1 , . . . , X k−1 , but not on X k , X k+1 , . . . .
This implies that I k is independent of X k . Consequently, E(X k I k ) = E(X k )E(I k )
for all k ≥ 1. Since E(I k ) = P {I k = 1} and P {I k = 1} = P {N ≥ k}, we obtain
(A.9) from (A.6) and
N ∞

E X k = E(X 1 )P {N ≥ k}.
k=1 k=1
For the general case, treat separately the positive and negative parts of the X i .
The assumption E(N) < ∞ is essential in Wald’s equation. To illustrate this,
consider the symmetric random walk {S n , n ≥ 0} with S 0 = 0 and S n = X 1 +
· · · + X n , where X 1 , X 2 , . . . is a sequence of independent random variables with
1
P {X i = 1} = P {X i = −1} = for all i. Define the random variable N as
2
N = min{n ≥ 1 | S n = −1}, that is, N is the epoch of the first visit of the random
walk to the level −1. Then E(X 1 + · · · + X N ) = −1. Noting that E(X i ) = 0, we