Page 439 - A First Course In Stochastic Models

434 APPENDICES

using the generating-function technique to be discussed in Appendix C. Omitting
the details, we state
1 − p n
E(L n ) = −pn + (2 (1 − p)) − 1 , n ≥ 1,
1 − 2p
1
E(A n ) = E(L n ), n ≥ 1.
1 − 2p
1
Indeed E(L n )/E(A n ) = 1 − 2p = in accordance with the fact that the house
37
percentage in European roulette is 2.702%.

Convolution formula
Let X 1 and X 2 be two independent, non-negative random variables with respective
probability distribution functions F 1 (x) and F 2 (x). For ease assume that F 2 (x)
has a probability density f 2 (x). Then, by a direct application of the law of total
probability, we have the convolution formula
x
P {X 1 + X 2 ≤ x} = F 1 (x − y)f 2 (y) dy, x ≥ 0.
0

Moments of a non-negative random variable
Let N be a non-negative, integer-valued random variable. A useful formula is
∞

E(N) = P {N > k}. (A.6)
k=0
To verify this result, write
∞ P {N > k} =
∞
∞ P {N = j} and
k=0 k=0 j=k+1
interchange the order of summation. The relation (A.6) can be generalized. For
any non-negative random variable X with probability distribution function F(x),
∞

E(X) = [1 − F(x)] dx. (A.7)
0
A probabilistic proof of (A.7) is as follows. Imagine that X is the lifetime of a
machine. Define the indicator variable I (t) by I (t) = 1 if the machine is still
working at time t and by I (t) = 0 otherwise. Then, by E[I (t)] = P {I (t) = 1}
and P {I (t) = 1} = P {X > t}, it follows that
∞ ∞ ∞

E(X) = E I (t) dt = E [I (t)] dt = P {X > t} dt,
0 0 0
which proves (A.7). The interchange of the order of expectation and integration is
justified by the non-negativity of I (t). The result (A.7) can be extended to

∞
k
E(X ) = k x k−1 [1 − F(x)] dx, k = 1, 2, . . . . (A.8)
0