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(A7)
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A probabilistic proof of (A7) is as follows Imagine that X is the lifetime of a machine De ne 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
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I (t) dt =
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E [I (t)] dt =
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P {X > t} dt,
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which proves (A7) The interchange of the order of expectation and integration is justi ed by the non-negativity of I (t) The result (A7) can be extended to E(Xk ) = k
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x k 1 [1 F (x)] dx,
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k = 1, 2,
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(A8)
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To see this, note that (A7) implies E(Xk ) =
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P {Xk > t} dt =
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P {X > t 1/k } dt
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and next use the change of variable t = x k Mean and variance of a random sum of random variables Let X1 , X2 , be a sequence of independent and identically distributed random variables whose rst two moments are nite Also, let N be a non-negative and integer-valued random variable having nite rst two moments If the random variable N is independent of the random variables X1 , X2 , , then
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k=1 N
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= E(N )E(X1 ),
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(A9)
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= E(N )var(X1 ) + var(N )E 2 (X1 ),
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(A10)
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where E 2 (X1 ) is the shorthand notation for [E(X1 )]2 The proof uses the law of total expectation By conditioning on N , we nd
N N
n=0
k=1 n
Xk | N = n P {N = n}
Xk P {N = n} =
nE(X1 )P {N = n},
which veri es (A9) Note that the second equality uses that the random variables X1 , , Xn are independent of the event {N = n} Similarly, E
=
n=0
E
| N = n P {N = n}
2 [nE(X1 ) + n(n 1)E 2 (X1 )]P {N = n}
2 = E(N )E(X1 ) + E[N (N 1)]E 2 (X1 )
(A11)
Using 2 (S) = E(S 2 ) E 2 (S), we obtain (A10) from (A9) and (A11)
436 Wald s equation
APPENDICES
The result (A9) remains valid when the assumption that the random variable N is independent of the sequence X1 , X2 , is somewhat weakened Suppose that the following conditions are satis ed: (i) X1 , X2 , is a sequence of independent and identically distributed random variables with nite mean, (ii) N is a non-negative, integer-valued random variable with E(N ) < , (iii) the event {N = n} is independent of Xn+1 , Xn+2 , for each n 1 Then it holds that E
k=1 N
= E(X1 )E(N )
(A12)
This equation is known as Wald s equation It is a very useful result in applied probability To prove (A12), let us rst assume that the Xi are non-negative The following trick is used For n = 1, 2, , de ne the random variable Ik by Ik = Then
N k=1 Xk
1 if N k, 0 if N < k
k=1 Xk Ik N
and so
Xk Ik
E(Xk Ik ),
where the interchange of the order of expectation and summation is justi ed by the non-negativity of the random variables involved The random variable Ik can take on only the two values 0 and 1 The outcome of Ik is completely determined by the event {N k 1} This event depends on X1 , , Xk 1 , but not on Xk , Xk+1 , This implies that Ik is independent of Xk Consequently, E(Xk Ik ) = E(Xk )E(Ik ) for all k 1 Since E(Ik ) = P {Ik = 1} and P {Ik = 1} = P {N k}, we obtain (A9) from (A6) and
E(X1 )P {N k}
For the general case, treat separately the positive and negative parts of the Xi The assumption E(N ) < is essential in Wald s equation To illustrate this, consider the symmetric random walk {Sn , n 0} with S0 = 0 and Sn = X1 + + Xn , where X1 , X2 , is a sequence of independent random variables with P {Xi = 1} = P {Xi = 1} = 1 for all i De ne the random variable N as 2 N = min{n 1 | Sn = 1}, that is, N is the epoch of the rst visit of the random walk to the level 1 Then E(X1 + + XN ) = 1 Noting that E(Xi ) = 0, we