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have, however, that E(X1 + + XN ) is not equal to E(N )E(X1 ) The reason is that E(N ) = Example A3 A reliability problem To illustrate Wald s equation, consider the following reliability problem An electronic system has a built-in redundancy in the form of a standby unit to support an operating unit The two units are identical When the operating unit fails, its tasks are immediately taken over by the standby unit if available A failed unit immediately enters repair The system goes down when the operating unit fails while the other unit is still in repair The lifetime L of an operating unit is assumed to have a continuous probability distribution F (x) with nite mean The repair time of a failed unit is a constant > 0 The successive lifetimes of the operating unit are independent of each other A repaired unit is as good as new Both units are in perfect condition at time 0 What is the expected time until the system goes down for the rst time To solve this problem, denote by L0 the lifetime of the operating unit installed at time 0 and denote by L1 , L2 , the lifetimes of the subsequent operating units Then the time until the rst system failure is distributed as L0 + L1 + + LN , where the random variable N denotes the rst n 1 for which Ln is less than the nth repair time The random variables L1 , , Ln and the event {N = n} are mutually dependent, but the event {N = n} is independent of Ln+1 , Ln+2 , for each n 1 Hence we can apply Wald s equation This gives E(time until the rst system failure) = E(L0 ) + E(L1 )E(N ) = [1 + E(N )] To nd E(N ), note that N has a geometric distribution with parameter p = P {L < } Hence E(N ) = 1/F ( ) and so E(time until the rst system failure) = 1 + 1 F ( ).

In practical applications the mean lifetime will be much larger than the mean repair time In other words, the occurrence of a system failure is a rare event For those situations there is a deep but extremely useful result stating that the time until the rst system failure is approximately exponentially distributed ; see also Example 224 Coef cient of variation Let X be a positive random variable with nite mean E(X) and nite standard deviation (X) The coef cient of variation of X is de ned by cX = (X) E(X)

.

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Since this quantity is dimensionless, it is a very useful measure for the variability of the random variable X Usually one works with the squared coef cient of variation 2 2 cX rather than with cX For example, the deterministic distribution has cX = 0, 2 = 1 and the Erlang distribution with shape the exponential distribution has cX 2 parameter k has the intermediate value cX = 1/k Failure rate function Let X be a positive random variable with a probability distribution function F (t) and a continuous probability density f (t) For example, the random variable X represents the lifetime of some item The failure, or hazard, rate function of the random variable X is de ned by f (t) r(t) = 1 F (t) for those values of t with F (t) < 1 The failure rate has a useful probabilistic interpretation Think of the random variable X as the lifetime of an item The probability that an item of age t will fail in the next t time units is given by P {t < X t + t | X > t} = = P {t < X t + P {X > t} t} t 0

f (t) t + o( t) as 1 F (t)

Hence r(t) t gives approximately the probability that an item of age t will fail in the next t time units when t is small Hence the name failure rate Noting that r(t) is the derivative of the function ln[1 F (t)], it follows that the failure rate function r(t) determines uniquely the corresponding lifetime distribution function F (t) by 1 F (t) = exp

r(x) dx ,

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