Page 442 - A First Course In Stochastic Models
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A. USEFUL TOOLS IN APPLIED PROBABILITY           437

                have, however, that E(X 1 + · · · + X N ) is not equal to E(N)E(X 1 ). The reason is
                that E(N) = ∞.

                Example A.3 A reliability problem

                To illustrate Wald’s equation, consider the following reliability problem. An elec-
                tronic 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 immedi-
                ately 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 finite mean µ. The repair time of
                a failed unit is a constant α > 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 first time?
                  To solve this problem, denote by L 0 the lifetime of the operating unit installed
                at time 0 and denote by L 1 , L 2 , . . . the lifetimes of the subsequent operating units.
                Then the time until the first system failure is distributed as L 0 + L 1 + · · · + L N ,
                where the random variable N denotes the first n ≥ 1 for which L n is less than
                the nth repair time. The random variables L 1 , . . . , L n and the event {N = n} are
                mutually dependent, but the event {N = n} is independent of L n+1 , L n+2 , . . . for
                each n ≥ 1. Hence we can apply Wald’s equation. This gives

                        E(time until the first system failure) = E(L 0 ) + E(L 1 )E(N)
                                                      = µ [1 + E(N)] .

                To find E(N), note that N has a geometric distribution with parameter p = P {L <
                α}. Hence E(N) = 1/F(α) and so

                                                                   1

                           E(time until the first system failure) = µ 1 +  .
                                                                 F(α)
                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 first system failure is approximately exponentially distributed; see also
                Example 2.2.4.

                Coefficient of variation

                Let X be a positive random variable with finite mean E(X) and finite standard
                deviation σ(X). The coefficient of variation of X is defined by
                                                 σ(X)
                                            c X =     .
                                                 E(X)
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