Let X1, X2, . . . be the i.i.d. lifetimes of some component in some large machine. The simplest replacement policy is to change a component as soon as it fails. In this case it may be necessary to call a repairman at night, which might be costly. Another policy, called replacement based on age, is to replace at failure or at some given age, a, say, whichever comes first, in which case the interreplacement times are Wk = min{Xk, a}, k ≥ 1. Suppose that c1 is the cost for replacements due to failure and that c2 is the cost for replacements due to age. In addition, let Yk be the cost attached to replacement k, k ≥ 1, and let N(t) be the number of replacements made in the time interval (0, t], where {N(t), t ≥ 0} is a Poisson process, which is independent of X1, X2, . . . . This means that Z(t) = N X (t) k=1 Yk is the total cost caused by the replacements in the time interval (0, t] (with Z(t) = 0 whenever N(t) = 0).

(a) Compute E Y1 and Var Y1.

(b) Compute E Z(t) and VarZ(t).

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