Proof. \(P(T_1>t) = P(N(t) = 0) = e^{-\lambda t}\), so \(T_1\) is exponential with rate \(\lambda \). Given \(T_1 = s\), the event \(T_2>t\) means no events in \((s, s + t]\), which by independent and stationary increments has probability \(e^{-\lambda t}\) whatever \(s\) is; so \(T_2\) is exponential and independent of \(T_1\), and the same argument applies to each later \(T_n\).

For \(S_n\): the \(n\)th event has occurred by time \(t\) exactly when at least \(n\) events have occurred in \((0, t]\), so \[F_{S_n}(t) = P(N(t)\geq n) = \sum _{j=n}^\infty \frac {e^{-\lambda t}(\lambda t)^j}{j!} .\] Differentiating term by term, \[f_{S_n}(t) = \sum _{j=n}^\infty \left [\frac {\lambda e^{-\lambda t}(\lambda t)^{j-1}}{(j - 1)!} - \frac {\lambda e^{-\lambda t}(\lambda t)^j}{j!}\right ],\] a telescoping sum in which everything cancels except the first term, \(\frac {\lambda e^{-\lambda t}(\lambda t)^{n-1}}{(n - 1)!}\). □

Solution.

(a)

\(E(S_{10}) = \frac {10}{\lambda } = 10\) days.

(b)

The gap \(T_{11}\) is exponential with rate \(1\), so \(P(T_{11}>2) = e^{-2} = 0.135\).

(c)

\(N(7)\) is Poisson with mean \(7\), so \(P(N(7) = 3) = \frac {e^{-7}7^3}{3!} = 0.052\), and \(P(N(1)\geq 1) = 1 - e^{-1} = 0.632\).

Solution. The rate is \(\lambda = 2\), so \(P(X>\frac 12) = e^{-2\cdot \frac 12} = e^{-1} = 0.368\). By the memoryless property the answer to the second question is the same, \(0.368\): an exponential repair that is still going on is as good as new.