Solution.

(a)

A counting process \(\{N(t),\ t\geq 0\}\) is a Poisson process with rate \(\lambda >0\) if

(i)

\(N(0)=0\);

(ii)

it has independent increments: for any \(0<t_1<t_2<t_3<t_4\), \(N(t_2)-N(t_1)\) is independent of \(N(t_4)-N(t_3)\);

(iii)

it has stationary increments: the distribution of the number of events in an interval depends only on the length of the interval;

(iv)

the number of events in an interval of length \(t\) is Poisson with mean \(\lambda t\): \[P\left (N(t+s)-N(s)=k\right ) = \frac {(\lambda t)^{k}}{k!}e^{-\lambda t}, \qquad k=0,1,2,\dots \]

(b)

The first arrival comes after time \(t\) exactly when there is no event in \([0,t]\), so \[P(X_1>t) = P\left (N(t)=0\right ) = e^{-\lambda t}.\] Hence \(P(X_1\leq t) = 1-e^{-\lambda t}\) and \(f_{X_1}(t) = \lambda e^{-\lambda t}\) for \(t>0\): \(X_1\) is exponential with parameter \(\lambda \).

For \(X_2\), condition on the first arrival. By independent increments, and then by stationary increments, \begin {align*} P(X_2>t\mid X_1=s) &= P\left (\text {no arrival in }(s,s+t]\mid X_1=s\right )\\ &= P\left (\text {no arrival in }(s,s+t]\right )\\ &= P\left (\text {no arrival in }(0,t]\right )\\ &= P\left (N(t)=0\right ) = e^{-\lambda t}. \end {align*}

This does not depend on \(s\), so \(P(X_2\leq t) = 1-e^{-\lambda t}\) and \(X_2\) is exponential with parameter \(\lambda \), independent of \(X_1\). Conditioning on \(S_{k-1}\) in the same way gives the result for every \(X_k\), by induction.

(c)

The \(n\)th event has happened by time \(t\) exactly when at least \(n\) events have occurred in \([0,t]\): \(N(t)\geq n\) if and only if \(S_n\leq t\). Hence \[F_{S_n}(t) = P(S_n\leq t) = P\left (N(t)\geq n\right ) = \sum _{k=n}^{\infty }e^{-\lambda t}\frac {(\lambda t)^{k}}{k!}.\] Differentiating term by term, \begin {align*} f_{S_n}(t) &= \sum _{k=n}^{\infty }\lambda e^{-\lambda t}\frac {(\lambda t)^{k-1}}{(k-1)!} - \sum _{k=n}^{\infty }\lambda e^{-\lambda t}\frac {(\lambda t)^{k}}{k!}\\ &= \lambda e^{-\lambda t}\left [\frac {(\lambda t)^{n-1}}{(n-1)!} + \sum _{k=n+1}^{\infty }\frac {(\lambda t)^{k-1}}{(k-1)!} - \sum _{k=n}^{\infty }\frac {(\lambda t)^{k}}{k!}\right ]. \end {align*}

The two remaining sums are the same series term for term, so they cancel, and \[f_{S_n}(t) = \lambda e^{-\lambda t}\frac {(\lambda t)^{n-1}}{(n-1)!}, \qquad t>0 ,\] the gamma density with shape \(n\) and rate \(\lambda \). This agrees with (b): \(S_n = X_1+X_2+\cdots +X_n\) is a sum of \(n\) independent exponentials.

Solution. Condition on \(N(t)\). Given \(N(t)=n\), \[E\left [\sum _{k=1}^{N(t)}(t-S_k)\,\middle |\,N(t)=n\right ] = nt - E\left [\sum _{k=1}^{n}S_k\,\middle |\,N(t)=n\right ].\] Given \(N(t)=n\), the arrival times \(S_1,\dots ,S_n\) are distributed as the order statistics of \(n\) independent uniform variables on \((0,t)\). So with \(U_1,\dots ,U_n\) independent UNIF\((0,1)\), \[E\left [\sum _{k=1}^{n}S_k\,\middle |\,N(t)=n\right ] = tE\left [\sum _{k=1}^{n}U_{(k)}\right ] = tE\left [\sum _{k=1}^{n}U_k\right ] = t\sum _{k=1}^{n}\frac 12 = \frac {nt}{2},\] the middle step because putting the \(U_k\) in order does not change their sum. Hence \[E\left [\sum _{k=1}^{N(t)}(t-S_k)\,\middle |\,N(t)=n\right ] = nt-\frac {nt}{2} = \frac {nt}{2},\] and taking the expectation over \(N(t)\), which has mean \(\lambda t\), \[E\left [\sum _{k=1}^{N(t)}(t-S_k)\right ] = E\left [\frac {t}{2}N(t)\right ] = \frac {t}{2}E\left [N(t)\right ] = \frac {\lambda t^{2}}{2}.\]