Consider a Poisson process with rate $λ$ and let $L$ be the time of the last arrival in the interval $[0, t]$, with $L = 0$ if there was no arrival. (a) Compute $E(t − L)$. (b) What happens when we let $t → ∞$ in the answer to (a)?
When I compute
$E(t − L)$,
I condition on $N(L)$.
$$E(t − L)=\sum_{n=0}^{\infty}E(t-L\mid N(L)=n)P(N(L)=n)$$
By definition of $L$, $N(L)=N(t)$, $$E(t − L)=tP(N(t)=0)+\sum_{n=1}^{\infty}E(t-L\mid N(t)=n)P(N(t)=n)$$
For $N(t)=n$, $L=T_n$ where $T_n$ is $Gamma(n, λ)$ distributed, $$E(t − L)=t\exp(-\lambda t)+\sum_{n=1}^{\infty}E(t-T_n)P(N(t)=n)\\=t\exp(-\lambda t)+\sum_{n=1}^{\infty}(t-n/\lambda)\exp(-\lambda t)\frac{(\lambda t)^{n}}{n!}\\=t\exp(-\lambda t)+t\exp(-\lambda t)(\exp(\lambda t)-1)-t\exp(-\lambda t)\exp(\lambda t)=0$$
What's wrong?
The answer is $(1/λ)(1 − \exp
(−λt))$
How can I attained this answer?
Once you condition on event $N(t)=n$, you know that in the interval $[0,t]$ there are $n$ arrivals with all inter-arrival times distributed identically. The expected value of each of these $n+1$ inter-arrival times is therefore simply $\frac{t}{n+1}$. Note also that this value applies to the case $n=0$, since $\frac{t}{n+1}=t$ then. Therefore,
\begin{eqnarray*} E(t-L) &=& \sum_{n=0}^{\infty}{E(t-L\mid N(t)=n)P(N(t)=n)} \\ &=& \sum_{n=0}^{\infty}{\dfrac{t}{n+1}\dfrac{(\lambda t)^n}{n!}e^{-\lambda t}} \\ &=& \dfrac{1}{\lambda}\sum_{n=0}^{\infty}{\dfrac{(\lambda t)^{n+1}}{(n+1)!}e^{-\lambda t}} \\ &=& \dfrac{1}{\lambda}\left[\left( \sum_{n=0}^{\infty}{\dfrac{(\lambda t)^{n}}{n!}e^{-\lambda t}}\right) - e^{-\lambda t}\right] \\ &=& \dfrac{1}{\lambda}\left(1 - e^{-\lambda t}\right). \end{eqnarray*}
Now as $t\to \infty,\;$ clearly, $\;E(t-L)\to \dfrac{1}{\lambda}$. This is the mean inter-arrival waiting time for the process, and this makes sense: as $t\to\infty$, the chance of getting no arrivals approaches $0$ and $t-L$ approaches an inter-arrival waiting time.