# Average waiting time in a Poisson process

Sheldon Ross's Introduction to Probability Models, exercise 5.44.b: cars pass a certain street location according to a Poisson process with rate $\lambda$. A woman who wants to cross the street at that location waits until she can see that no cars will come by in the next $\tilde t$ time units. I'm trying to find her expected waiting time.

I am given a hint to condition on the time of the first car, which I'll call $T_1$. The process is noted $\{N(t) \;|\; t \geq 0\}$. Her waiting time can be defined as $$S := \min_{t\geq0}\{N(t+\tilde t) - N(t) = 0\} \geq 0\quad.$$

Clearly, $$\mathbb E[S\;|\;T_1 \geq \tilde t] = \mathbb E[0] = 0\quad.$$ If this approach is good, how should I proceed for $T_1 < \tilde t$?

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The exercise seems to ask for $E(S)$ where $S=\inf\{t\ge0;N(t+\tilde t)=N(t)\}$ hence $S\ge0$, in any case this is the convention used below.
Using the random time $T_1$ when the first car passes, as you suggested, one gets $S=0$ on $T_1> \tilde t$. On $T_1\le\tilde t$, the pedestrian must wait until time $T_1$ and then wait for the first suitable time to cross the street. The lack of memory of the Poisson process at time $T_1$ implies that this residual time is distributed like $S$ and is independent on $T_1$. Hence, $$E(S)=E(E(S|T_1);T_1\le\bar t)=E(T_1+E(S);T_1\le\bar t)=E(T_1;T_1\le\bar t)+E(S)P(T_1\le\bar t).$$ Solving for $E(S)$ and using the fact that $T_1$ is exponentially distributed with parameter $\lambda$, one gets $$E(S)=\frac{E(T_1;T_1\le\bar t)}{P(T_1\ge\bar t)}=\mathrm{e}^{\lambda\bar t}\int_0^{\bar t}\lambda s\mathrm{e}^{-\lambda s}\mathrm{d}s,$$ hence $$E(S)=\frac1\lambda(\mathrm{e}^{\lambda\bar t}-1-\lambda\bar t).$$ Post hoc checks: The distribution of $\lambda S$ should depend on $\lambda\bar t$ only. The random variable $S$ should converge in distribution to $0$ when $\lambda\to0$, for any given $\bar t$, and also when $\bar t\to0$, for any given $\lambda$.