Help with Poisson Stochastic Process

Question

Cars pass along the road in accordance with the Poisson process of intensity $\lambda$ . A pedestrian crosses the road at time $W$ as soon as he sees that there will be no cars during time $T$ (visibility is infinite). Find the $E[W]$.

Answer 1

Let's say $T_1 = $ time until first car passes. Then

$$\mathsf E[W] = \mathsf E[W|W=0]P(W=0) + \mathsf E[W|W>0]P(W>0) \\ \mathsf E[W|W= 0] = 0 \therefore \mathsf E[W] = \mathsf E[W|W>0]P(W>0)$$

Note here that the event $W>0$ is equivalent to the event $T_1 \le T.$ Note further that $\mathsf E[W|T_1 \le T] = \mathsf E[T_1 | T_1 \le T] + \mathsf E[W]$ because you wait for the first car to pass and then you start over. Lastly, remember that $T_1 \sim \mathrm{Exp}(\mathrm{rate}=\lambda)$. Then

$$\mathsf E[W] = (\mathsf E[T_1 | T_1 \le T] + \mathsf E[W])P(T_1 \le T) \tag{*}$$

Let's talk about the distribution of $T_1|T_1\le T. \ \ T_1 \sim \mathrm{Exp}(\mathrm{rate}=\lambda), P(T_1 \le T) = 1-e^{-\lambda T}$ so $f_{T_1 | T_1 \le T}(t) =\cfrac{\lambda e^{-\lambda t}}{1-e^{-\lambda T}} \text{ for } 0 \le t \le T$ is the pdf of this distribution. Thus

$$(*) = \mathsf{E}[W] = \left(\int_0^T t\cfrac{\lambda e^{-\lambda t}}{1-e^{-\lambda T}} dt + \mathsf{E}[W]\right)\left(1-e^{-\lambda T}\right) = \left(\cfrac{1-e^{-\lambda T}-\lambda e^{-\lambda T}}{\lambda - \lambda e^{-\lambda T}} + \mathsf E[W]\right) \left(1-e^{-\lambda T}\right)$$

$$\mathsf E[W] = \cfrac{1-e^{-\lambda T}-\lambda e^{-\lambda T}}{\lambda} + \mathsf E[W] - e^{-\lambda T} \mathsf{E}[W] \therefore \mathsf E[W] e^{-\lambda T} = \cfrac{1-e^{-\lambda T}-\lambda e^{-\lambda T}}{\lambda} \\ \therefore \mathsf E[W] = \cfrac{1-e^{-\lambda T}-\lambda e^{-\lambda T}}{\lambda e^{-\lambda T}}$$