A pedestrian wishes to go across a single-lane road where the cars arrive according to a Poisson process with the rate λ. The time needed for him to cross the road safely is denoted as k. He will have to wait until he sees a gap of at least k between the coming cars. If the gap between the car that just arrived and the next one is at least k, he begins to cross the road. Let T denote the random time he needs to wait by the road.
What is the expected time the pedestrian needs to cross this particular road?
I've encountered several similar problems on here, but none of them gives a detailed, step-by-step explanation, so I am really confused. I would appreciate any help! Thank you in advance.
Best Answer
This question could be open to interpretation but my reading of it is that when the pedestrian arrives at the road he can immediately see the road from where he stands to $k$ time units along the road, so that he can identify a coming time gap of $k$ units. So, if this section of road is empty at that time, he crosses immediately so that $T=0$ and the total time taken for the crossing is $T+k=k$.
Let $X$ be the time of the first car arrival and condition on the event $X\lt k$, meaning that the next car lies in that $k$-length section of road:
\begin{align} E(T) &= P(X\lt k)E(T\mid X\lt k) + P(x\geq k)E(T\mid X\geq k) \\ & \\ &= P(X\lt k)\left[E(T) + E(X\mid X\lt k)\right] + P(x\geq k)\cdot 0 \\ & \qquad\qquad\text{since, if $X\lt k$, we count $X$ and then re-start the wait} \\ & \qquad\qquad\text{and if $X\geq k$, there is no waiting required} \\ & \\ \therefore\quad E(T) &= \dfrac{1}{1-P(X\lt k)} P(X\lt k) E(X\mid X\lt k) \\ & \\ &= \dfrac{1}{e^{-\lambda k}} \int_{x=0}^{k} \lambda x e^{-\lambda x}\;dx \\ & \\ &= e^{\lambda k} \left[ e^{-\lambda x}\left( -x-1/\lambda \right)\right]_{x=0}^{k} \\ & \\ &= \dfrac{1}{\lambda}\left( e^{\lambda k} - 1 \right) - k. \end{align}
Therefore, the mean time to complete the crossing is:
$$E(T) + k = \dfrac{1}{\lambda}\left( e^{\lambda k} - 1 \right).$$