1
$\begingroup$

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.

$\endgroup$
5
$\begingroup$

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:

$$E(T) + k = \dfrac{1}{\lambda}\left( e^{\lambda k} - 1 \right).$$

$\endgroup$
10
  • $\begingroup$ Thank you so much for the swift response! I'm not familiar with the Poisson processes yet, so how did you set the first equation? Thank you in advance! $\endgroup$ Jan 11 '16 at 15:36
  • $\begingroup$ Hi @PineApples. The first equation is not specific to Poisson processes. It is a use of the Total Law of Expectation. See the last part of the intro of en.wikipedia.org/wiki/Law_of_total_expectation : One special case states that if $A_1,A_2,\ldots ,A_n$ ... Here we have $n=2$ with $A_1="X\lt k"$ and $A_2$ its complement. $\endgroup$
    – Mick A
    Jan 11 '16 at 16:11
  • $\begingroup$ I read the entry but I'm still confused. I understand that it is the sum of two possible probabilities, when we have to wait for the car and when we can cross the street but don't know how E(T ∣ X<k) was obtained and why the case is special, is it possible that you explain that, please? Sorry for all the trouble I've caused, I'm really grateful for the time you've taken out of your day to help me. $\endgroup$ Jan 11 '16 at 16:34
  • $\begingroup$ @PineApples Do you mean you don't understand why $E(T|X\lt k)=E(T)+E(X|X\lt k)$? If $X\lt k$ we have to wait for this next car that we see to pass by. That car is $X$ time away from us so that's how long we have to wait for it to pass. Once the car passes, we start over again as though we just arrived so the remaining expected waiting time is $T$ again, and that value is independent of the $X$ value for the car that just passed by. We add these two amounts together for the total waiting time. $\endgroup$
    – Mick A
    Jan 11 '16 at 20:06
  • $\begingroup$ Okay, I think I understand that part now, thank you! How did we get the expression in line 3? Where the fraction is included, since both P(X<k) and [E(T)+E(X∣X<k)] are still included in the equation as well? And what is the relationship between step 3 and 4, where did we get the integral? Is it some sort of formula? I am completely new to Poisson processes. $\endgroup$ Jan 11 '16 at 20:45

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.