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A chicken wishes to cross a single lane of traffic on a long-straight road. The arrival process of cars is a Poisson process of rate $\lambda$. She needs a car-free interval of length $c$ in order to safely cross the road.

How long does it take her to cross the road? How long on average does it take the chicken to cross two similar lanes (each with traffic at rate $λ)$ if $(a)$ she must walk straight across, (assume that the chicken will not cross if at any time during the crossing there is a car in either direction), or $(b)$ there is a traffic island, and she may stop between the lanes. Show from your answers that we expect it to take longer in scenario $(a)$.

  • Firstly, I want to try to calculate the expected time for the chicken to cross the road, say $H$. I'm slightly unsure how this is derived?
  • For $(a)$, I need the answer above but noting the question says the chicken is affected by cars in both lanes, then the rate should be $2\lambda$. $i.e$ looking to find $H(2\lambda,2c)$.
  • For $(b)$, If she can stop then we will need $2\times H(\lambda,c)$, since we should assume the chicken is not affected by a car in the other lane when crossing the first lane.

Ultimately, it is the beginning of the question I am unsure on - the rest should follow trivially$(?)$.

Community
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Btzzzz
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1 Answers1

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Each arrival is independent. The probability that the next arrival will take longer than $c$ is some probability $p(\lambda,c)$. Because the trials are independent, the number of car arrivals that you will need to wait for is a geometric random variable $N(\lambda,c)$, and you sum that many independent Exponential($\lambda$) random variables $T_i$ to determine how long the chicken waits before beginning to cross. Because the times and $N$ are not independent, you cannot simply compute the expectation of this sum by using Wald's identity. You will need to actually work with the distribution or use conditioning to make things look independent.

(As an aside, a general principle on these kinds of forums is to reduce the problem to the minimal case that is sufficient for your needs, so we really only needed to see the first part, provided that the second and third parts are trivial for you given the first.)

Ian
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  • Okay makes sense - from there my thought process would be that we could have $H=cP(T_i>c) + P(T_i<c)(E(T_i|T_i<c)+c)$ ? and we compute the final expectation as you say? – Btzzzz Mar 29 '18 at 20:11
  • @Btzzzz I think that's not quite right. Either $T_1<c$ or $T_1>c$; if $T_1>c$ then the time is just $c$ (which you have correctly included). Otherwise you have to wait again. For the time assuming you have to wait again, that's $E[T \mid T_1<c]=E[T]+E[T_1 \mid T_1<c]$, where the first term comes because after you wait for $T_1$ it's as if nothing happened. The second term is easily calculated, so you get a simple algebraic equation for $E[T]$. – Ian Mar 29 '18 at 20:18
  • Ah sorry, I think it should be adding $H$ not $c$ at the end because it will start over once we have an arrival before $c$? – Btzzzz Mar 29 '18 at 20:21
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    That's right. The contribution of $c$ is in the first term (assuming that you want the time of the whole trip, not the time to wait before starting to cross). In general this is a nice "renewal" method for computing the expectation without having to work with this sum directly. – Ian Mar 29 '18 at 20:21