# Probability question with interarrival times

(Why did the chicken cross the road?) A chicken wants to cross a single-lane road where the cars arrive according to a Poisson process with rate $\lambda$. She needs at least $k$ minutes to cross the road safely, so she will have to wait until she sees a gap of at least $k$ between the oncoming cars. If the gap between the car that just arrived and the next one is at least $k$ then she starts crossing the road immediately. Let $T$ denote the random time she needs to wait by the road.

Find the expected time needed to cross the road.

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@Sasha You're confusing "chicken" and "hen" – crf Sep 14 '12 at 5:28

The chicken waits between the passages of car $i-1$ and car $i$ without crossing the road if and only if no gap between car $n-1$ and car $n$ is greater than $k$, for any $n\leqslant i$. Thus, $$T=\sum_{i=1}^{+\infty}D_i\cdot[D_1\leqslant k,\ldots,D_i\leqslant k],$$ where $(D_i)_{i\geqslant 1}$ is i.i.d. with exponential distribution of parameter $\lambda$. By independence, $$\mathrm E(D_i;D_1\leqslant k,\ldots,D_i\leqslant k)=a^{i-1}b,\quad a=\mathrm P(D\leqslant k),\quad b=\mathrm E(D;D\leqslant k),$$ hence $\mathrm E(T)=b/(1-a)$. One knows that $1-a=\mathrm P(D\gt k)=\mathrm e^{-\lambda k}$, and $$b=\int_0^kx\cdot\lambda\mathrm e^{-\lambda x}\cdot\mathrm dx=\frac{1-(1+\lambda k)\mathrm e^{-\lambda k}}\lambda.$$ Finally, $$\mathrm E(T)=\frac{\mathrm e^{\lambda k}-1-\lambda k}\lambda.$$

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What does $D_i$ represent in therms of the problem? Is it the arrival of car $i$ ? – audiFanatic Mar 24 '14 at 5:31
@audiFanatic The gap between two successive arrivals. – Did Mar 24 '14 at 6:24
Ok, thanks. Could you take a look at my question? It is somewhat similar to this one, but the approach I took is different. math.stackexchange.com/questions/724228/… – audiFanatic Mar 24 '14 at 6:32

Hint: condition on the size of the first gap (i.e. the time until the first car arrives).

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The question is ambiguous.

1) If the question is how much time the chicken has to wait till it decides to cross the road, proceed as follows. The time of waiting can be represented as $$T=\sum_{i=0}^N Y_i$$ Where $Y_0, Y_1, \dots$ are the arrival times of cars (i.i.d $exp(\lambda)$, being a Poisson process with parameter $\lambda$), and $N$ is the number of cars until (not including) the first one to arrive after at least $k$ minutes. $N\sim\mathrm{Geom}(P(5<Y_0))$. To calculate $E(T)$, use the formula of Total Expectation, conditioning on $N$. In the course of your calculations, you will have to use basic facts about infinite series of functions.

2) If the question is how much time it would take the chicken to cross the road once it started doing so, then again the data is ambiguous:

a. If it takes the chicken exactly k minutes to cross the road, then the expected time is k.

b. If it takes the chicken at least k minutes to cross the road, then we need to know the distribution of the length of time it takes the chicken to cross the road in order to answer the question what will come first: the chicken crossing the road or it getting hit by a car.

Note that in all cases, it is assumed the chicken is clairvoyant and can predict with certainty that the next car will arrive in no less than k minutes, which is quite an accomplishment, whether or not you're a chicken.

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If it's a straight road with no nearby entrance or exit points, and the cars maintain constant speed, and $k$ is small enough, then the chicken can already see the cars that will arrive within the next $k$ minutes, so the prediction is not that hard. You should avoid crossing a road if you can't predict whether a car will arrive before you finish crossing. – Robert Israel Sep 15 '12 at 0:43
@RobertIsrael: What if the car comes from the future, "Back to the Future" style, out of the blue? The poor chicken doesn't stand a chance, no matter how vigilant it is! – Evan Aad Sep 15 '12 at 6:33

Nothing new. I just want to upload for no reason ^_^

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