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Am currently working on a Stochastic Poisson process on my project. I have thought and settled on the below scenario which I think is appropriate. However, solving it am not getting what I expect.

I want to cross a road at a spot where cars pass according to a Poisson process with a rate of lambda. I will begin to cross as soon as I see there will be no cars passing for the next time C units. I have taken N=Number of cars that pass before I cross and T=Time I begin to cross the road.

I want to determine the E(N) and Also E(T). What I know is that to find the E(T), I will have to condition on N

Ben
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2 Answers2

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I am not sure, what you are trying to model here. Maybe if you give some more details, I may be able to provide a better answer.

From what you are saying, it seem to me like you are doing the "gambler's fallacy" here, just for Poisson processes instead of a binomial process.

Now the gambler's fallacy arises, because a predictor is used to predict a value, that can by definition of the underlying process be shown to be independent from each other. For example if I throw a fair die, the probability of the die showing a 6 is by definition of the underlying binomial process independent of the number of 6 that I have seen before (or rather have not seen before). If I assume the probability I will see a 6, because I have not seen a 6 on the die for a long time, I will overestimate it.

The same is true for a Poisson process. By definition of a Poisson process, the expected number of cars in the next time interval (e.g. time it takes to cross the road), is by definition of a Poisson process independent of the number of cars that have passed before. Therefore, if you use the number of cars that have passed before to predict anything, you can only succumb to the gambler's fallacy and you must overestimate your chances, just as a gambler looking at the lack of 6es in the run before must overestimate his chances.

So for the question about when to cross the road, if the cars really follow a Poisson process? Just pick any time. The Poisson process guarantees that the probability of at least one car in the next time interval is independent of the number of previous cars. Therefore, my chances will be the same for all T and I should not worry about prediction and just see to get across as fast as possible.

Alexis
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LiKao
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  • Thank you for you feedback. I may need some clarification on: When I am calculating the Expected number of cars that will pass before I cross (E(N)), Will I assume since the mean of a Poisson is lambda that the E(N) = lambda. Or does this assumption hold in this case? – Ben Feb 10 '17 at 09:56
  • @Ben Yes, your E(N) will be lambda, as this is the expected value of the poisson process independently of everything that happened before. – LiKao Feb 10 '17 at 10:00
  • Knowing that, am now left with out troubling question. E(T). Will this be Lambda multiplied by N Since I will cross after N cars pass? – Ben Feb 10 '17 at 10:07
  • @Ben Not sure, what you mean exactly by "E(T)"? Expected number that pass, when you cross at time T? Or the expected value of the time at which you pass (meaning that T is some random variable, for which you never gave any distribution). The main point above is, that all calculation that yo can do to predict the number of cars that pass or the best time etc, will always have the gambler's fallacy, i.e. will be wrong. – LiKao Feb 10 '17 at 10:15
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    @Ben Or more specifically: Let $E(T)$ be the unconditional expected value of the number of cars that pass, while you are crossing the road for a given point in time $T$. Then $E(T)=\lambda$ for all $T$. Now let, $E(T|N_T)$ the conditional expected value conditioned that in some interval before $T$ exactly $N_T$ cars passed. Then we still have $E(T|N_T)=\lambda$ as these two things are independent. That means, we cannot predict anything, since with the best mode of prediction we always get the same value. – LiKao Feb 10 '17 at 10:19
  • I now get you. I highly appreciate your help @LiKao. – Ben Feb 10 '17 at 10:31
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You might have a check at this post: Poisson Process, which dealt with $E(T)$. Concerning $E(N)$, the situation is similar, you still need to condition upon the first passage, that is $$E(N)= E(E(N|J_1)),$$ where $J_1$='when the first car will pass'. Then $$E(N)= E(N|J_1>a)P(J_1>C)+ \int_0^C E(N|J_1=u)f_{J_1}(u) du$$ $$= 0+ \int_0^C (1+ E(N)) \lambda e^{-\lambda u} du .$$ It remains to solve the equation to find $E(N).$

LiKao, in your answer $E(T), E(N)$ do not depend at all on the constant $C$, while if it increases also $E(T), E(N)$ should increase.

KarloKik
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