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Here is an exercise.

Suppose that people arrive at a bus stop in accordance with a Poisson process with rate $\lambda$. The bus departs at time $t$. Let $X$ denote the total amount of waiting time of all those who get on the bus at time $t$. Let $N(t)$ denote the number of arrivals by time $t$.

Now we want to calculate $E[X\;|\;N(t)]$.

The solution says that the waiting time of each person, $T$, is uniformly distribute in $(0,t)$. So $E[X\;|\;N(t)=n]=n\int_0^t(t-s)\frac{1}{t}ds$.

My question is, why $T$ is uniformly distribute? If people arrivals obeying Poisson process, shouldn't $T$ obeying exponential distribute with parameter $\lambda$?

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    No because $T$ depends on buses, not people.2012-12-01

2 Answers 2