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Let $X_n$ be the interarrival times for a Poisson process $\{N_t; t \geq 0\}$ with rate $\lambda$. Is it possible to calculate the probability $P\{ X_k \leq T \text{ for } k \le n, \sum_{k=1}^{n}{X_k} = t, X_{n+1}>T\}$ for given $t$ and $T$ (suppose $t$ and $T$ are compatible), i.e., how to calculate expectation of the first time that the next interarrival time is larger than or equal to T?

Thank you!

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Do you mean $\sum_{k=1}^nX_k\le t$? You can then differentiate to get the density. –  robjohn Oct 4 '11 at 20:14

2 Answers 2

The interarrival times of a Poisson process have absolutely continuous distributions hence the event $A_n(t)=[X_1+\cdots+X_n=t]$ has probability zero, for every $t$. The event the OP is considering is included in $A_n(t)$ hence its probability is zero as well.

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Good point. I thought he was intending $\sum_{k=1}^nX_k\le t$ (and then possible computing a density from that), but it does say $=$ in the question. –  robjohn Oct 4 '11 at 20:13

Hint: Are the $X_i$ independent and identically distributed? If so, would $P(X_k\le T)$ be the same for all $k$?

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Thank you for your reply. I don't know how to deal with $\sum{X_k} = t$. –  epsilon Oct 4 '11 at 19:01
    
@epsilon: yes, it is still ugly. To get the cdf, you would need to integrate the product of the density functions ($\lambda x_k\;e^{-\lambda x_k}$) over the cube $0\le x_k\le T$ minus the area where $\sum_{k=1}^n x_k>t$. I am not sure there is a simpler way. –  robjohn Oct 4 '11 at 20:29

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