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I need to compute the following joint distribution in a Poisson process:

$f_{S_A S_{A+B}}(t_1, t_2), t_2\ge t_1$

$S_A$ and $S_{A+B}$ are the arrival epochs of the $A^{th}$ and ${A+B}^{th}$ arrivals respectively. I tried the following:

\begin{align} P(S_A\ge t_1, S_{A+B}\ge t_2)&=P(N_{t_1}\le A, N_{t_2}\le A+B)\\ &=P((N_{t_1}\le A)P(N_{t_2-t_1}\le B)\\ &=\sum_{k=0}^A \frac{e^{-\lambda t_1} (\lambda t_1)^k}{k!}\sum_{l=0}^B \frac{e^{-\lambda (t_2-t_1)} (\lambda (t_2-t_1))^l}{l!} \end{align}

To find the density, I will need to differentiate this wrt $t_1$ and $t_2$. However the expressions are unwielfy. Is the approach correct? Is there a simpler way to arrive at the solution?

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  • $\begingroup$ Are $A$ and $B$ the inter-arrival times of the first two events of the Poisson process? If so, they follow an exponential distribution, and not a Poisson distribution. If you use that then you will get a much nicer probability. $\endgroup$ – QQQ Dec 18 '14 at 17:11
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Hint: The random variables $S_A$ and $S_{A+B}-S_A$ are independent and $S_{A+B}-S_A$ is distributed like $S_B$. Now, you ought to know the distribution of each $S_A$ hence you ought to be able to finish this.

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    $\begingroup$ Thanks Did. I guess it would be $f_{S_A}(t_1)\times f_{S_B}(t_2-t_1)$, with the $S_A$ being Erlang. Correct? $\endgroup$ – Bravo Dec 20 '14 at 17:37
  • $\begingroup$ Indeed. $ $ $ $ $\endgroup$ – Did Dec 20 '14 at 20:16

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