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I am trying to calculate the density of $(T_1,T_2)$ where $T_1$ is the time of the first event and $T_2$ is the time of the second event. I have been looking at the Wiki article on Poisson process and while it has been helpful, I haven't been able to figure out how to apply it to the non homogeneous case.

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If the density of $T_1$ is $\lambda e^{-\lambda t_1}I[0\le t_1]$ and the density of $T_2$ given $T_1=t_1$ is $\lambda e^{-\lambda (t_2-t_1)}I[t_1 \le t_2]$ then the density of $(T_1,T_2)$ is $\lambda^2 e^{-\lambda t_2} I[0\le t_1 \le t_2]$.

If the rate $\lambda$ varies over time then you have to adjust these, but the principle of multiplying the two together still applies.

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  • $\begingroup$ Interesting. I got that result but I assumed it was incorrect because the $t_1$ term was not present in the formula outside of the indicator function. Thank you! $\endgroup$ – DumbQuestion Nov 30 '11 at 10:20
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    $\begingroup$ +1. Simple and clear and correct. Sanity check: the integral of the density should be 1 (it is). $\endgroup$ – Did Nov 30 '11 at 10:59
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    $\begingroup$ To be complete, when the rate at time $t$ is $\lambda(t)$, the density of $(T_1,T_2)$ at $(t_1,t_2)$ is $$\lambda(t_1)\lambda(t_2)\mathrm e^{-\Lambda(t_2)}\cdot[0\leqslant t_1\leqslant t_2],$$ where, for every $t\gt0$, $\Lambda(t)$ denotes the integral of $\lambda(\ )$ on $(0,t)$. (Hence some $\lambda$'s are missing from @Henry's answer.) $\endgroup$ – Did Dec 2 '11 at 12:50

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