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The arrival times of the first and second event are $S_1$ and $S_2$, and the number of arrivals follow a poisson process. How would I compute the Joint PDF of $S_1$ and $S_2$?

I have found the PDF of $S_1$ and $S_2$: $f_1(s)=\lambda e^{-\lambda s}$ and $f_2(s)=\lambda^2 se^{-\lambda s}$.

I also know the variables $S_1$ and $S_2-S_1$ are independent.

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  • $\begingroup$ Sorry, I differentiated the CDF incorrectly, but I will correct the original post. $\endgroup$ – Timothy Hedgeworth Apr 23 '15 at 15:27
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I also know the variables $S_1$ and $S_2-S_1$ are independent.

Yes, and thus: $$\begin{align} f_{S_1,S_2}(s_1, s_2) & = f_{S_1, S_2-S_1}(s_1, s_2-s_1) \\[1ex] & = f_{S_1}(s_1)\cdot f_{S_2-S_1}(s_2-s_1) \end{align}$$

Now, do you know what $\;f_{S_2-S_1}(t)\;$ is?

Hint: A Poisson process is memoriless.

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  • $\begingroup$ I am pretty sure: $f_{S_2-S_1}(t) = \lambda e^{-\lambda t}$. So would $f_{S_1,S_2}(s_1, s_2)=f_{S_1, S_2-S_1}(s_1, s_2-s_1)=\lambda^2 e^{-\lambda s_1} e^{-\lambda (s_2-s_1)}=\lambda^2 e^{-\lambda s_2}$? $\endgroup$ – Timothy Hedgeworth Apr 23 '15 at 15:51
  • $\begingroup$ @TimothyHedgeworth Yes, indeed -- for $0 < s_1 < s_2$ ; it's $0$ elsewhere. $\endgroup$ – Graham Kemp Apr 23 '15 at 16:18

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