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Let $N_t, t\geq0$, be a Poisson process with intensity $\lambda$. Let $0<t_1<t_2$ be times and $0\leq n_1\leq n_2$ be nonnegative integers. Find the conditional probability $$P(N_{t_1}=n_1 \lvert N_{t_2}=n_2)$$ Identify this as a standard distribution, and find the parameters of this distribution.

My working:

$P(N_{t_1}=n_1 \lvert N_{t_2}=n_2)=\frac{P(N_{t_1}=n_1 \cap N_{t_2}=n_2)}{P(N_{t_2}=n_2)}$
$=\frac{P(N_{t_1}=n_1) \cap P(N_{t_2}=n_2)}{P(N_{t_2}=n_2)}$
$=\frac{e^{-\lambda}\frac{\lambda^{n_1}}{n_1!}e^{-\lambda}\frac{\lambda^{n_2}}{n_2!}}{e^{-\lambda}\frac{\lambda^{n_2}}{n_2!}}$
$=P(N_{t_1}=n_1)$

where the event $N_{t_1}=n_1$ and $N_{t_2}=n_2$ are independent. And the parameter is still $\lambda$?

But I doubt whether my answer is correct. There are some few points that I don't quite understand, since $t_1<t_2$, technically $N_{t_1}=n_1$ happened first, and if given $N_{t_2}=n_2$, then automatically $N_{t_1}=n_1$ must have already happened, but how can $P(N_{t_1}=n_1 \lvert N_{t_2}=n_2)=P(N_{t_1}=n_1)$? So the occurrence of $N_{t_2}=n_2$ does not affect at all?

I understand that conditional probability does not mean which event happened first, it just our knowledge of a given event has happened. But somehow the result $P(N_{t_1}=n_1 \lvert N_{t_2}=n_2)=P(N_{t_1}=n_1)$ seems a bit weird for me. But I'm not sure what went wrong in my working.

Would anyone please kindly give some help and show what went wrong and how to correct it?

EDITED

Thanks to @joriki. So we have $P(N_{t_2-t_1}=n_2-n_1)=e^{-\alpha}\frac{\alpha^{n_2-n_1}}{(n_2-n_1)!}=e^{-\alpha}\frac{\alpha^{n_2}}{\alpha^{n_1}}\times\frac{1}{(n_2-n_1)!}$.

Then I am not sure how to proceed to get $P(N_{t_1}=n_1 \lvert N_{t_2}=n_2)$?

Thanks!

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  • $\begingroup$ For Poisson process, you have independent increment - so the increments $N_{t_1} - N_{0}$ and $N_{t_2} - N_{t_1}$ are independent. $\endgroup$ – BGM Apr 10 '16 at 15:04
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What went wrong is that you assumed that the events $N_{t_1}=n_1$ and $N_{t_2}=n_2$ are independent. This already implies $P(N_{t_1}=n_1 \mid N_{t_2}=n_2)=P(N_{t_1}=n_1)$, independent of the distribution. They are in fact dependent. The clue to the problem is the memorylessness of the Poisson process. The fact that $N_{t_1}=n_1$ does not affect the probability that there are $n_2-n_1$ points in the remaining time $t_2-t_1$. Thus $P(N_{t_2}=n_2 \mid N_{t_1}=n_1)=P(N_{t_2-t_1}=n_2-n_1)$, and from this you can obtain $P(N_{t_1}=n_1 \mid N_{t_2}=n_2)$.

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  • $\begingroup$ Thanks @joriki. If $N_{t_1}=n_1$ and $N_{t_2}=n_2$ are not independent then how can we write the formula for $P(N_{t_1}=n_1|N_{t_2}=n_2)$? since we cannot multiply each pdf. $\endgroup$ – user71346 Apr 10 '16 at 14:00
  • $\begingroup$ @user71346: I wrote $P(N_{t_2}=n_2 \mid N_{t_1}=n_1)$. From the work you showed, it seems that you know how to get $P(N_{t_2}=n_2 \cap N_{t_1}=n_1)$ from this, and how to get $P(N_{t_1}=n_1 \mid N_{t_2}=n_2)$ from that. $\endgroup$ – joriki Apr 10 '16 at 14:47

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