Poisson Process proof that

Question

For a Poisson process show, for $s<t$ that $$P(N(s)=k\mid N(t)=n)={n\choose k}\left(\frac{s}{t}\right)^k\left(1-\frac{s}{t}\right)^{n-k},\space > k=0,1,\dots,n$$

I tried a few things but I could not prove

$$P(N(s)=k\mid N(t)=n)=\frac{P(N(s)=k\cap N(t)=n)}{P(N(t)=n)}$$ since $s<t$ I did $N(t)=N(t-s)+N(s)$ $$P(N(s)=k\mid N(t)=n)=\frac{P(N(s)=k\cap N(t-s)+N(s)=n)}{P(N(t)=n)}$$ $$=\frac{P(N(s)=k\cap N(s)=n-N(t-s))}{P(N(t)=n)}=\frac{P(N(t-s)=n-k)}{P(N(t)=n)}$$

I know that $N(t-s)>0\sim \mathrm{Poisson}(\lambda(t-s))$ and $N(t)\sim \mathrm{Poisson}(\lambda t)$

$$P(N(s)=k\mid N(t)=n)=\frac{e^{-\lambda(t-s)}(\lambda(t-s))^{n-k}}{(n-k)!}\frac{n!}{e^{-\lambda t}(\lambda t)^n}$$

that's what I tried

Answer 1

$N(t)$ is not equal to $N(t-s)+N(s)$, but $$ N(t) = \Big(N(t) - N(s)\Big) + \Big(N(s)\Big) $$ and the two expressions inside the $\Big(\text{big parentheses}\Big)$ are independent of each other (whereas $N(t-s)$ and $N(s)$ are not independent of each other). So \begin{align} \Pr(N(s) = k \mid N(t) = n) & = \frac{\Pr(N(s)=k\ \&\ N(t) =n) }{\Pr(N(t)=n)} \\[10pt] & = \frac{\Pr(N(s)=k\ \&\ N(t)-N(s) = n-k)}{\Pr(N(t)=n)} \\[10pt] & = \frac{\Pr(N(s)=k)\Pr(N(t)-N(s) = n-k)}{\Pr(N(t)=n)}. \end{align} The last line is where independence is used. If you find these three probabilities then routine simplifications will do the rest.

Notice that the answer is the same regardless of the value of $\lambda$. The lack of dependence on $\lambda$ means the conditional distribution of arrival times before $t$, given the value of $N(t)$ does not depend on $\lambda$. Since their distribution does not depend on $\lambda$, those arrival times are irrelevant to the problem of estimating $\lambda$ given then data, once $N(t)$ has been taken into account. In other words, all information about arrival times relevant to estimation of $\lambda$ is contained in $N(t)$, provided you can be certain that it's a Poisson process with a constant rate. But that arrival time data might contain information that could reasonably make you suspect that it does not follow such a model.

Answer 2

By Bayes,$$P(N(s)=k\mid N(t)=n)=P(N(t)=n\mid N(s)=k)\frac{P(N(s)=k)}{P(N(t)=n)}$$Your identity gives:$$=P(N(t-s)=n-k)\frac{P(N(s)=k)}{P(N(t)=n)}$$Sub in Poisson distributions, rest is algebra:$$=e^{-\lambda(t-s)}\frac{(\lambda(t-s))^{n-k}}{(n-k)!}\frac{e^{-\lambda s}(\lambda s)^k n!}{e^{-\lambda t}(\lambda t)^n k!}\\={n \choose k}\frac{(t-s)^{n-k}s^k}{t^n}\\={n \choose k}\frac{s^k}{t^k}\frac{(t-s)^{n-k}}{t^{n-k}}$$