Proving the general formula for conditional expectation of a Poisson Process

Question

I am studying a course on Stochastic Processes and encountered the following proof exercise on Poisson Processes:

If $N$ is a Poisson Process with intensity $\lambda$, then for $0<s<t$ where $k \leq n$ are non-negative integers, then prove the following: $$ P(N(s) = k | N(t) = n) = \binom{n}{k} \Big{(} \frac{s}k \Big{)}^k \Big{(} 1 - \frac{s}t \Big{)}^{n-k}$$

I know that $N(t)$ (and analogously, $N(s)$) will follow the Poisson distribution with parameter $\lambda t $. This allows me to use Bayes’ Formula for the conditional expectation to see that:

$$ P(N(s) = k | N(t) = n) = \frac{P(N(s) = k \text{ and } N(t) = n)}{ P(N(t) = n)}$$ $$ = \frac{n!}{e^{-\lambda t}(\lambda t)^n} P(N(s) = k \space \text{and} \space N(t) = n) $$ $$ = \frac{n!}{e^{-\lambda t}(\lambda t)^n} \Big{(}P(N(t) = n | N(s) = k)\frac{e^{-\lambda s}(\lambda s)^k}{k!} \Big{)}$$

Then simplifying the above expression yields the following:

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

Assuming this is right, this doesn't seem to be particularly close to the desired form. Is there another approach or perhaps a way to complete my attempt so far that I am not seeing? One thing worth noting is that I haven't really used any of the properties of a Poisson Process yet, really I've just treated $N(s)$ and $N(t)$ as two Poisson distributions - which may be my mistake.

I would be grateful for any assistance with this exercise.

Answer 1

Continuing where I left off, using the hint given in the comments, I left off at the following point:

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

Note that the following result holds: $$P(N(t)=n|N(s)=k)=P(N(t−s)=n−k)$$ due to the fact that Poisson Processes satisfy the independence property that $N_t - N_s$ is independent of $\mathscr{F}_s$ for any $0<s<t< \infty$.

Therefore the above simplifies as follows:

$$P(N(s)=k|N(t)=n) = \frac{n! \space e^{-\lambda (s - t)}s^k}{k! \lambda ^{n-k}t^n}P(N(t−s)=n−k)$$ $$ = \frac{n! \space e^{-\lambda (s - t)}s^k}{k! \lambda ^{n-k}t^n} \times \frac{e^{- \lambda (t-s)}(\lambda (t-s))^{n-k}}{(n-k)!}$$ $$ = \frac{n! }{k!(n-k)! } \times \frac{s^k((t-s))^{n-k}}{t^n}$$ $$ = \frac{n! }{k!(n-k)! } \times \Big{(}\frac{s}{t}\Big{)}^k \times \Big{(} \frac{t-s}{t}\Big{)}^{n-k}$$ $$ = \binom{n}{k} \times \Big{(}\frac{s}{t}\Big{)}^k \times \Big{(} 1 - \frac{s}{t}\Big{)}^{n-k} \quad \quad \square$$