Finding $P\{N(s) = k\,|\,N(t) = n\}$

Question

For a Poisson process with rate $λ$, find $P\{N(s) = k\,|\,N(t) = n\}$ when $s < t$.

My work:

$N(s)\cap N(t)=k$ since $s<t$, thus $$P\{N(s) = k\,|\,N(t) = n\}=\frac{P\{N(s) = k, N(t) = n\}}{P\{N(t) = n\}}=\frac{P\{N(s) = k\}}{P\{N(t) = n\}}=\frac{e^{-\lambda}\lambda^k/k!}{e^{-\lambda}\lambda^n/n!}=\frac{n!}{k!}\lambda^{k-n}$$ Is this correct?

Answer 1

Note that $$P(N(s)=k, N(t)=n) \ne P(N(s)=k)$$

For the numerator, we have

\begin{align}P(N(s)=k, N(t)=n)& =P(N(s)=k, N(t)-N(s)=n-k) \\& = P(N(s)=k)P( N(t)-N(s)=n-k)\\ &= P(N(s)=k)P( N(t-s)=n-k)\\\end{align}