For a Poisson process show, for $s < t$, that $P(N(s)=k|N(t)=n) = \binom{n}{k} (\frac{s}{t})^k (1-\frac{s}{t})^{n-k}$
How to prove this property in a Poisson process?
2 Answers
The joint probability mass function for $(N(s), N(t))$ for $s
Let
$\ \ \ \ A$ be the number of events occurring in the interval $[0,s]$,
$\ \ \ \ B$ be the number of events occurring in the interval $[s,t]$,
and
$\ \ \ \ C$ be the number of events occurring in the interval $[0,t]$.
Using the fact that in a Poisson process, events occurring in one time frame are independent of events occurring in a different time frame, we may write the sought after probability as $\tag{1} \eqalign{P ( A =k \mid C= n) &={P\bigl((A=k)\cap ( C=n) \bigr)\over P(C=n)}\cr &={P\bigl((A=k)\cap ( B=n-k) \bigr)\over P(C=n)}\cr &={P (A=k) P ( B=n-k) \over P(C=n)} .} $
Now, if the Poisson process has parameter $\lambda$, then
$\ \ \ \ A$ has Poisson distribution with parameter $s\lambda$,
$\ \ \ \ B$ has Poisson distribution with parameter $(t-s)\lambda$,
and
$\ \ \ \ C$ has Poisson distribution with parameter $t\lambda$.
So, using equation $(1)$: $\eqalign{ P( A =k \mid C= n) &={ {(s\lambda)^k e^{-s\lambda}\over k!} {( (t-s)\lambda)^{n-k} e^{-(t-s)\lambda}\over (n-k)!} \over {(t\lambda)^n e^{-t\lambda}\over n!}}\cr &={n\choose k} {s^k(t-s)^{n-k} \lambda^k \lambda^{n-k} e^{-t\lambda}\over t^n\lambda^ne^{-t\lambda}}\cr &={n\choose k} {s^k(t-s)^{n-k}\over t^kt^{n-k}}\cr &={n\choose k} {\Bigl({s\over t}\Bigr)^k \Bigl(1-{s\over t}\Bigr)^{n-k}}.\cr } $