Given that $n$ events have occurred in a Poisson process $N(t)$ with rate $\lambda$ until the time $a$ . Find the probability distribution of $N(t)$ until the time $b$ where $0<b<a$
My attempt : $$P(N(a)=n)=\frac{e^{-\lambda a}(a\lambda)^n}{n!}$$ We have stationary increment. $$P(N(b)=k|N(a)=n) = \frac{P(N(b)=k,N(a)=n)}{P(N(a)=n)}$$ $$=\frac{P(N(a)-N(b)\leq n-k)}{P(N(a)=n)}$$
Is this correct and how to proceed further ?
Thanks
You are on the right track, but the last equality has some problem, here is the correct way to handle $P[N(b) = k, N(a) = n]$: \begin{align} & P[N(b) = k, N(a) = n] \\ = & P[N(b) = k, N(a) - N(b) = n - k] \\ = & P[N(b) = k] P[N(a) - N(b) = n - k] \quad \text{by increment independence} \\ = & P[N(b) = k] P[N(a -b) = n - k] \quad \text{by stationarity} \\ = & e^{-\lambda b}\frac{(\lambda b)^k}{k!} \times e^{-\lambda (a - b)}\frac{(\lambda (a - b))^{n - k}}{(n - k)!} \end{align} Simplification gives the final result (binomial distribution)...