How the number of arrivals in the future effects the number of arrivals in the past in a Poisson process?

Question

Let $\{N(t), t \geq 0\}$ be a Poisson process with rate $\lambda$, for $s < t$ find

(a) $E[N(t)|N(s) = 4]$

(b) $E[N(s)|N(t) = 4]$

I solved (a) as $E[N(t)|N(s) = 4] = 4 + E[N(t - s)] = 4 + \lambda(t- s)$ but I do not know how can I solve (b).

P.S. Answer is $4s/t$ according to my notes...

Answer 1

For (a), by independence of increments we have \begin{align} \mathbb E[N(t)\mid N(s) = 4] &= \mathbb E[N(t)-N(s) + N(s)\mid N(s) = 4]\\ &= \mathbb E[N(t)-N(s)\mid N(s)=4] + \mathbb E[N(s)\mid N(s) = 4]\\ &= \mathbb E[N(t)-N(s)] + 4\\ &= \lambda(t-s) + 4. \end{align}

For (b), we "reverse" the conditioning as follows: \begin{align} \mathbb E[N(s)\mid N(t)=4] &= \sum_{j=0}^4 j\cdot\mathbb P(N(s)=j\mid N(t)=4)\\ &= \sum_{j=0}^4 \frac{\mathbb P(N(s)=j,N(t)=4)}{\mathbb P(N(t)=4)}\\ &=\sum_{j=0}^4 \frac{\mathbb P(N(t)=4\mid N(s)=j)\mathbb P(N(s)=j)}{\mathbb P(N(t)=4)}\\\\ &=\sum_{j=0}^4 \frac{\mathbb P(N(t-s)=4-j)\mathbb P(N(s)=j)}{\mathbb P(N(t)=4}\\ &=\sum_{j=0}^4 \frac{e^{-\lambda(t-s)}(\lambda(t-s))^{4-j}/(4-j)!e^{-\lambda s}(\lambda s)^j/j!}{e^{-\lambda t}(\lambda t)^4/4!}\\ &= \frac{4s}t. \end{align}