Conditional Probability in Poisson Process

Question

Suppose that $N(t), t\ge 0$ is a Poisson process such the $E[N(9)]=6$.

(i) Find the mean and variance of $N(8)$.

(ii) Find $P(N(2)\le3)$

(iii) Find $P(N(4)\le5|N(2)\le3)$ - How do I solve this?

Attempt

(i) $\lambda t$ = 6, implies $\lambda =\frac{2}{3}$.

Mean = $E[N(8)] = \lambda 8 = \frac{2}{3}8 = 5.33$ Variance = Mean = $5.33$

(ii) $P[N(2) \le3]$ = sum of poisson probability with $\lambda t = \frac{2}{3}2 = \frac{4}{3}$ for $x = \{0,1,2,3\}.$ Answer = $0.9535$

(iii) NO attempt at this, please how can I solve the conditional probability.

Thanks

Answer 1

(i) and (ii) are correctly reasoned and the calculations are okay.

(iii) use conditional probability, and the independence of Poisson counts in disjoint intervals. That is: $(N_4{-}N_2)$ is i.i.d. to $N_2$ .

$$\begin{align} \mathsf P(N_4\leq 5\mid N_2\leq 3)~=~&\dfrac{\mathsf P(N_4\leq 5, N_2\leq 3)}{\mathsf P(N_2\leq 3)} \\[1ex] =~& \dfrac{\sum_{n=0}^3\mathsf P(N_2=n)~\mathsf P((N_4{-}N_2)\leq 5-n)}{\mathsf P(N_2\leq 3)} & \text{independence} \\[1ex] =~& \dfrac{\sum_{n=0}^3\sum_{m=0}^{5-n}\mathsf P(N_2=n)~\mathsf P(N_2=m)}{\mathsf P(N_2\leq 3)} & \text{identical distribution} \end{align}$$

$$\begin{align}\mathsf P(N_4\leq 5\mid N_2\leq 3)~=~& \dfrac{\mathsf e^{-4/3}\sum_{n=0}^3\sum_{m=0}^{5-m} (4/3)^{n+m}/(n!m!)}{\sum_{n=0}^3 (4/3)^n/n!} \\[1ex] =~& \dfrac{48373}{13185~\mathsf e^{4/3}}\\[1ex] \approx~& 0.9670{\small 8}\end{align}$$