Poisson: $P(N(4)-N(2)=5|N(4)=8)$

Question

Let {N(t), t ≥ 0} be a Poisson process of rate 2. Determine:

a) $P(N(4)-N(2)=5|N(4)=8)$

Attempt: The conditional poisson distribution is uniformly distributed between the interval [0,4]. Therefore, this would be binomial distributed with probaiblity $p=1/2$

$P(N(2)=3|N(4)=8)=$ $8\choose3$$(1/2)^8$

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b) $E(N(4)-N(2)|N(3)=1)$

Attempt: This part I really not sure. I break it into two cases. The first case is when the even $N(3)=1$ happens between [0,2], this has probaiblty 2/3.
If it happens between [2,3], it has probaiblity 1/3.

Therefore, $E(N(4)-N(2)|N(3)=1)=(1/3)(\lambda +1)+(2/3)(2\lambda)$

Answer 1

1. Yes.   We could also use the definition of Conditional Probability, and the fact that the count of Poisson events occurring in disjoint intervals are independent.

$$\begin{align} \mathsf P(N(4)-N(2)=3\mid N(4)=8) ~=~& \dfrac{\mathsf P(N(2)=5)~\mathsf P(N(4)-N(2)=3)}{\mathsf P(N(4)=8)} \\ =~& \dfrac{((2\lambda)^5\mathsf e^{-2\lambda}/5!)~((2\lambda)^3\mathsf e^{-2\lambda}/3!)}{(4\lambda)^8\mathsf e^{-4\lambda}/8!} \\ =~& \binom{8}{3}\frac 1{2^8} \end{align}$$

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2. No.   For this you need to use Linearity of Expectation:

$$\begin{align}\mathsf E(N(4)-N(2)\mid N(3)=1) ~=~ & \mathsf E(N(4)\mid N(3)=1)-\mathsf E(N(2)\mid N(3)=1)\end{align}$$

So, when you are given that one event occurs in the first three unit-times:

• How many do you expect will occur in the next unit-time? $~\mathsf E(N(4)\mid N(3)=1)$
• How many do you expect occurred in the first two unit-times? $~\mathsf E(N(2)\mid N(3)=1)$