1. ## Poisson Proofs

Let $(N_t)$ be a Poisson process with rate $\lambda$ and $N_0 = 0$.

Prove the following:

(a) for $0 \leq u \leq t$ and $\ j = 0,1,2,...n, Pr(N_u = j | Nt_t = n) = {n \choose j} \left(\frac{u}{t}\right)^2 \left(1 - \frac{u}{t}\right)^{n-j}$
i.e. the conditional distribution of $N_u$ given $N_t = n$ is Bin $(n,\frac{u}{t})$

(b) for $s,t \geq 0, \mathbb{E}[N_t \ \ N_{t+s}] = \lambda t(1+\lambda(t+s))$

2. Originally Posted by BrooketheChook
Let $(N_t)$ be a Poisson process with rate $\lambda$ and $N_0 = 0$.

Prove the following:

(a) for $0 \leq u \leq t$ and $\ j = 0,1,2,...n, Pr(N_u = j | Nt_t = n) = {n \choose j} \left(\frac{u}{t}\right)^2 \left(1 - \frac{u}{t}\right)^{n-j}$
i.e. the conditional distribution of $N_u$ given $N_t = n$ is Bin $(n,\frac{u}{t})$

(b) for $s,t \geq 0, \mathbb{E}[N_t \ \ N_{t+s}] = \lambda t(1+\lambda(t+s))$
What did you try for (a)? If you write the definition of conditional probability, this should go easily.

For (b), a hint (yes it is) is: $N_{t+s}=N_t+(N_{t+s}-N_t)$.

3. For (a) I get

$Pr(N_u = j | N_t = n) = \frac{(\lambda u)^j e^{-\lambda u}(\lambda (t-u))^{n-j}e^{-\lambda(t-u)}n!}{j!(n-j)!(\lambda t)^n e^{-\lambda t}}$

Maybe its because im really tired, but I cant see how this leads to

${n \choose j} \left(\frac{u}{t}\right)^2 \left(1 - \frac{u}{t}\right)^{n-j}$

4. Originally Posted by BrooketheChook
For (a) I get

$Pr(N_u = j | N_t = n) = \frac{(\lambda u)^j e^{-\lambda u}(\lambda (t-u))^{n-j}e^{-\lambda(t-u)}n!}{j!(n-j)!(\lambda t)^n e^{-\lambda t}}$

Maybe its because im really tired, but I cant see how this leads to

${n \choose j} \left(\frac{u}{t}\right)^{\color{red}j} \left(1 - \frac{u}{t}\right)^{n-j}$
Many a simplification happens here. The exponentials simplify, and so do the terms with lambda. It remains $\frac{u^j (t-u)^{n-j}n!}{t^n j!(n-j)!}=\frac{u^j t^{n-j}(1-u/t)^{n-j}}{t^n} {n\choose j}$ $=\left(\frac{u}{t}\right)^j \left(1-\frac ut\right)^{n-j}{n \choose j}$.