Independent poisson processes conditional

Question

Consider two Poisson processes, independent $N_1 (t)$ and $N_2(t)$ of parameters $\mu_1$ and $\mu_2$ and $G(t) = N_1 (t) + N_2(t)$.

I have to find the conditional distribution of $N_1 (t)$ given $G(t)$ = k, with $k = 0, 1 , \ldots$

$P(N_1(t) = n_1 | G(t) = k) = P(N_2(t) = k - n_1) = e^{-\mu_2 \cdot t} \cdot \frac{(\mu_2t)^{k-n_1}}{(k-n_1)!}$

and

$F(a) = \sum_{i=0}^{a} P(N_1(t) = i | G(t) = k)$

it's correct ?

Answer 1

Guide:

Use Bayes rule, \begin{align} P(N_1(t)=n_1 |G(t)=k) &= \frac{P(N_1(t)=n_1 )}{P(G(t)=k)} \cdot P(G(t)=k|N_1(t)=n_1) \\ &= \frac{P(N_1(t)=n_1 )}{P(G(t)=k)} \cdot P(N_2(t)=k-n_1) \end{align}

Hopefully, you can evaluate the expression above.

Remark: Correction to your mistake.

$$P(N_1(t) = n_1 | G(t) = k) = P(N_2(t) = k - n_1 \color{red}{|G(t)=k})$$