# Independent poisson processes conditional

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 ?

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.

$$P(N_1(t) = n_1 | G(t) = k) = P(N_2(t) = k - n_1 \color{red}{|G(t)=k})$$
• Ok perfect, but I can not understand why can we write that $P(G(t)=k|N_1(t)=n_1) = P(N_2(t)=k-n_1)$ and not $P(N_1(t) = n_1 | G(t) = k) = P(N_2(t) = k - n_1)$ – Gabriele Picco Dec 17 '17 at 17:57
• $P(G(t)=k|N_1 = n_1) = P(N_2(t)=k-n, N_1(t)=k|N_1(t)=k)=P(N_2(t)=k-n,|N_1(t)=k) = P(N_2(t)=k-n)$ because $N_2$ and $N_1$ are independent. For the other case $P(N_1(t)=n_1|G(t)=k) = P(N_2(t)=k-n_1|G(t)=k)$ and we can't drop the condition because $G(t)$ and $N_2(t)$ are not independent. – Siong Thye Goh Dec 17 '17 at 18:02