find the condition distribution of $Y_1 \mid Y_1 + Y_2$ when $Y_1,Y_1$ are Poisson distributed

Question

Suppose that we have to independent Poisson RV:s. $Y_1 \sim Po(\lambda_1), Y_2 \sim Po(\lambda_1c) $ for some constant $c$

I want to find the distribution $Y_1\mid Y_1 +Y_2 $ : here is my attempt but I don't see how this leads to some known distribution.

First:

$P(Y_1 =y_1,Y_1+Y_2 = t) = P(Y_1 = y_1, Y_2 = t-y_1) = P(Y_1 = y_1) P(Y_2 = t-y_1) $

$P(Y_1=y_1|Y_1 + Y_2 = t) = \dfrac{P(Y_1 = y_1) P(Y_2 = t-y_1)}{\sum_{s \leq t } P(Y_1 = y_1) P(Y_2 = t-s)} = \dfrac{e^{-\lambda_1}\dfrac{\lambda_1^{y_1}}{y_1!}e^{-\lambda_1c}\dfrac{(\lambda_1c)^{t-y_1}}{(t-y_1)!}}{\sum_{s\leq t}e^{-\lambda_1}\dfrac{\lambda_1^{s}}{s!}e^{-\lambda_1c}\dfrac{(\lambda_1c)^{t-s}}{t-s!} } $

Answer 1

$Y_1+Y_2\sim\mathcal P\text{ois}((1+c)\lambda_1)$ because you have arrivals occuring independently at constant rates $\lambda_1$ and $c\lambda_1$ in two independent Poison processes; so arrivals are occuring independently at constant rate $(1+c)\lambda_1$ in a Poison process.

Therefore:

$$\sum_{s=0}^{t} \dfrac{e^{\lambda_1}\lambda_1^s}{s!}\dfrac{e^{c\lambda_1}(c\lambda_1)^{t-s}}{(t-s)!}~=~ \dfrac{e^{(1+c)\lambda_1}((1+c)\lambda_1)^t}{t!}$$

Answer 2

The situation can be viewed as one process with parameter $(1+c)\lambda_1$ in which the elements are independently randomly marked as members of $Y_1$ with probability $\frac 1{1+c}$ and as members of $Y_2$ with probability $\frac c{1+c}$. Knowing how many elements the overall process produced doesn't change this fact. Thus, if you know that $Y_1+Y_2=t$, then $Y_1$ is distributed according to $\mathsf{Binomial}\left(t,\frac1{1+c}\right)$.