Hi I am doing a mock exam and have encounted following question that I don't know how to answer.
Given two independent Poisson distributions
$ Y_1 \sim Po(\lambda)$ and $Y_2 \sim Po(c\lambda)$
Let $u=Y_1+Y_2$ and $v=Y_2$, what is the conditional distribution f(v|u)?
I get the joint density function
$f(Y_1,Y_2;\lambda) = \frac{1}{Y_1! Y_2!} e^{-\lambda(1+c)} c^{Y_2} \lambda^{Y_1+Y_2}$
and know $f(v|u)=\frac{f(u,v)}{f(u)}$
where
$f(u,v) = \frac{1}{(u-v)!v!} +\lambda^u c^v e^{-\lambda(1+c)}$
But what is $f(u)$?
Best Answer
Using the definition of conditional probability (that you correctly stated) you get
$$\mathbb{P}[V|U=u]=\frac{\mathbb{P}[Y_2=v]\cdot\mathbb{P}[Y_1=u-v]}{\mathbb{P}[Y_1+Y_2=u]}$$
Now observing that $Y_1+Y_2\sim Po[(1+c)\lambda]$
$$\mathbb{P}[V|U=u]=\frac{e^{-c\lambda}(c\lambda)^v e^{-\lambda}\lambda^{u-v}}{e^{-(1+c)\lambda}[(1+c)\lambda]^u}\cdot\frac{u!}{v!(u-v)!}=\binom{u}{v}\left( \frac{c}{1+c} \right)^v\cdot\left( \frac{1}{1+c} \right)^{u-v}$$
for $v=0,1,2,\dots,u$
that is a binomial $B\left( u; \frac{c}{1+c}\right)$