Let $X$ be a Poisson random variable with parameter $λ$ and $Y$ be a Poisson random variable with parameter $μ$.
$X, Y$ are assumed to be independent.
Prove that $X|X + Y$ is a Binomial random variable with the success probability $\frac{\lambda}{\lambda + \mu}$.
A Poisson variable with parameter $\lambda$ is the number of events occuring in one time unit in a Poisson process with rate $\lambda$. Whenever the $X+Y$ events occurred, since with probability $1$ they occurred at different times, we can form intervals around them containing only one event each. The a priori probability for an interval of length $\mathrm dt$ to contain an event of type $X$ was $\lambda\mathrm dt$, and likewise $\mu\mathrm dt$ for type $Y$. Thus, the posterior probability, given that an event occurred in the interval, that it was of type $X$ is
$$ \frac{\lambda\mathrm dt}{\lambda\mathrm dt+{\mu\mathrm dt}}=\frac\lambda{\lambda+\mu}\;. $$
Since the probabilities for all intervals in a Poisson process are independent, and $X$ and $Y$ are assumed to be independent, each event is an independent Bernoulli experiment with success probability $\frac\lambda{\lambda+\mu}$, so adding them yields a binomial variable with that success probability.
By the definition of conditional probability, we have for each positive integer $n$ and nonnegative integer $k\leqslant n$: $$ \mathbb P(X=k\mid X+Y=n) = \frac{\mathbb P(X=k,X+Y=n)}{\mathbb P(X+Y=n)} = \frac{\mathbb P(X=k)\mathbb P(Y=n-k)}{\mathbb P(X+Y=n)}. $$ Now, \begin{align} \mathbb P(X+Y=n) &= \sum_{k=0}^n \mathbb P(X+Y=n\mid X=k)\mathbb P(X=k)\\ &= \sum_{k=0}^n \mathbb P(X=k)\mathbb P(Y=n-k)\\ &= \sum_{k=0}^n e^{-\lambda}\frac{\lambda ^k}{k!} e^{-\mu}\frac{\mu^{n-k}}{(n-k)!}\\ &= \frac{e^{-(\lambda+\mu)}}{n!}\sum_{k=0}^n \frac{n!}{k!(n-k)!}\lambda^k \mu^{n-k}\\ &=\frac{e^{-(\lambda+\mu)}}{n!}\sum_{k=0}^n\binom nk\lambda^k \mu^{n-k}\\ &=\frac{e^{-(\lambda+\mu)}}{n!}(\lambda+\mu)^n, \end{align} so that $X+Y$ has Poisson distribution with mean $\lambda+\mu$. It follows that \begin{align} \frac{\mathbb P(X=k)\mathbb P(Y=n-k)}{\mathbb P(X+Y=n)} &= \frac{e^{-\lambda}\frac{\lambda^k}{k!}e^{-\mu}\frac{\mu^{n-k}}{(n-k)!}}{e^{-(\lambda+\mu)}\frac{(\lambda+\mu)^n}{n!}}\\ &= \frac{n!}{k!(n-k)!}\frac{\lambda^k \mu^{n-k}}{(\lambda+\mu)^k(\lambda+\mu)^{n-k}}\\ &= \binom nk\left(\frac\lambda{\lambda+\mu}\right)^k\left(\frac\mu{\lambda+\mu}\right)^{n-k}, \end{align} and hence conditioned on $\{X+Y=n\}$, $X$ has $\mathrm{Bin}\left(n,\frac\lambda{\lambda+\mu}\right)$ distribution.
