Sum of Poisson variables through factorizing

Question

This might be similar to my previous question, however I don't see how to use the same logic here. I've recently came across the fact that the sum of $X\sim \text{Po}(\gamma)$ and $Y\sim \text{Po}(\delta)$ (if they are independent) has the distribution $\text{Po}(\gamma + \delta)$ and also that $$P(X= k | X+Y = n) = \text{Bi}( n, \gamma/(\gamma+\delta); k) \quad n\in \mathbb N_0, k=0, \dots, n$$ which was proven by showing that $$P(X=k, X+Y = n) = \text{Po}(\gamma + \delta; n) \text{Bi}(n, \gamma/ (\gamma + \delta); k)$$ i.e. "splitting" the joint probability into $P(X+Y=n)$ and $P(X=k|X+Y=n)$.

My main question is: How did we prove statement? That is how do we know that what we really found is the factorization into $P(X+Y=n)P(X=k|X+Y=n)$ and haven't just "accidentally" stumbled across the expression for a Poisson and Binomial probability in the expression.

edit: Possibly a (much) better way of asking is: we know that $$P(X+Y=n)P(X=k|X+Y=n)=\text{Po}(\gamma + \delta; n) \text{Bi}(n, \gamma/ (\gamma + \delta); k)$$ But how do we know that $$P(X+Y=n) = \text{Po}(\gamma + \delta; n)$$ $$P(X=k|X+Y=n)= \text{Bi}(n, \gamma/ (\gamma + \delta); k)$$

Answer 1

This answer is based on BGM's comment.

We start with this equality $$P(X+Y=n)P(X=k|X+Y=n)=\text{Po}(\gamma + \delta; n) \text{Bi}(n, \gamma/ (\gamma + \delta); k)$$

Since one of the expressions on each side does not depend on $k$, we can sum both sides w.r.t. $k$ to obtain that $$P(X+Y=n) \sum^{n}_{k=0} P(X=k|X+Y = n) = \text{Po}(\gamma + \delta; n) \sum^n_{k=0} \text{Bi}(n, \gamma/(\gamma + \delta); k)$$ Both sums equal one since we're summing over the whole distribution, yielding $$P(X+Y=n) = \text{Po}(\gamma + \delta; n)$$ Dividing both sides of the original expression by $\text{Po}(\gamma + \delta; n)$ yields $$P(X = k | X+Y = n) = \text{Bi}(n, \gamma/ (\gamma + \delta); k)$$

Answer 2

To calculate $P(X+Y=n)$ we use the convolution formula:

$P(X+Y=n)=\sum_{x=0}^n f_X(x)\cdot f_Y(n-x)$

$$P(X+Y=n)=\sum_{x=0}^n e^{-\gamma} \cdot \frac{\gamma^x}{x!}\cdot e^{-\delta} \cdot \frac{\delta^{n-x}}{(n-x)!} $$

$$P(X+Y=n)=\sum_{x=0}^n e^{-\gamma-\delta} \cdot \gamma^x \cdot \delta^{n-x} \cdot \frac{1}{x!(n-x)!} $$

$e^{-\gamma-\delta}$ can be factored out.

If $\frac{1}{x!(n-x)!}$ is multiplied by $n!$ then it becomes the binomial coefficient $n \choose x$

$$n!\cdot P(X+Y=n)=e^{-\gamma-\delta}\cdot \sum_{x=0}^n {n \choose x}\cdot \gamma^x \cdot \delta^{n-x} $$

Applying the binomial theorem the sum becomes $(\gamma + \delta)^n$

Thus $n!\cdot P(X+Y=n)=e^{-\gamma-\delta}\cdot (\gamma + \delta)^n$

$ P(X+Y=n)=e^{-\gamma-\delta}\cdot \frac{(\gamma + \delta)^n}{n!}$

$P(X+Y=n)=Poi(\gamma+\delta, n)$

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My main question is: How did we prove statement? That is how do we know that what we really found is the factorization into $P(X+Y=n)P(X=k|X+Y=n)$

Let $A$ be the event that $X+Y=n$. And $B$ the event that $X=k$

You want to split $P(A\cap B)$. For this purpose you use the multiplication rule for dependent events.

$P(A\cap B)=P(A)\cdot P(B|A)$