Sum of independent binomials is binomial My question is related to the question 
Sum of Independent Binomials
How is it possible to sum X from 0 to w? I see two potential problems:


*

*Since $X\sim$Binomial(n,p), $X$ can maximally be $n$, and if $w=n+m$, how can then $x$ be $n+m$?

*If $w> n$, how do we calculate ${n \choose w}$?  This means that $\sum_{x=0}^w{m\choose x}{n\choose w-x}$ does not make sense when $w>n$? There is also a problem if $w>m$, since then ${m\choose x}$ does not make sense when $x=w$.
Based on 1 and 2, it seems like the proof given in   Sum of Independent Binomials
is only correct when $w\leq \min(n,m)$. Is this correct?
 A: First, note that for two independent binomial variables $X \sim B(n,p)$ and $Y \sim B(m,p)$ and $W=X+Y$, we have
$$P(W = w) = \mathop{ \sum \sum}_{y+x = w} p (y) p(x)$$
and the convolution formula you see on the other post is the next step from here, i.e. put $ y = w - x$ and sum over all values of $x$. That formula might be a little confusing if you forget where it comes from so it's worth keeping the above double sum in mind.
Your question now is about the values of these variables that we sum over. While it is true that $X$ cannot be larger than $n$ and similarly $Y$ cannot be larger than $m$, you are correct here, this does not mean that $w$ needs to be smaller than both $n$ and $m$. The random variable $W$ is the sum of two nonnegative random variables and as such, it can assume all integer values between $0$ and $n+m$. 
For each of these integer values we count the combinations of $X$ and $Y$ that yield this value and add their probabilities but we always respect their boundaries while doing so. To give you an example, suppose we are looking for the probability of $P\left( W= n+m \right)$. The only way this can happen is if $X$ and $Y$ assume their maximal values. It makes no sense to compute things like $P(X = n+ 1)$ etc. No, this probability must come from
$$P\left( W= n+m \right) = P\left( X = n \right) P\left( Y = m \right) $$
and only from here.
A: Let $X \sim \text{Bin}(m,p)$ and $Y \sim \text{Bin}(n,p)$. You want to find the distribution of $W = X + Y$.
Let's ask what that even means. For example, in what circumstances does $W = 2$? It is when $X + Y = 2$, e.g. $X = 1, Y = 1$.
Now is it clear what your confusion is?
