Question about Proof of Sum of Independent Binomials

Question

While leafing through "Introduction to Probability" (Hwang, Blitzstein), I encountered the following theorem and proof.

If $X \sim \texttt{Bin}(n, p)$, $Y \sim \texttt{Bin}(m, p)$, and $X$ is independent of $Y$, then $X + Y \sim \texttt{Bin}(n + m, p)$.

The proof in the book begins with

$$ P(X+Y=k) = \sum_{j=0}^{k} P(X+Y=k\vert X=j)P(X=j) $$

I understand the summation $\sum_{j=0}^{k}$ if $k\leq n$, but, what if $k > n$? Then wouldn't $P(X+Y=k\vert X=k)P(X=k)$ be undefined because $P(X+Y=k\vert X=k)$ is undefined? After all, $P(X=k) = 0$, since the support of $X$ is $\{0, \dots, n\}$ and $k> n$.

Answer 1

The term in the sum may be written $$ P(X+Y=k, X=j) $$ This is clearly zero if $j>n$ and $X\sim \text{Bin}(n , p)$. So I would not say it was problematic.