If $X $~ $Bin(n, p)$ and $Y$ ~ $Bin(m, p)$ I want to show that $X-Y$ is not binomial.

I tried to use the technique of this proof given in lecture that Sum of two independent binomial variables is binomial with $X+Y$ ~ $Bin(n+m, p)$ analogous for the situation $X-Y$.

I suspect that $X-Y$ should be ~ $Bin(n-m, p)$, and therefore not binomial for the case that $m$ > $n$ as that would a negative amount of independent Bernoulli trials with a positive probability p; by definition of the Binomial distribution this makes no sense. Probability of "success" in zero or less Bernoulli trials should be zero, not positive

This would seem to agree with the solution given by the book:

"*A Binomial can’t be negative, but X −Y is negative with positive probability.*" But is $X-Y$ a negative, or is it only negative for $m > n$?

Is this the right train of thought?

If so, is there a mathematical way to express $X-Y$ ~ $Bin(n-m, p)$?