Finding the conditional probability $p(X-Y|X+Y)$ for two binomial distributions

Question

Suppose that $X \sim \text{binomial}(n, p)$ and $Y \sim \text{binomial}(m, p)$ are independent random variables. How can I find the conditional distribution $p(X-Y|X+Y)$?

I have tried the following:

$$p(X-Y|X+Y) = \frac{p(X-Y,X+Y)}{p(X+Y)}$$

I know that the sum of two binomial random variables with similar success probability is a binomial, so $X + Y \sim \text{binomial}(n+m, p)$. However, I'm having difficulty finding $p(X-Y, X+Y)$.

I also can find the distribution $p(X, X+Y)$. We can write

$$p(X, X+Y) = p(X=x) p(X+Y=k|X=x) = p(X=x) p(Y=k-x) = {n \choose x}p^x (1-p)^{n-x} {m \choose k-x}p^{k-x} (1-p)^{m-k+x} = {n \choose x} {m \choose k-x}p^{k} (1-p)^{m+n-k}$$

However, I don't think I can use it to compute $p(X-Y, X+Y)$. I've also tried writing $p(X-Y, X+Y)$ by conditioning over $Y$:

$$p(X-Y, X+Y) = \sum_{y} p(X-Y, X+Y|Y = y)p(Y=y)$$

But I don't know how to proceed from here.

Answer 1

You can use $$\mathbb P(X-Y=d, X+Y=s) = \mathbb P\left(X=\frac{s+d}{2}, Y=\frac{s-d}{2}\right)$$

to give $$\mathbb P(X-Y=d \mid X+Y=s) \\= \frac{{n \choose (s+d)/2}p^{(s+d)/2}(1-p)^{n-(s+d)/2 }{m \choose (s-d)/2}p^{(s-d)/2}(1-p)^{m-(s+d)/2 } }{{n+m \choose s}p^{s}(1-p)^{n+m-s } }$$ where all the $p$ and $(1-p)$ terms cancel if neither are $0$ to give $$= \frac{{n \choose (s+d)/2}{m \choose (s-d)/2} }{{n+m \choose s} }$$

assuming $s$ and $d$ have the same parity and $0 \le s \le n+m$ and $-m \le d \le n$

Answer 2

Guide:

\begin{align} P(X-Y = h|X+Y = k) &=\frac{P(X-Y = h,X+Y = k)}{P(X+Y=k)} \\ &= \frac{P(X= \frac{h+k}2,Y = \frac{k-h}2)}{P(X+Y=k)} \\ \end{align}