Conditional expectation $\mathbb{E}(X|X+Y)$ for 2 different Binomial distributions

Question

Let X,Y be independent random variables, where X have Binomial distribution $B(n,p)$ and Y have Binomial distribution $B(m,p)$. Find $\mathbb{E}(X|\sigma(X+Y))(\omega)$.

What I know is that X+Y have distribution $B(n+m,p)$, so

$\mathbb{E}(X|\sigma(X+Y))(\omega)= \sum_{k=0}^{n+m}\mathbb{E}(X|X+Y=k)\mathbf{1}_{X+Y=k}(\omega)$

and generally I have to find $\mathbb{E}(X|X+Y=k)$, but I totally stuck because X and Y aren't i.i.d.

If they will be, then by symmetry $\mathbb{E}(X|X+Y=k)=\mathbb{E}(Y|X+Y=k)$, so $2\mathbb{E}(X|X+Y=k)$ will be equal to $\mathbb{E}(X|X+Y=k) + \mathbb{E}(Y|X+Y=k)= \mathbb{E}(X+Y|X+Y=k)=X+Y$, so in this case $\mathbb{E}(X|X+Y=k)=\frac{X+Y}{2}$, but I don't know if I can make use of this in my situation.

I know also that $\mathbb{E}(X|X+Y=k)= \frac{1}{\mathbb{P}(X+Y=k)}\int_{X+Y=k}Xd\mathbb{P}$ and I know, that this integral will become a sum because we deal with discrete distribution, but I have know idea how to calculate this (and did't find any useful hint on internet so far). How can I calculate such thing (the problem for me is that I'm integrating with respect to X and Y)?

Answer 1

You can use your symmetry idea by decomposing $X$ and $Y$ into sums of IID Bernoulli random variables. Explicitly, one can write

$$ X = \sum_{i=1}^nZ_i, \qquad Y = \sum_{i=n+1}^{n+m}Z_i,$$

where $\{Z_i\}$ are IID $B(1,p)$ random variables. Now you can use your symmetry idea: for each $i\in\{1,\ldots,n+m\}$, one has

$$\mathbb E(Z_i|X+Y) = \mathbb E\left(Z_i\,\Big|\,\sum_{j=1}^{n+m}Z_j\right) = \mathbb E\left(Z_1\,\Big|\,\sum_{j=1}^{n+m}Z_j\right)$$

and so

$$X+Y = \sum_{i=1}^{n+m}\mathbb E\left(Z_i\,\Big|\,\sum_{j=1}^{n+m}Z_j\right) = (n+m)\mathbb E\left(Z_1\,\Big|\,\sum_{j=1}^{n+m}Z_j\right).$$

It follows that

$$\mathbb E(X|X+Y) = \sum_{i=1}^n\mathbb E(Z_i|X+Y) = \sum_{i=1}^n\frac{X+Y}{n+m} = \frac{n}{n+m}(X+Y).$$

Answer 2

Your second approach $$\mathbb{E}(X\mid X+Y=k)= \frac{\int_{\{X+Y=k\}}Xd\mathbb{P}}{\mathbb{P}(X+Y=k)}$$ works as well. For the numerator, write $X=\sum _\limits{i=1}^nI(\text{trial $i$ is success})$ as a sum of indicators and compute $$ \begin{align} \int_{\{X+Y=k\}}Xd\mathbb{P}&=\mathbb E \left[X\,I(X+Y=k) \right]\\&= \mathbb E\sum I(\text{trial $i$ is success and $k$ successes in $n+m$ trials})\\ &=\sum \mathbb{P}(\text{trial $i$ is success and $k-1$ successes in remaining $n+m-1$ trials})\\ &\stackrel{(*)}=\sum \mathbb P(\text{trial $i$ is success})\mathbb P(\text{$k-1$ successes in remaining $n+m-1$ trials})\\ &=n\cdot p \cdot{n+m-1\choose k-1}p^{k-1}(1-p)^{n+m-1-(k-1)} \end{align} $$ using independence in step (*). Divide this by the denominator $\mathbb P(X+Y=k)={n+m\choose k}p^k(1-p)^{n+m-k}$, simplify the algebra, and obtain $$ \mathbb E(X\mid X+Y=k)=\frac{n k}{n+m}.$$