# Finding joint conditional expectation of binomial

In this question, the author finds the probability mass function and then finds the expectation because it happens to be a hypergeometric and therefore the answer is well known.

In general, how do you find the expectation by first finding the probability mass function and/or moment functions?

If I wanted to use conditional expectation, would this be correct : $$E(X)=\sum_{k=0}^mE[(X|X=m-k)]P(X=m-k) = \sum E(X)P(X=m-k)$$ I am not sure if I have to sum over all y, or how this should be set up in general.

Also, if I wanted to solve this with moment generating functions, am I right that it would become $$M_{x+y}=M_xM_y=(pe^t+1-p)^n(pe^t+1-p)^n=(pe^t+1-p)^{2n}$$ Then, to get the expectation of x, evaluate $M_{x+y}'(0)$, but how would I condition the moment function? $(pe^t+1-p)^{2n}|x+y=m$ ? Shortcut:

• $\mathbb E[X\mid X+Y=m]+\mathbb E[Y\mid X+Y=m]=\mathbb E[X+Y\mid X+Y=m]=m$.
• $\mathbb E[X\mid X+Y=m]=\mathbb E[Y\mid X+Y=m]$ on base of symmetry.

From this it follows directly that: $$\mathbb E[X\mid X+Y=m]=\frac{m}2$$

The first equality in your question is correct (and quite useless), but the second is not correct.

This because $\mathbb E(X\mid X=m-k)=m-k$ (not $\mathbb EX$).

• I love this shortcut. I want more shortcuts like this. – Frank May 22 '18 at 7:49