0
$\begingroup$

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? $(pe^t+1-p)^{2n}|x+y=m$ ?

Question + Answer

$\endgroup$
0
$\begingroup$

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 $\mathbb EX$).

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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.