1
$\begingroup$

Suppose x and y are discrete random variable, I can write them in summation. And it seems like they are equal. Any ideas?

| improve this question | | | | |
$\endgroup$
0
$\begingroup$

It's $E[y|x<y]* P(x<y)$ on the second expression but otherwise fine yes. We can prove it using the law of total expectation without independence of $x,y$.

| improve this answer | | | | |
$\endgroup$
  • $\begingroup$ He also states the variables are discrete, so one of the signs should be $\leq$ or $\geq$. $\endgroup$ – Raskolnikov Apr 27 '19 at 10:33

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.