How $E(X/X+Y)=E(Y/X+Y)$ when $X,Y$ are i.i.d's

Question

How $E(X/X+Y)=E(Y/X+Y)$ when $X,Y$ are i.i.d's

I have recently started studying probability and statistics on my own. I am presently watching harvard lectures. In that professor told that the above hold by symetry. I did not understand how? kindly elaborate

At 49.01 in https://www.youtube.com/watch?v=PgawcWisb0I&list=PL2SOU6wwxB0uwwH80KTQ6ht66KWxbzTIo&index=26

Did you mean $E\left[\frac{X}{X+Y}\right]$ or $E[X\mid X+Y]$? Looks like the second one but still you accepted an answer that uses the first one. — StubbornAtom, Feb 22 '20 at 10:43
@StubbornAtom sir in that youtube link professor is talking about 1st one right? pls correct me if i am wrong. I am a beginner. Started with harvard lectures. How the proof varies in case of 2nd one. Can you pls elaborate? — Nascimento de Cos, Feb 22 '20 at 20:20
@StubbornAtom sir i have gone through the link u sent. Thank you — Nascimento de Cos, Feb 22 '20 at 20:20
I am no sir, but you linked to a lecture about conditional expectations and added the relevant tag as well. That tells me it is the second one. — StubbornAtom, Feb 22 '20 at 20:51
@StubbornAtom If it is conditional expectation(2nd case as you have pointed) E(X|T=X+Y) = E(Y|T) (by exchangeability of i.i.d s). Is this argument suffice. But the way without using exchangeability, how to prove this? kindly explain — Nascimento de Cos, Feb 22 '20 at 21:15

quester · Accepted Answer · 2020-02-22T08:29:44.877

2

since $X, Y$ are iid then for joint distribution we have:

$$\forall_{(x,y) \in \Omega} f_{XY}(x, y) = f_{XY}(y,x)$$

then:

$$E[\frac{Y}{X+Y}] = \iint_{\Omega} \frac{y}{x+y}f_{XY}(x,y)dxdy=\iint_{\Omega} \frac{y}{x+y}f_{XY}(\underline{y,x})dxdy=^{Fubini Theorem}\iint_{\Omega} \frac{y}{x+y}f_{XY}(y,x)\underline{dydx}=^{x<->y}\iint_{\Omega} \frac{x}{x+y}f_{XY}(x,y)dxdy=E[\frac{X}{X+Y}]$$

edited Feb 22 '20 at 08:29

answered Feb 22 '20 at 08:02

quester

472
3
12

It is not given that $X,Y$ have densities. – StubbornAtom Feb 22 '20 at 10:46

score 2 · Answer 2 · answered Feb 22 '20 at 08:12

If $X$ is independent of $Y$ and $X \sim Y$ then $X$ and $Y$ are exchangeable. That means that for any measurable function $f$, one has $f(X,Y) \sim f(Y,X)$, and consequently $E\bigl[f(X,Y)\bigr] = E\bigl[f(Y,X)\bigr]$ when this expectation exists. Apply this result to $f(x,y) = \frac{x}{x+y}$.

score 0 · Answer 3 · answered Feb 22 '20 at 08:36

Just an intuitive addition: when two RVs are iid, symmetry rules govern. For example, there is no reason that one RV is greater than the other with higher probability, i.e. $P(X>Y)=P(Y>X)$. Or, it's not expected that their expectations in the same form to differ. That means you can always switch variables, e.g. $E[X^2/Y]=E[Y^2/X]$.

How $E(X/X+Y)=E(Y/X+Y)$ when $X,Y$ are i.i.d's

3 Answers3