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

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

Answer 1

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:

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

Answer 2

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 $f(x,y) = \frac{x}{x+y}$.

Answer 3

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. $E[X^2/Y]=E[Y^2/X]$.