# If $X$ and $Y$ are iid, is $E(X\mid X+Y)=E(Y\mid X+Y)$?

Just like question asked, my thought is they have the same mapping.

My classmate gives me a counterexample: if $$X$$ and $$Y$$ are equal, then $$X+Y=2X$$. However, $$\operatorname{Var}(2X) \neq \operatorname{Var}(X+Y)$$. So $$2X$$ and $$X+Y$$ don't have the same distribution? I am confused.

• Yes. This is essentially because the distribution of $(X, Y)$ is the same as $(Y, X)$. Apr 24, 2019 at 3:55
• Why would $\text{Var}(2X) \neq \text{Var}(X+Y)$ in your classmate's "counterexample"? It's not as though the variance of the sum is the sum of variances here. The answer to your question is yes, because you are conditioning two i.i.d. random variables on the same $\sigma$-algebra.
– snar
Apr 24, 2019 at 4:09
• @snar The conditional expectations of two iid variables on the same $\sigma$-algebra are not necessarily (or generally) equal. Apr 24, 2019 at 4:59
• @spaceisdarkgreen You are of course correct, as the example $X = E[X | X] \neq E[Y | X] = E[Y]$ shows.
– snar
Apr 24, 2019 at 15:11

If $$X$$ and $$Y$$ are exactly equal, the only way for them to be independent is if they're constant (i.e. have 0 variance). Hence we'd actually have $$\text{Var}(2X) = \text{Var}(X + Y) = 0$$ in this case.

Like Sangchul Lee said, the way to prove your claim is to note that $$(X ,Y)$$ and $$(Y, X)$$ have the same joint distribution.

• iid doesn't mean $X=Y$. Suppose you roll two identical dice. Apr 24, 2019 at 4:24
• @S.PhilKim That's my point, if $X$ and $Y$ are equal then they can't be i.i.d. unless they're constant. Hence his classmate's counterexample is not a counterexample. Apr 24, 2019 at 4:27

iid: Independent and Identical Distribution.

If you roll two identical dice, you don't get two identical number($$2X$$) every time.
$$2X$$ is not the same as $$X+Y$$(Sum of two dice). Not even in distribution.

$$2X$$ will be either 2,4,6,8,10,12, and the probabilities are all the same $$\frac{1}{6}$$.
However, $$X+Y$$ will be one of 2,3,4,5,6,7,8,9,10,11,12 and the probabilities are all different.
Pr($$X+Y=12$$) = $$\frac{1}{36}$$ but Pr($$X+Y=6$$) = $$\frac{5}{36}$$
Indeed this example is famous as an example of the central limit theorem.

Therefore, Var($$2X$$) $$\ne$$ Var($$X+Y$$) is what it is supposed to be.

In fact, Let Var($$X$$) = $$\sigma^2$$, then Var($$2X$$) = 4$$\sigma^2$$, and Var($$X+Y$$) = 2$$\sigma^2$$.

Since we don't distinguish these two dice (identical), $${E}(X|X+Y) ={E}(Y|X+Y)$$