For i.i.d random variables $X$ and $Y$ , is $E [X ∣ σ (X + Y)] = \frac{X + Y}{2}$ ?

Question

Let $X$ and $Y$ be i.i.d. random variables; we want to calculate the conditional expectation with respect to the $σ$ -algebra generated by $X + Y$ :

E [X ∣ σ (X + Y)]

Now, generally for random variables $X, Y \in L^{1}$ , if

E [X 1_{A} (X) 1_{B} (Y)] = E [Y 1_{A} (Y) 1_{B} (X)] (A, B \in B (R))

then

E [X 1_{C} (X + Y)] = E [Y 1_{C} (X + Y)] (C \in B (R))

So here is my solution so far: the above holds for i.i.d. random variables $X, Y$ , so

E [X ∣ σ (X + Y)] = E [Y ∣ σ (X + Y)]

, and then we have

E [X ∣ σ (X + Y)] = \frac{1}{2} E [X + Y ∣ σ (X + Y)] = \frac{X + Y}{2}

I feel like I am missing something here...

Related: math.stackexchange.com/q/78546/321264. – StubbornAtom Jul 28 '20 at 7:49 — StubbornAtom, Jul 28 '20 at 7:49

Kavi Rama Murthy · Accepted Answer · 2020-07-28 05:18:05Z

The third line in your proof is proved rigorously as follows:

E [X 1_{C} (X + Y)] = \int x 1_{{(x, y) : x + y \in C}} d F_{X, Y} (x, y) .

Applying the transformation

(x, y) \to (y, x)

and noting that

F_{X, Y} = F_{Y, X}

we see that

\int x 1_{{(x, y) : x + y \in C}} d F_{X, Y} (x, y) = \int y 1_{{(x, y) : x + y \in C}} d F_{X, Y} (x, y)

. Hence

E [Y 1_{C} (X + Y)] = E [X 1_{C} (X + Y)]

.

1 Answer 1