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

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

$$E [X mid sigma(X+Y)]$$

Now, generally for random variables $X, Y in L^1$, if
$$E[X1_A(X)1_B(Y)] = E[Y1_A(Y)1_B(X)] quad (A, B in mathcal{B}(mathbb{R}))$$

then$$E[X1_C(X+Y)] = E[Y1_C(X+Y)] quad (C in mathcal{B}(mathbb{R}))$$

So here is my solution so far: the above holds for i.i.d. random variables $X, Y$, so
$$E[X mid sigma(X+Y)] = E[Y mid sigma(X+Y)]$$, and then we have
$$E[X mid sigma(X + Y) ] = frac{1}{2} E[X + Y mid sigma(X + Y) ] = frac{X+Y}{2}$$

I feel like I am missing something here…

Mathematics Asked on December 27, 2020

1 Answers

One Answer

The third line in your proof is proved rigorously as follows: $$E[X1_C(X+Y)]=int x1_{{(x,y): x+y in C }} dF_{X,Y}(x,y).$$ Applying the transformation $(x,y)to (y,x)$ and noting that $F_{X,Y}=F_{Y,X}$ we see that $$int x1_{{(x,y): x+y in C }} dF_{X,Y}(x,y) =int y1_{{(x,y): x+y in C }} dF_{X,Y}(x,y)$$. Hence $$E[Y1_C(X+Y)]=E[X1_C(X+Y)]$$.

Correct answer by Kavi Rama Murthy on December 27, 2020

Add your own answers!

Related Questions

Finding $AC$ given that $ABCD$ is inscribed in a circle

1  Asked on February 7, 2021 by questionasker


Proofing Metropolis Hastings

0  Asked on February 7, 2021 by nestroy


Integrate $2x(2x-3)^frac{1}{2} dx$.

5  Asked on February 5, 2021 by pretending-to-be-a-calculator


Rule of inference. Is the argument valid?

2  Asked on February 5, 2021 by blue


Ask a Question

Get help from others!

© 2022 All rights reserved.