0
$\begingroup$

Let $X,Y$ be bounded random variables on $(\Omega, \mathcal{F}, \mathbb{P})$. I want to show that each implication implies the next, but the converses are all false:

(i) $X, Y$ are independent

(ii) $\mathbb{E}[X|Y] = \mathbb{E}[X]$ a.s.

(iii) $\mathbb{E}[XY] = \mathbb{E}[X] \mathbb{E}[Y]$

I find it confusing that we have the statement (ii) in the middle, and that's what is causing me trouble to figure out the proofs and counterexamples.

$\endgroup$
4
  • 1
    $\begingroup$ Hint. Tower property. $\endgroup$
    – Kurt G.
    Nov 15, 2022 at 4:11
  • $\begingroup$ @KurtG. How do I apply the Tower Property in this context? What $\sigma$-algebras do I need to introduce to use it? $\endgroup$
    – struggler
    Nov 15, 2022 at 7:28
  • $\begingroup$ The tower property, $\mathbb E[\mathbb E[X|Y]]=\mathbb E[X]$, also known as the "law of iterated expectations," is valid in general, even for correlated random variables, so I'm not sure how it will help here. $\endgroup$
    – 3rdMoment
    Nov 15, 2022 at 8:11
  • 1
    $\begingroup$ The tower property will help to go from (ii) to (iii). OP was supposed to show that each implication implies the next and wrote that (ii) in the middle is confusing. $\endgroup$
    – Kurt G.
    Nov 15, 2022 at 11:46

1 Answer 1

Reset to default
0
$\begingroup$

Statement (ii) means that the conditional mean of $X$ given $Y$ (i.e. the nonlinear regression function of $X$ on $Y$) does not depend on Y, but is constant at the unconditional mean of $X$.

An example where (ii) holds but not (i) is where one of the following values for $(X,Y)$ is selected with equal probability: (-1,0),(1,0),(-2,1)(2,1).

Statement (iii) means that $X$ and $Y$ are uncorrelated.

An example where (iii) holds but not (ii) is where one of the following points is selected with equal probability: (-1,-1),(2,0),(-1,1)

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .