2
$\begingroup$

To be more precise, fix a probability space $(\Omega,\mathcal{F},\mathbb{P})$, and let $X:\Omega\to\mathbb{R}$ be an $L^1$ random variable. Suppose $\Sigma\subset\mathcal{F}$ is such that $\mathbb{E}[X|\Sigma]=0$, then is $X$ independent of $\Sigma$?

This is true in the trivial cases where $\Sigma=\{\emptyset,\Omega\}$ or $\mathcal{F}$.

$\endgroup$
2
$\begingroup$

Let $X\sim N(0,1)$ and $Y\ge 1$ a.s. is any r.v. independent of $X$. Then $$ \mathsf{E}[(X/Y)\mid \sigma(Y)]=0=\mathsf{E}[X/Y] \quad\text{a.s.} $$ but $(X/Y)$ is not independent of $\sigma(Y)$.

$\endgroup$
  • $\begingroup$ The division could've been multiplication (allowing you to drop the $Y \geq 1$ requirement). $\endgroup$ – Ian Sep 24 '17 at 3:49
  • $\begingroup$ What do you mean by "a.s."? $\endgroup$ – James Sep 24 '17 at 3:51
  • 1
    $\begingroup$ @James "a.s." is a standard abbreviation for "almost surely". $\endgroup$ – Ian Sep 24 '17 at 3:52
  • $\begingroup$ @James Conditional expectations are defined up to a null set. $\endgroup$ – d.k.o. Sep 24 '17 at 4:00
  • 1
    $\begingroup$ Sure, but you could have just said $Y$ is bounded and then used multiplication. There is no "secret" to the division. $\endgroup$ – Ian Sep 24 '17 at 15:29
4
$\begingroup$

Consider the example I gave here:

Flip a fair coin to determine the amount of your bet: if heads, you bet \$1, if tails you bet \$2. Then flip again: if heads, you win the amount of your bet, if tails, you lose it. (For example, if you flip heads and then tails, you lose \$1; if you flip tails and then heads you win \$2.) Let $X$ be the amount you bet, and let $Y$ be your net winnings (negative if you lost).

Then $X,Y$ are not independent; for instance we have $P(X=2) = 1/2$, $P(Y=1)=1/4$, and $P(X=2, Y=1) = 0$. But taking $\Sigma = \sigma(X)$ we have $E[Y \mid \Sigma] = E[Y \mid X] = 0$. You can see that $P(Y=1 \mid X=1) = P(Y=-1 \mid X=1) = 1/2$, so $$E[Y \mid X=1] = 1 \cdot \frac{1}{2} + (-1) \cdot \frac{1}{2} = 0.$$ Likewise $E[Y \mid X=2] = 0$. The idea is that the value of $X$ does affect the possible values for $Y$, but no matter how the first flip came up, you are still making a (conditionally) fair bet.

This is similar in spirit to the example given by d.k.o. but perhaps a little more elementary.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.