1
$\begingroup$

I have a random variable $X$ and its conditional expectation $E(X | G) = Y$.

In the context of some homework, I have to reach the conclusion that $P(X = 0 \text{ and } Y \neq 0) = 0$ But isn't this always the case? I think we have $P(X = 0 \text{ and } Y \neq 0) = P(Y \neq 0 | X=0) P(X=0)$ and $P(Y \neq 0 | X=0) = 0$ since when $X = 0$, its conditional expectation is $0$ too

Am I missing something here? thanks!

  • 0
    @Firefeather Neither is correct.2012-10-12

1 Answers 1

1

First, you probably misunderstood the question: $G$ denotes a sigma-algebra, not an event. Note that if $B$ is an event (of nonzero probability) then $\mathbb E(X\mid B)$ is a real number, not a random variable, and you are asked to consider the situation where $\mathbb E(X\mid G)$ coincides with the random variable $Y$.

Second, the claimed property does not hold without some additional hypotheses: assume that the sigma-algebra is $G=\{\varnothing,\Omega\}$, then $Y=\mathbb E(X)$, hence every random variable $X$ such that $\mathbb E(X)\ne0$ and $\mathbb P(X=0)\ne0$ is a counterexample.