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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!

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    "when $X=0$ its conditional expectation is 0 too"... whose conditional expectation?2012-10-10
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    E(X | G), for any G possible?2012-10-10
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    For each event $G$, $X|G$ is a different random variable. If by "when $X=0$" you mean that you are considering the random variable $X|[X=0],$ that is a "constant random variable" (i. e. it takes only one value) and indeed, its expectation is zero. But this does not mean that $E(X)=0,$ nor that these considerations will help you in evaluating $P[Y\ne 0|X=0]$.2012-10-10
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    alright I understand my mistake, I though that P(X = 0) meant "probability that X is the constant 0"2012-10-10
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    The conditional probability part doesn't always work. You can only split $P(X = 0 \textrm{ and } Y \neq 0)$ into $P(Y \neq 0 | X = 0) P(X = 0)$ if $P(X = 0) \neq 0$, because, by definition, $P(Y \neq 0 | X = 0) \equiv \frac{P(X = 0 \textrm{ and } Y = 0)}{P(X = 0)}$, so $P(X = 0) = 0$ results in division by zero, making $P(Y \neq 0 | X = 0)$ undefined.2012-10-11
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    @lezebulon Can you clarify what you are representing with the letter $G$? Is $G$ an event (such as "$G$ is the event that $X > 0$")? Or, if $G$ is a random variable, by "for any $G$ possible", do you mean "for any selection of random variable $G$", or do you mean "allowing $G$ to be any of the values in its support" (thus distinguishing $E(X|G)$ from $E(X|G=g)$)?2012-10-11
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    @Firefeather Neither is correct.2012-10-12

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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.