I mean, is it true that $E(Y|X) = \phi(X)?$
if so, how should we derive the form of X?
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What is $\phi$ in this case? – Clarinetist Oct 24 '14 at 04:35
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@Clarinetist I wonder if $E(Y|X)$ is a function of X, and $\phi $is that function, is that true? or under what condition it's true – annimal Oct 24 '14 at 04:36
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Yes, this is a consequence of the Doob-Dynkin lemma, which states that any $\sigma(X)$-measurable random variable is of the form $\phi(X)$ for some (usually not unique) measurable function $\phi$.
How to compute $\phi$ depends on what you know about $X$ and $Y$. For instance, if $X$ has a discrete distribution, you may define $$\phi(x) := \begin{cases} E[Y \mid X = x],& \text{if $P(X=x)>0$} \\ 42, & \text{else.} \end{cases}$$
Nate Eldredge
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Hi! I was wondering if you knew of a proof that (in the general case) $\phi(x) = \frac{\mathbb{E}[Y\mathbf{1}_{X=x}]}{P(X=x)}$ for all $x$ such that $P(X=x)>0$? This holds when $X$ is discrete, as you note, and it’s not hard to prove.
Does it hold more generally e.g., when $X$ is not discrete, but the induced measure $\mu_{X}$ happens to have a point mass at ${x}$?
– MathematicsStudent1122 Oct 11 '21 at 15:09 -
@AlohaSine: Pretty much the same proof works. By properties of conditional expectation we have $E[\phi(X) 1_{X=x}] = E[E[Y|X] 1_{X=x}] = E[Y 1_{X=x}]$. On the other hand, $\phi(X) 1_{X=x} = \phi(x) 1_{X=x}$ almost surely, so $E[\phi(X) 1_{X=x}] = E[\phi(x) 1_{X=x}] = \phi(x) E[1_{X=x}] = \phi(x) P(X=x)$ because $\phi(x)$ is not random. So in fact $\phi(x) P(X=x) = E[Y 1_{X=x}]$ is true no matter what. If $P(X=x) = 0$ it's trivial, and if $P(X=x) > 0$ you can divide by $P(X=x)$ to get your claim. – Nate Eldredge Oct 11 '21 at 17:59