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Same as here. Let $X$ be a square integrable random variable on $(\Omega,\mathcal{F},P)$. Let $\mathcal{G}$ be a sub-$\sigma$-algebra of $\mathcal{F}$. Define the conditional variance of $X$ given $\mathcal{G}$ by $$\operatorname{var}(X\mid\mathcal{G})=\Bbb{E}[(X-\Bbb{E}[X\mid\mathcal{G}])^2\mid\mathcal{G}]$$ Prove the formula

$$\operatorname{var}(X)=\operatorname{var}(\Bbb{E}[X\mid\mathcal{G}])+\Bbb{E}[\operatorname{var}(X\mid\mathcal{G})]$$

1 I don't understand part of the presented solution.

2 Is my attempt also correct?

3 Alternate solutions?

1 I don't understand this part of the solution wherein

$$\operatorname{var}(X\mid \mathcal{G})=\Bbb{E}[(X-\Bbb{E}[X\mid\mathcal{G}])^2 \mid \mathcal{G}]$$

$ =\mathbb E[ X^2 \mid \mathcal G] - \mathbb E [X \mid \mathcal G]^2 $. I think it is making use of the fact that $\mathbb E [X \mid \mathcal G]$ is $\mathcal G$-measurable in saying:

$\mathbb E[XE[X\mid\mathcal G]\mid\mathcal G] = \mathbb E[X\mid\mathcal G] \ E[X\mid\mathcal G] = E[X\mid\mathcal G]^2$, but isn't $E[X\mid\mathcal G]$ supposed to be bounded? How to show this?

My attempt:

$E(X^2) < \infty$

$\to E(X) < \infty$

$\to E(X\mid\mathcal G) < \infty$, but I don't think this step is correct.

2 My own attempt: I tried evaluating $$E(\operatorname{var}(X\mid\mathcal{G}))=E(\Bbb{E}[(X-\Bbb{E}[X\mid\mathcal{G}])^2\mid\mathcal{G}])$$

Is it correct to say that $E(\Bbb{E}[(X-\Bbb{E}[X\mid\mathcal{G}])^2\mid\mathcal{G}]) = E([(X-\Bbb{E}[X\mid\mathcal{G}])^2])$?

After expanding the square I encounter $E[XE[X\mid\mathcal G]]$ and rewrite

$= E[E[XE[X\mid\mathcal G]]\mid\mathcal G]$

$ = E[E[X\mid\mathcal G] \ E[X\mid\mathcal G]]$ (*)

$ = E[E[X\mid\mathcal G]^2]$

(*) Assuming this is all correct, I am still unsure about this part. Is it really bounded?

3 Is there a way to go about this without saying $E[X\mid\mathcal G]$ is bounded?

4 Is this wrong? It's from Stochastic Calculus class notes, though the problem is for an Advanced Probability class (and the notes don't say anything about bounded weirdly).

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1 Answer 1

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  1. What is used in this solution is called pull out and works for any integrable random variable (it does not need to be bounded):

    Let $X \in L^1$ and $Y \in L^1(\mathcal{G})$ (i.e. integrable and $\mathcal{G}$-measurable) such that $X \cdot Y \in L^1$. Then $$\mathbb{E}(X Y \mid \mathcal{G}) = Y \cdot \mathbb{E}(X \mid \mathcal{G}).$$

    Proof: Let us assume that we know the result for bounded $Y$, i.e. if there exists $n$ such that $|Y(\omega)| \leq n$ for all $\omega \in \Omega$. For $Y \in L^1(\mathcal{G})$, we define approximations $$Y_n := (-n) \vee Y \wedge n.$$ Then, by definition, $Y_n$ is $\mathcal{G}$-measurable and bounded. Moreover, $Y_n \to Y$ almost surely and $|XY_n| \leq |XY|$. Therefore, it follows from the (conditional) dominated convergence theorem that $$\begin{align*} \mathbb{E}(X \cdot Y \mid \mathcal{G}) &= \lim_{n \to \infty} \mathbb{E}(X Y_n \mid \mathcal{G}) \\ &= \lim_{n \to \infty} Y_n \mathbb{E}(X \mid \mathcal{G}) \\ &= Y \mathbb{E}(X \mid \mathcal{G}).\end{align*}$$ This finishes the proof. Regarding your own thoughts: $\mathbb{E}(X^2) < \ \infty$ implies that $\mathbb{E}(X \mid \mathcal{G})< \infty$ almost surely, but this is something different from $\mathbb{E}(X \mid \mathcal{G})$ being bounded. $\mathbb{E}(X \mid \mathcal{G})$ is bounded if, and only if, we can find $n \in \mathbb{N}$ such that $$|\mathbb{E}(X \mid \mathcal{G})(\omega)| \leq n$$ for all $\omega \in \Omega$ (the bound does not depend on $\omega$).

  2. First of all: Yes, the equality $$\mathbb{E}\big( \mathbb{E}(Y^2 \mid \mathcal{G}) \big) = \mathbb{E}(Y^2)$$ holds for any square integrable random variable $Y$ (by tower property). Nevertheless, your attempt fails: You are trying to show that the expectations of the random variables $\text{var}(X \mid \mathcal{G})$ and $\text{var}(\mathbb{E}(X \mid \mathcal{G}))+\mathbb{E}(\text{var}(X \mid \mathcal{G}))$ coincide. But this does not imply that the random variables are the same.
  3. Yes, see the first part.
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    $\begingroup$ @BCLC $\wedge$ ($\vee$) denote the minimum (maximum), i.e. $$Y_n := \min\bigg\{ \max\{-n,Y\}, n \bigg\}.$$ And what do you mean by your fourth question? The statement there is simply less general. If it holds for $Y$ such that $XY \in L^1$, $Y \in L^1$, then it holds in particular for bounded $Y$. $\endgroup$
    – saz
    Sep 23, 2014 at 5:27
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    $\begingroup$ @BCLC Yes, if we replace "$Y$ bounded" by "$Y \in L^1$ and $XY \in L^1$", the statement remains valid. (But note, however, that we need in the proof above the statement for bounded random variables, i.e. one has to prove it first for bounded random variables and then consider the more general case.) $\endgroup$
    – saz
    Sep 23, 2014 at 5:36
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    $\begingroup$ @BCLC Yes. It is a well-known fact that $L^2(\mathbb{P}) \subseteq L^1(\mathbb{P})$, i.e. $Y:= \mathbb{E}(X \mid G) \in L^2$ implies $Y \in L^1$. This follows for example from Jensen's inequality. $\endgroup$
    – saz
    Sep 24, 2014 at 5:13
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    $\begingroup$ @BCLC Where did I claim that $Y$ is square integrable? I just mentioned that the equality $$\mathbb{E}(\mathbb{E}(Y^2 \mid G)) = \mathbb{E}(Y^2)$$ holds for any square-integrable rv $Y$. $\endgroup$
    – saz
    Sep 25, 2014 at 5:50
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    $\begingroup$ @BCLC Yes, this follows from the very definition of conditional expectation. $\endgroup$
    – saz
    Sep 25, 2014 at 7:47

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