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I have a question about conditional expectation and conditional variance. It's a very general question. We defined conditional variance by

$$ \operatorname{Var}(X|\mathcal{F}):=E((X-E(X|\mathcal{F}))^2|\mathcal{F}) $$

For a random variable $ X $ and a $\sigma$-algebra $ \mathcal{F}$. Are there any inequalities such that

$ \operatorname{Var}(X|\mathcal{F})\le \operatorname{Var}(X)$ or $ \operatorname{Var}(X|\mathcal{F})\ge \operatorname{Var}(X)$ and the same question for the conditional expectation:

$E(X|\mathcal{F}) \le E(X)$ or $ E(X|\mathcal{F}) \ge E(X)$

Are any one of them true in general, or what further assumption have to bee assumed that a conclusion as above is true? Often such an inequality would be very usful. Thanks in advance.

hulik

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You might want to have a look at the (small and excellent) textbook Probability with martingales by David Williams. –  Did Jan 4 '12 at 9:41
    
@ Didier Piau: Thanks for your comment, but unfortunately the book is not free available or I didn't find it. –  user20869 Jan 4 '12 at 10:24
    
Which textbook do you use? –  Did Jan 4 '12 at 13:42

2 Answers 2

up vote 1 down vote accepted

None of these inequalities are true in general.

Suppose you have

  • $\Pr(X=0,Y=0) = \frac{1}{5}$
  • $\Pr(X=0,Y=1) = \frac{1}{5}$
  • $\Pr(X=2,Y=1) = \frac{1}{5}$
  • $\Pr(X=-5,Y=2) = \frac{1}{5}$
  • $\Pr(X=3,Y=2) = \frac{1}{5}$

Then defining $\mathcal{F}$ on values of $Y$

we have $E[X]=0$, $E[X|Y=1]=1$, $E[X|Y=2]=-1$,

and $Var(X)=7.6$, $Var(X|Y=0)=0$, $Var(X|Y=2)=16$.

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@ Henry: Thanks for your answer. Though I have an additional question. In fact I would like to bound $ Var(X|\mathcal{F}) $ by $ \operatorname{const}\cdot E(X^2)$. Is this possible? –  user20869 Jan 4 '12 at 10:33
    
No: you can exceed any given ratio –  Henry Jan 4 '12 at 15:09
    
@ Henry: Thanks for your quick answer. As mentioned below (see comment after Didier Piau's answer) the motiviation behind this question can be found in link. I wondered if in a more general setting something would be true. Because then I would be able to prove 4.) in the related question –  user20869 Jan 4 '12 at 16:08

All these inequalities fail in general, and for good reasons.

Assume for example that $\mathcal F$ is the sigma-algebra generated by $X$. Then $\mathrm E(X\mid\mathcal F)=X$ almost surely hence $\mathrm E(X)\leqslant \mathrm E(X\mid\mathcal F)$ almost surely or $\mathrm E(X)\geqslant \mathrm E(X\mid\mathcal F)$ almost surely are impossible unless $\mathrm E(X)= \mathrm E(X\mid\mathcal F)$ almost surely, that is, $X$ almost surely constant.

Here is another example. Assume that $X=YZ$ with $Y$ and $Z$ independent and $\mathrm E(Z)=0$ and $\mathrm E(Z^2)=1$, and that $\mathcal F$ is the sigma-algebra generated by $Y$. Then $\mathrm E(X\mid\mathcal F)=\mathrm E(Z)Y=0$ almost surely and $\mathrm{Var}(X\mid\mathcal F)=\mathrm E(X^2\mid\mathcal F)=E(Z^2)Y^2=Y^2$ almost surely while $\mathrm E(X)=0$ and $\mathrm E(X^2)=\mathrm E(Y^2)$ hence no almost sure comparison can hold.

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Thanks for your answer. See link it's a related question, in fact it's 4. in the other question (see link). This was the reason for asking this question in a more general setting. –  user20869 Jan 4 '12 at 16:02

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