# Almost sure pointwise inequality between conditional variance and variance or between conditional expectation and expectation

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

• 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

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

• @ 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, unless $X$ is almost surely constant.

Here is another example. Assume that $X=YZ$ with $Y$ and $Z$ independent, $\mathrm E(Z)=0$, $\mathrm E(Z^2)=1$, 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.

• 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