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

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.
