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Is that true that if $ F_{(k-1)}$ is a sigma algebra generated by random variables $X_1,X_2,\ldots,X_{k-1}$, then

$$ E(X_k \mid F_{k-1}) \le E(X_k)\quad\text{a.s. ?}$$

Michael Hardy
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Anna
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    The only case when this happens is when $E(X_k\mid F_{k-1})=E(X_k)$ almost surely, that is, when $E(X_k\mid F_{k-1})$ is deterministic. – Did Dec 16 '14 at 18:37

2 Answers2

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Conditional expectation can not be compared to an unconditional one with $<,>$. Because in the case of any condition you restrict your sample space to some subset of the original sample space. Outside this subset your probability measure becomes zero. If one chooses a suitable subset (interval) then the expectation in the case of condition can be made larger than the one you get for the unconditional case.

Seyhmus Güngören
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It is not true. One of the simplest counterexamples is any instance in which $X_1,\ldots,X_n$ are independent. Another simple case is one in which $n=3$ and the support of each random variable is $\{0,1\}$ and $X_3$ is the mod $2$ sum of $X_1$ and $X_2$. In that case, if $X_1=0$ and $X_2=1$, then $X_3$ must be $1$, which is greater than $\operatorname{E}(X_3)=1/2$.

But generally, letting $Y=\operatorname{E}(X_k\mid F_{k-1})$, we have $$ \operatorname{E}(Y)=\operatorname{E}(\operatorname{E}(X_k\mid F_{k-1})) = \operatorname{E}(X_k). $$ One cannot have $\operatorname{E}(Y)=\operatorname{E}(X_k)$ if $Y\le\operatorname{E}(X_k)$ almost surely, except in cases where $Y=\operatorname{E}(X_k)$ almost surely.

Michael Hardy
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