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Consider $X$ a nonnegative random variable on some probability space $(\Omega, \mathcal{F}, \mathbb{P})$. We are given that

$$ A = \left\{ \omega : X(\omega) > 0 \right\} $$

and that $\mathcal{G}$ is a sub-sigma-field of $\mathcal{F}$. Can we conclude that $\mathbb{E}(X \ |\ \mathcal{G}) > 0$ for almost all $\omega \in A$?

I think the answer is no, but I'm having trouble coming up with a counterexample. One thing I tried was

$$ \Omega = \left\{ a, b , c \right\}, \; \mathcal{G} = \left\{ \emptyset, \Omega, \{a, b\}, \{c\}\right\} $$ with $X(\omega) = \mathbf{1}_{\{c\}}$ and $\mathbb{P}(\{c\}) = 0$, $\mathbb{P}(\{ a \}) = \mathbb{P}(\{ b \}) = 1/2$. Here, obviously $A = \{ c \}$ and

$$ \int_A \mathbb{E}(X | \mathcal{G}) \text{d}\mathbb{P} = \int_A X \text{d} \mathbb{P} = 0, \text{ since } \mathbb{P}(A) = 0$$

However, we can set $\mathbb{E}(X \ |\ \mathcal{G})$ equal to any positive number above for $\omega \in A$, so we cannot disprove the claim. Any help?

More generally, any pointer to resources with "pathological" cases and counterexamples in conditional probability and conditional expectation would be more than helpful.

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  • $\begingroup$ There is no need to assume $X \geq 0$. Defining $A = \{\omega : X(\omega)>0\}$ is enough to conclude $E[X|\mathcal{G}]>0$ for all $\omega \in A$ except for a set of probability measure 0. $\endgroup$ – Michael May 21 '18 at 19:00
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Let $X$ be any random variable (possibly one that can take negative values) and let $A = \{\omega : X(\omega) >0\}$. You can indeed conclude $E[X|\mathcal{G}]>0$ for all $\omega \in A$ except possibly a set of probability measure 0 (note that I have replaced your "almost all" with "except for a set of probability measure 0").

Suppose $B \subseteq A$, $P[B]>0$, and $E[X|\mathcal{G}]\leq 0$ for all $\omega \in B$. We want to reach a contradiction. Note that $B \cap\{\omega : X(\omega)>1/n\} \nearrow B$. So there is a positive integer $m$ such that $$P[B \cap \{\omega : X(\omega)>1/m\}] >0 \quad (Eq. *)$$
So \begin{align} 0&\overset{(a)}{\geq}\int_B E[X|\mathcal{G}]dP \\ &\overset{(b)}{=} \int_B X dP \\ &\overset{(c)}{\geq} \int_B (1/m)1_{\{X>1/m\}}dP \\ &= (1/m)P[B \cap \{\omega : X(\omega)>1/m\}] \\ &\overset{(d)}{>}0 \end{align} where (a) holds because $E[X|\mathcal{G}]\leq 0$ for all $\omega \in B$; (b) holds by definition of conditional expectation; (c) holds because $X(\omega) \geq (1/m)1_{\{X>1/m\}}$ for all $\omega \in B \subseteq A$; (d) holds by (Eq. *). The conclusion $0>0$ is the desired contradiction. $\Box$


However, a "counter-example" that relates to expectation is that it is not always the case: $$ E[X+Y]=E[X]+E[Y]$$ This equation is true whenever the right-hand-side does not lead to an undefined case of $\infty - \infty$. It fails, for example, if $E[X]=\infty$ and $Y=1-X$.

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First of all, recall that $X \geq 0$ implies $\mathbb{E}(X \mid \mathcal{G}) \geq 0$ almost surely. Moreover, we know that

$$\forall G \in \mathcal{G}: \quad \int_G X \, d\mathbb{P} = \int_G \mathbb{E}(X \mid \mathcal{G}) \, d\mathbb{P};$$

in particular, we find for $G := \{\mathbb{E}(X \mid \mathcal{G}) = 0\}$

$$0 = \int_G \mathbb{E}(X \mid \mathcal{G}) \, d\mathbb{P} = \int_G X \, d\mathbb{P}.$$

As $X \geq 0$ this implies

$$X=0 \quad \text{almost surely on $\{\mathbb{E}(X \mid \mathcal{G})=0\}$}$$

i.e.

$$\{\mathbb{E}(X \mid \mathcal{G})=0\} \subseteq \{X=0\} \quad \text{up to a null set}.$$

Taking the complement, we conclude that

$$\{X>0\} \subseteq \{\mathbb{E}(X \mid \mathcal{G})>0\} \quad \text{up to a null set},$$

and this proves the assertion.

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    $\begingroup$ (+1, but minor comment on first line): $X\geq 0 \implies E[X|\mathcal{G}]\geq 0$ [for all $\omega$ except possibly a set of measure 0]. $\endgroup$ – Michael May 21 '18 at 19:05
  • $\begingroup$ @Michael Thanks for your upvote and the comment; I've just fixed it. $\endgroup$ – saz May 21 '18 at 19:08

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