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Let $X \in L_1$ be an integrable random variable and $\mathcal{F}$ a $\sigma$-algebra. Suppose $$\mathbb{E}[X|\mathcal{F}]\geq 0.$$ Let $Z\geq 0 $ be a non-negative ranfom variable. Do we have
$$\mathbb{E}[ZX|\mathcal{F}]\geq 0?$$

My idea is to show for any $A \in \mathcal{F}$, $$\int_{A}\mathbb E[ZX|\mathcal{F}]dP = \int_{A}ZXdP\geq 0$$ using the fact that $$\int_{A}\mathbb E[X|\mathcal{F}]dP = \int_{A}XdP\geq 0.$$ But it is hard for me to make a rigorous argument.

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    You might make the problem easier by looking for counter-examples with random variables $Z, X$ with $Z\geq 0$, $E[X]=0$, $E[XZ]<0$. In other words, there is no reason to be distracted by sigma algebras. – 2017-02-23
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    Imagine $Z = 0$ for $X \geq 0$ and $Z=c$ for $X < 0$ for a suitable $c$. – 2017-02-23

1 Answers 1

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Let $\Omega=\{1,2,3,4\}$, $\mathcal F=\{\emptyset,\{1,2\},\{3,4\},\Omega\}$.

Let $$X(1)=-1,X(2)=1,X(3)=0,X(4)=0$$ and

$$Z(1)=100,Z(2)=Z(3)=Z(4)=0.$$

Finally let $p_1=P(\{1\})=\frac1{10},\ p_2=P(\{2\})=\frac2{10}, \ p_3=P(\{3\})=\frac2{10}, \ p_4= P(\{4\})=\frac5{10}$.

Now,

$$E[X\mid \mathcal F](\omega)=\begin{cases}\frac{p_2-p_1}{p_1+p_2}=\frac13&\text{ if } &\omega\in A\\ 0&\text{ if } &\omega\in B. \end{cases}$$

At the same time

$$E[ZX\mid \mathcal F](\omega)=\begin{cases}-100\frac{p_1}{p_1+p_2}=-\frac{100}3&\text{ if } &\omega\in A\\ \ \ \ \ 0&\text{ if } &\omega\in B. \end{cases}.$$

That is, $Z$ is able to attenuate the negative part of $X$.