Inequality with Expectations of an Indicator Function

Question

Let $X$ be a random variable. We have:

$$E\left[\mathbb{1}\left\{X>0 \right\}\mid A\right]>E\left[\mathbb{1} \left\{X>0 \right\}\mid B\right]$$

where $\mathbb{1}\left\{\cdot\right\}$ represents an indicator function.

Can I show that this implies $E\left[X\mid A\right]>E\left[X\mid B\right]$ ?

My intuition suggests that this implication doesn't hold, but I'm not sure how to go about showing it.

Answer 1

Here's a counterexample: Suppose $X\mid A$ is always equal to $1$, so that $E[1\{X>0\}\mid A]=1$ and $E[X\mid A]=1$, but $X\mid B$ is distributed such that it is zero with probability 0.9 and 100 with probability 0.1. Then $E[1\{X>0\}\mid B]=0.1$ and $E[X\mid B] = 10$

Answer 2

If you let $A = C\cup D$, with $C$ and $D$ disjoint,
$$ x(w) = \begin{cases} -1 & w\in B \\ -5 & w\in C\\ 1 & 2\in D \end{cases} $$

Then clearly $\mathbb{E}[1\{X>0\}\vert B] = \mathbb{P}(X>0\vert B) = 0$.

Now if you assume that $\mathbb{E}[1\{X>0\}\vert C] = 0.5 = \mathbb{E}[1\{X>0\}\vert B]$, we have $$\mathbb{E}[X\vert A] = -1*0.5 + -100*0.5 = -2 < \mathbb{E}[X\vert B] = -1 $$