# Probabilistic inequality for an antisymmetric function

Let $$X$$ be a random variable with $$\mathbb P(-1 \leq X \leq 1) = 1$$. Does $$\mathbb E(X) \geq 0$$ imply $$\mathbb E[X(1-|X|)] \geq 0?$$ The function is antysymmetric around $$0$$ and has more probability mass on the positive side. Intuitively, it should be correct.

If not, is it true if I assume $$\mathbb E(X) > 0$$?

Let $$X$$ take the values $$-\frac 1 2$$ with probability $$\frac 1 4$$ and $$1$$ with probability $$\frac 3 4$$. You can check that $$EX >0$$ but $$E(X(1-|X|) <0$$.

• Ok thank you! Would have been nice, if the inequality is true :/ – Danijel Jan 14 at 12:40

The answer is no.

A simple counter-example is letting $$X\equiv 1$$.
meaning, $$X$$ is a constant random variable equal to 1.

We get $$\mathbb E[X(1-\vert X\vert)] = \mathbb E[X(1-\vert 1\vert)] = 0$$

• Ok sorry, I was a bit imprecise. I don't need strict inequalities. What if I replace all $<$ by $\leq$?. – Danijel Jan 14 at 12:32
• Again a bit imprecise... I meant all $>$ by $\geq$. I will edit the question. – Danijel Jan 14 at 12:36