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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 $\mathbb E(X) > 0$?

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2 Answers 2

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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 $E(X(1-|X|) <0$.

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  • $\begingroup$ Ok thank you! Would have been nice, if the inequality is true :/ $\endgroup$
    – Danijel
    Commented Jan 14, 2020 at 12:40
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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 $$\mathbb E[X(1-\vert X\vert)] = \mathbb E[X(1-\vert 1\vert)] = 0$$

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  • $\begingroup$ Ok sorry, I was a bit imprecise. I don't need strict inequalities. What if I replace all $<$ by $\leq$?. $\endgroup$
    – Danijel
    Commented Jan 14, 2020 at 12:32
  • $\begingroup$ Again a bit imprecise... I meant all $>$ by $\geq$. I will edit the question. $\endgroup$
    – Danijel
    Commented Jan 14, 2020 at 12:36

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