Let $X$ be real valued random variable such that $\mathbb{E}(X) = 0$. Denote by $X^{+}$ the positive part of $X$, i.e. $X^{+} = \max(0, X)$. I am trying to show that $$\mathbb{E}(X^{+}) \leq \sqrt{\text{Var}(X)}$$.

Is my reasoning correct/could it use any improvements?

Step 1: $\mathbb{E}(X^{+}) \leq \mathbb{E}(|X|)$, and therefore it suffices to show that $\mathbb{E}(|X|)^{2} \leq \text{Var}(X)$.

Step 2: Since $\mathbb{E}(X) = 0$ we have $\text{Var}(X) = \mathbb{E}(X^{2})$ so it suffices to show that $\mathbb{E}(|X|)^{2} \leq \mathbb{E}(X^{2})$.

Step 3: By Jensen's inequality, $\mathbb{E}(|X|)^{2} \leq \mathbb{E}(|X|^{2}) = \mathbb{E}(X^{2})$

  • 1
    $\begingroup$ Just to add on: The inequality is strict since $x^2$ is strictly convex. $\endgroup$
    – rookie
    Jul 19, 2017 at 13:48

1 Answer 1


$\color{green}{\checkmark}$ Looks good to me.

Perhaps show Step $0$: $\left|\begin{align}\Bbb E(X^+) & = \Bbb E(X\mid X\geq 0)\;\Bbb P(X\geq 0)\;+\;\;\Bbb E(0\mid X<0)\;\;\Bbb P(X<0) \\ \Bbb E(|X|) & = \Bbb E(X\mid X\geq 0)\;\Bbb P(X\geq 0)+\Bbb E(-X\mid X<0)\;\Bbb P(X<0) \\ \hline \therefore \Bbb E(X^+) & \leq \Bbb E(|X|)\end{align}\right.$


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