Markov's inequality is as follows.

Theorem (Markov's Inequality) . Let $\psi: \mathbf{R} \to [0, +\infty]$ be a Borel measurable function and let $\psi_\ast(A) = \inf_{y \in A} \psi(y)$, for any Borel set $A$. Then for any random variable $X$, $$\psi_{\ast}(A) \mathbf{P}(X \in A) \leq \mathbf{E}[\psi(X)\mathbf{1}_A] \leq \mathbf{E} \psi(X).$$

The common, special case is when $\psi(x) = \max(x, 0)$, and $A = [a, \infty)$, where $a > 0$. Then $\psi_\ast(A) = a$, and the outer inequality reads as $\mathbf{E} X_+\geq a \mathbf{P}(X \geq a)$. If $X$ is nonnegative, then $\mathbf{P}(X \geq a) \leq \mathbf{E}(X \geq a)/a$.

Now here's the exercise.

Let $X$ be a non-negative random variable with $\mathrm{Var}(X) \leq 1/2$. Then show $\mathbf{P}(-1 + \mathbf{E}X \leq X \leq 2 \mathbf{E}X) \geq 1/2$. (Hint: use Markov's inequality)

If the assumption implies $\mathbf{E}X \geq 1$, then $$\mathbf{P}(-1 + \mathbf{E}X \leq X \leq 2 \mathbf{E}X) \geq \mathbf{P}(-1 + \mathbf{E}X \leq X \leq 1 + \mathbf{E}X ) \geq 1/2,$$ with the last inequality due to Markov (the first inequality is due to $u \leq 1 + \mathbf{E}X$ implies that $u \leq 2\mathbf{E}X$).

So I would like either a hint to show that $\mathbf{E} X \geq 1$, if that must hold, or a hint to another solution approach (but not the entire solution, of course!)


If $EX < 1$, then $-1+EX < 0$, so \begin{align*} P(-1+EX\le X\le 2EX) = P(X\le 2EX) = 1-P(X>2EX)\,\ge\,1-\frac{EX}{2EX} = \frac 1 2. \end{align*}

  • $\begingroup$ I see, so you're suggesting I split it into two cases, $\mathbf{E}X \geq 1$, and $\mathbf{E} X < 1$. $\endgroup$ – Drew Brady Aug 29 '18 at 5:43
  • $\begingroup$ @DrewBrady Exactly. $\endgroup$ – amsmath Aug 29 '18 at 15:54

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