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Here you can find the probabilistic proof of the Chebyshev Inequality. I don't understand Step 3 which uses the following inequality:

$$ \mathbb{E}\left[\mathbf{I}_{\{ X^2 >1\}} \right] \leq \mathbb{E}\left[X^2\right] $$

Can you explain why this is true?

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Note that $$X^2 = X^2 I_{\{X^2 > 1\}} + X^2 I_{\{X^2 \leq 1\}} \geq X^2 I_{\{X^2 > 1\}} \geq I_{\{X^2 > 1\}}.$$ Then take expectations on both sides.

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  • $\begingroup$ Thank you. This only works if the bound is 1, right? Imo this is not so easy to see and a small comment in the wikipedia article would help. $\endgroup$
    – stollenm
    Feb 19, 2018 at 10:34
  • $\begingroup$ @stollenm Yes, I use that the lower bound is $1$ in passing to the final inequality. $\endgroup$
    – aduh
    Feb 19, 2018 at 10:38
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You need to recall that \begin{equation} \mathbb{E}\left[\mathbf{I}_{A} \right] = P(X\in A) = \int_A dP \end{equation}

Therefore, \begin{equation} \mathbb{E}\left[\mathbf{I}_{\{X^2>1\}} \right] = \int_{\{X^2>1\}} dP\leq \int_{\{X^2>1\}} X^2dP \leq \int_{\{X^2>1\}\cap\{X^2<1\}} X^2dP =\int_\mathbb{R} X^2 dP \end{equation} because $X^2$ is greater than 1.

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