# Probabilistic Proof of Chebyshev Inequality - Step 3

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?

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.
• @stollenm Yes, I use that the lower bound is $1$ in passing to the final inequality.
You need to recall that $$\mathbb{E}\left[\mathbf{I}_{A} \right] = P(X\in A) = \int_A dP$$
Therefore, $$\mathbb{E}\left[\mathbf{I}_{\{X^2>1\}} \right] = \int_{\{X^2>1\}} dP\leq \int_{\{X^2>1\}} X^2dP \leq \int_{\{X^2>1\}\cap\{X^2<1\}} X^2dP =\int_\mathbb{R} X^2 dP$$ because $X^2$ is greater than 1.