# Expectation Inequality with indicator function

When I read proof of Chebyshev's inequality, I came across the problem.

At first, the proof is : \begin{align*} P(|X-r|\geq k\sigma) &= E(\chi_{|X-\mu| \geq k\sigma}) \\ &= E(\chi_{\left( \frac{X-\mu}{k\sigma} \right)^2 \geq 1}) **from\_here** \\ &\leq E\left( \left( \frac{X-\mu}{k\sigma} \right)^2 \right) **to\_ here** \\ &= \frac{1}{k^2} \frac{E((X-\mu)^2)}{\sigma^2} \\ &= \frac{1}{k^2} \end{align*}

As I make a point above, I cannot understand why the inequality holds.

Could you explain it? Thank you in advance.

• This question has a purely visual answer posted under "Chebyshev's Theorem" at quantdec.com/envstats/notes/class_06/properties.htm. – whuber Jul 19 '18 at 13:06
• This is a little difficult for me... But Thank you very much. I will read it until I understand. – Jourd Jul 19 '18 at 13:13

The inequality depends on the fact that for any $y \in \mathbb{R}$, $$\chi_{y^2 \geq 1} \leq y^2.$$ The proof is by cases. Suppose $y^2 < 1$. Then $$\chi_{y^2 \geq 1} = 0 \leq y^2.$$ Now suppose $y^2 \geq 1$. Then $$\chi_{y^2 \geq 1} = 1 \leq y^2.$$ To obtain the inequality in your question, simply substitute $y = \frac{X - \mu}{k\sigma}$. Since this inequality holds for all values of $X$, it certainly holds in expectation.
To simplify slightly the good answer above (+1 by the way), it's because, for any nonnegative random variable $Y$, $$Y \ge Y \mathbb{1}(Y > 1) \ge \mathbb{1}(Y > 1).$$ Then just take expectations on both sides of both inequalities.
Sorry to change your notation. I don't like the $\chi$ thing.