Chebyshev's Inequality proof explanation - squared values of probability

Question

In the proof of Chebyshev's Inequality we do the following:

Claim: for some random variable $Y$ and some value $a > 0$

$Pr[|Y - E[Y]| \geq a] = \frac{E[(Y - E[Y])^2]}{a^2}$

Let's refer to $E[Y]$ as $\mu$

\begin{align*} Pr[|Y - \mu| \geq a] =&\ Pr[|Y - \mu|^2 \geq a^2] \\ =& \ Pr[X \geq a^2]\ \ \ \ if \ X = (Y - \mu)^2\\ \leq& \ \frac{E[X]}{a^2}\ \ \ \ by \ Markov's \ inequality\\ =& \ \frac{E[(Y - E[Y])^2]}{a^2} \end{align*}

My question is, how can we just assert:

$Pr[|Y - \mu| \geq a] =\ Pr[|Y - \mu|^2 \geq a^2]$

What is the intuition or line of thought behind this?

Answer 1

Claim: $|Y-\mu|\ge a$ holds $\Leftrightarrow$ $|Y-\mu|^2\ge a^2$ holds.

Proof:

In general, if $g$ is a non-decreasing real valued function then applying $g$ to both sides of an inequality preserves the inequality.

"$\Rightarrow$": Let $g(x) = x^2$ for $x\ge0$ and $g(x)=x$ for $x<0.$

"$\Leftarrow$": Let $g(x) = \sqrt x$ for $x\ge0$ and $g(x)=x$ for $x<0.$

This shows that the events are identical, so their probabilities are equal.

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Here's another way to think about this:

Imagine the probabilities were not the same. Suppose $P(|Y-\mu|\ge a)>P(|Y-\mu|^2\ge a^2).$ Then with positive probability, $|Y-\mu|\ge a$ holds but $|Y-\mu|^2\ge a^2$ does not hold, which is a contradiction. Similiarly, the second probability can't be greater. Therefore they must be equal.

Answer 2

Another way to see it is that $$\mathbb{P}(|X|\geq a)=\mathbb{E}\left[1_{|X|\geq a}\right]$$ Now look at the indicator function: $$1_{|X|\geq a}=\begin{cases} 1 & \text{if }|X|\geq a\\ 0 & \text{if}\,|X|<a \end{cases}=\begin{cases} 1 & \text{if }|X|^{2}\geq a^{2}\\ 0 & \text{if}\,|X|^{2}<a^{2} \end{cases}$$ As it preserves inequalities in this setting. Meaning: $1_{|X|\geq a}=1_{|X|^{2}\geq a^{2}}$. It is the same analogy as Epiousios, but just specifying that the indicator function is equal in both cases.