By Chebyshev's inequality it is known that $$\mathbb{P}\left(|X-\mu|<k\sigma\right) \geq 1-\frac{1}{k^2}\,.$$ Then does it follow that $$\mathbb{P}\left((\mu-k\sigma)^2 \leq X^2 \leq (\mu +k\sigma)^2\right) \geq 1-\frac{1}{k^2}$$ assuming $\mu > k\sigma$?

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    $\begingroup$ Welcome to CV, Irna. What do the rules of algebra tell you? Have you considered plugging in some simple values for $\mu,$ $\sigma,$ and $k$ to check? Try a negative value for $\mu$ :-). $\endgroup$
    – whuber
    Apr 18, 2023 at 15:24
  • $\begingroup$ Hi Whuber, my intuition tells me yes - if $f$ is a monotonically increasing function and $1-1/k^2$ elements take on values $x \in [L, U]$, then those same elements will satisfy $f(x) \in [f(L), f(U)]$. In this case $f(x) = x^2$. I’m wondering if my intuition is correct. $\endgroup$
    – Irna Mosa
    Apr 18, 2023 at 15:39
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    $\begingroup$ Don't use intuition: use mathematics. $\endgroup$
    – whuber
    Apr 18, 2023 at 15:41
  • $\begingroup$ Yikes. Square rooting the second inequality gives the first one. This is not a good first impression I’m making on this forum. $\endgroup$
    – Irna Mosa
    Apr 18, 2023 at 15:48
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    $\begingroup$ Okay, I will plug some numbers in for you. Suppose $X$ has a $1/2$ chance to equal $1$ and a $1/2$ chance to equal $3,$ so that $\mu=2$ and $\sigma=1.$ Let $k = 6$ so that $(\mu-k\sigma)^2=(2-(6)(1))^2= 16.$ Thus, $\Pr((\mu-k\sigma)^2\le X^2)=0.$ $\endgroup$
    – whuber
    Apr 18, 2023 at 18:09

1 Answer 1


Consider the graph of $|X-\mu|$ which is like a triangle, and the graph of $k\sigma$ which is like a straight line. In the example image below we use $k=3$

example of the two graphs together

The inequality

$$|X-\mu| < k\sigma$$

translates into $$\mu-k\sigma <X < \mu+k\sigma$$

If you take the square (which is a non-injective, ie. many-to-one or non-invertible, function) then this covers two regions because the cases $\mu-k\sigma <-X < \mu+k\sigma$ will also satisfy the squared inequality.

So, if $\mu > k\sigma$ then

$$\mathbb{P}\left((\mu-k\sigma)^2 < X^2 < (\mu +k\sigma)^2\right) = \mathbb{P}\left(\mu-k\sigma < -X < \mu +k\sigma \right) +\mathbb{P}\left(\mu-k\sigma < X < \mu +k\sigma \right) \geq \mathbb{P}\left(\mu-k\sigma < X < \mu +k\sigma \right) = \mathbb{P}\left(|X-\mu|<k\sigma\right) \geq 1-\frac{1}{k^2}$$

If $X>0$ then the inequality in the middle becomes an equality.

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    $\begingroup$ This derivation strongly depends on $\mu > k\sigma$ which ensures that both bounds $\mu-k\sigma$ and $\mu+k\sigma$ are positive. If the lower bound $\mu-k\sigma$ is negative, then the region also contains points like $X=0$, and these will 'dissapear' when you square all terms. You will have to use instead $$\mathbb{P}\left(0 < X^2 < (\mu +k\sigma)^2\right) $$ $\endgroup$ Apr 19, 2023 at 6:36

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