# Variance of the modulus of a random variable

Let $X$ be a random variable with mean $\mu$ and variance $\sigma^2$. What is the upper-bound on the variance of $Y=\left|X\right|$?

My gut feeling says that $\operatorname{Var}(Y) \leq \operatorname{Var}(X)$ because 'modulus' is a many-to-one function.

Note :- It is easy to see that if $X$ takes only positive values, $\operatorname{Var}(Y) = \operatorname{Var}(X)$

• stats.stackexchange.com/questions/5782/… - generalizes the above question. – Vivek Bagaria Jun 27 '16 at 5:06
• What do you mean by "the" upper bound? Why would there only be one such bound? Which kind of a bound are you seeking? There's a simple upper bound to be found (and rather easily found), but it's not clear if it's what you seek. How does this problem arise? – Glen_b Jun 27 '16 at 7:21
• Glen_b: I am seeking a upper-bound in terms of $\mu$, $\sigma$ and other possible quantities. – Vivek Bagaria Jun 27 '16 at 7:38
• Clearly. That doesn't doesn't really deal with what I was getting at. $1$. Why "the" rather than "a"? What makes it unique? $2$. Why do you need such an upper bound? Where does the problem come from? – Glen_b Jun 27 '16 at 8:00
• 1) "the"/"a" was due ti my bad grammar. 2) The problem is only from observations with some real data and I don't have any source for this. – Vivek Bagaria Jun 27 '16 at 8:50

So $$\def\var{\text{var}} \var\bigl( |X| \bigr) = E\left(X^2\right) - E\bigl( |X| \bigr)^2.$$ You know how to write $E(X^2)$ in terms of $\mu$ and $\sigma$.

Now define a new random variable $X^+$ by $X^+ = X$ if $X>0$, and $X^+=0$ if $X\le 0$; similarly let $X^- = X$ if $X < 0$ and $X^-=0$ if $X\ge 0$.

Assuming both $E\left(X^+\right)$ and $E\left(X^-\right)$ exist, show that $$\var\bigl( |X| \bigr)= \var(X) + 4E\left(X^+\right)E\left(X^-\right).$$

Show that this is $\le \var(X)$, and check that the bound is tight.

• this is pretty clearly (or at least very likely to be) a self-study question (which the OP is being oddly cagey about), and you should probably stick to the guidelines for them (i.e. hints and guidance rather than explicit solutions) until it's clear that it isn't. – Glen_b Jun 27 '16 at 8:04
• @Glen_b I did not take it as such, and was truly intrigued by the question (as I am fully autodidact in stats there are many classical exercises that I don’t know), that’s why I started to reply. I edited the answer. – Elvis Jun 27 '16 at 8:44
• I've deleted mine; your answer is more complete. – Glen_b Jun 27 '16 at 9:32

We know that $$\;\;\;\;X \leq |X|\\ \Rightarrow E\big(X\big) \leq E\big(|X|\big)\\ \Rightarrow E\big(X\big)^2 \leq E\big(|X|\big)^2$$

Using the above in $$\def\var{\text{var}} \var\bigl( |X| \bigr) = E\left(X^2\right) - E\bigl( |X| \bigr)^2.$$

we get

$$\var\bigl( |X| \bigr) \leq E\left(X^2\right) - E\bigl( X \bigr)^2 = \var(X)$$