Given a random variable $X$, from the definition of the variance, $\mathbb{E}[X^2] \geq \mathbb{E}[X]^2$. It seems intuitive that $\mathbb{E}[|X|] \geq |\mathbb{E}[X]|$ should be true. Is this the case?
$\begingroup$
$\endgroup$
5
-
2$\begingroup$ That is almost true : It might be "Equal" too. Change $\gt$ to $\ge$. $\endgroup$– PremMay 11 at 15:05
-
1$\begingroup$ Try and look up Jensen's Inequality. If your intuition tells you that expectation behaves just like the "integration" operator then that is because it is an integration operator $\endgroup$– Mr.Gandalf SauronMay 11 at 15:08
-
3$\begingroup$ Apply expectation to $-|X| \leq X \leq |X|$. $\endgroup$– azif00May 11 at 15:26
-
2$\begingroup$ Expectation inequality involving absolute values is a duplicate question. $\endgroup$– Sarvesh Ravichandran IyerMay 11 at 15:44
-
$\begingroup$ I made the change > to ≥. $\endgroup$– user3679030May 11 at 21:01
Add a comment
|
2 Answers
$\begingroup$
$\endgroup$
They could be equal, but other than that you're basically right. The intuition is that $E[X]$ is a sum with both positive and negative terms, some of which cancel each other out, but $E[|X|]$ has the same terms but all made positive. (This assumes a discrete random variable; if your random variables are continuous, replace sums with integrals.)
More generally: These are both examples of Jensen's inequality, which says that $E(f(X)) \ge f(E(X))$ for a random variable X and a convex function $f$. When we have $f(x) = x^2$ this becomes $E(X^2) \ge E(X)^2$ (the variance is positive); when we have $f(x) = |x|$ this becomes $E(|X|) \ge |E(X)|$, as you observed.
$\begingroup$
$\endgroup$
If $f(x)$ is the pdf of $X$ you have
$$\mathbb{E}[|X|]=\int_Df(x)|x|\mathrm{d}x$$ and $$|\mathbb{E}[X]|=\left|\int_D f(x)x\mathrm{d}x\right|$$ Appying the absolute value properties: $$|\mathbb{E}[X]|=\left|\int_D f(x)x\mathrm{d}x\right|\leq \int_D |f(x)\cdot x|\mathrm{d}x=\int_D |f(x)|\cdot|x|\mathrm{d}x=\int_D f(x)|x|\mathrm{d}x=\mathbb{E}[|X|]$$ So $$|\mathbb{E}[X]|\leq\mathbb{E}[|X|]$$
Note that a $f(x)$ is always non negative, so $|f(x)|=f(x)$
