# Relation between two statistical measures

Consider a random variable $$X$$ with the associated density function $$f_X(x)$$ and "zero" mean.

Define the following quantities:

(1) $$E[X^2] := \int_{-\infty}^{+\infty} x^2 f_X(x) dx$$

(2) $$E[ |X| ] := \int_{-\infty}^{+\infty} |x| f_X(x) dx$$

I see that $$E[X^2]$$ is the variance (noting that the mean is zero). But I have no idea if $$E[|X|]$$ is already well-known and useful in the probability context.

Anyway, here is the question: I wonder if there is some functional (in)equality between $$E[X^2]$$ and $$E[|X|]$$. Something of the following form: the existence of a mapping $$\rho : \mathbb{R}_{\geq 0} \to \mathbb{R}_{\geq 0}$$ such that $$E[|X|] \leq \rho(E[X^2])$$.

You may make a fair assumption on the density function $$f_X(x)$$ if required. The zero mean assumption is made to make the life easier. You may also drop it if necessary.

• This is a special case of stats.stackexchange.com/questions/244202. – whuber Apr 30 '19 at 17:44
• $E(|X-\mu|)$ is often called the mean deviation or the mean absolute deviation. The relationship between mean deviation and standard deviation is discussed in several questions on site. One example -- though in answer to a narrower question -- is here; note that the result there uses the standard (convex) version of Jensen's inequality. – Glen_b -Reinstate Monica Apr 30 '19 at 23:39

By Jensen's inequality (for concave functions instead of convex functions), we have $$E[|X|] = E[\sqrt{X^2}] \leq \sqrt{E[X^2]}.$$ Alternatively, the Cauchy-Schwarz inequality can be used to yield the same ineuqality: $$E[|X|]^2 = E[|X|\cdot 1]^2 \leq E[|X|^2] E[1^2] = E[X^2].$$ This makes no assumptions on the existence of a probability density or having mean zero (or any mean for that matter).