Given a positive random variable $x$ with continuous probability density $f(x)$. What is the main difference between the ordinary mean value
$$\bar{x} = \int_0^\infty\,x\,f(x)\ dx$$
and the root mean squared expression
$$\tilde{x}= \sqrt{\int_0^\infty\,x^2\,f(x)\ dx}.$$
Are there any relations or inequalities between both of them?
Best Answer
It is well known that the variance $\sigma^2$ of $x$ is defined as follows:
$$\sigma^2 = \tilde{x}^2 - \bar{x}^2.$$
The variance is always non-negative. Therefore:
$$\sigma^2 \geq 0 \Rightarrow \tilde{x}^2 - \bar{x}^2 \geq 0 \Rightarrow \tilde{x} \geq |\bar{x}|.$$