While playing around with random variables, I noticed a following pattern:
If $X \sim Bin(n, p)$, then $E[X] = np$, $\mu(X) = \lfloor np \rfloor$, $|E[X] - \mu(X)| = \{np\} \leq \sqrt{np(1-p)} = \sqrt{Var[X]}$
If $X \sim Geom(p)$, then $E[X] = \frac{1}{p}$, $\mu(X) = - \lfloor \frac{1}{\log_2(1-p)} \rfloor$, $|E[X] - \mu(X)| = |\frac{1}{p} + \lfloor \frac{1}{\log_2(1-p)} \rfloor| \leq \frac{\sqrt{(1-p)}}{p} = \sqrt{Var[X]}$
If $X \sim U[a; b]$, then $E[X] = \mu(X) = \frac{a + b}{2}$ $|E[X] - \mu(X)| = 0 \leq \frac{a-b}{2\sqrt{3}} = \sqrt{Var[X]}$
If $X \sim N(\mu, \sigma)$, then $E[X] = \mu(X) = \mu$ $|E[X] - \mu(X)| = 0 \leq \sigma = \sqrt{Var[X]}$
If $X \sim Exp(\lambda)$, then $E[X] = \frac{1}{\lambda}$, $\mu(X) = - \frac{\ln(2)}{\lambda}$, $|E[X] - \mu(X)| = \frac{\ln(2)-1}{\lambda} \leq \frac{1}{\lambda} = \sqrt{Var[X]}$
If $X \sim Pareto(x, \alpha)$, then $E[X] = \frac{\alpha x}{\alpha - 1}$, $\mu(X) = \alpha 2^{\frac{1}{\alpha}}$, $|E[X] - \mu(X)| = |\frac{\alpha x}{\alpha - 1} - \alpha 2^{\frac{1}{\alpha}}| \leq \frac{x}{\alpha - 1} \sqrt{\frac{\alpha}{\alpha - 2}} = \sqrt{Var[X]}$
It left me wondering, whether it is a part of some general pattern or just a coincidence.
My question is:
Is it true for any random variable $X$ with finite second moment, that $|E[X] - \mu(X)| \leq \sqrt{Var(X)}$?
I tried applying Markov and Chebyshev inequalities to prove this inequality in general, but without success (no idea how to apply them to median).
