The map $g(t)=(E|X|^t)^{\frac 1t}$ is monotonic.

Question

Let $X$ be a random variable.
Prove that $g:(0,\infty)\rightarrow[0,\infty]$ which is defined by $g(t)=(E|X|^t)^{\frac 1t}$ ($E$ marks the expected value) is monotonic.

I tried experimenting with Markov's inequality but I had a hard time dealing with it since $X$ is not in a specific class of random variables.

Answer 1

Fix $s\lt t$ and define $p:=t/s \gt 1$. The map $x\mapsto |x|^p$, is convex, hence for each non-negative random variable $Y$, we have $$\left(\mathbb E[Y]\right)^p\leqslant \mathbb E[Y^p].$$
Using this with $Y=|X|^s$, we derive that $$\left(\mathbb E[|X|^s]\right)^{t/s} \leqslant \mathbb E[|X|^{s\cdot t/s} ],$$ from which the wanted inequality follows.