# Expand the inequality $E(|X|^r)^\frac{1}{r}\le E(|X|^s)^\frac{1}{s}$ to $-\infty<r<s<\infty$

For a random variable $$X$$ and numbers $$0

$$E(|X|^r)^\frac{1}{r}\le E(|X|^s)^\frac{1}{s}$$

This almost immediately follows from Holder's inequality for random variables $$Y,Z$$

$$E(|ZY|)\le E(|Z|^p)^\frac{1}{p}E(|Y|^q)^\frac{1}{q}$$ Setting $$Y=1$$ we end up with

$$E(|Z|)\le E(|Z|^p)^\frac{1}{p}.$$

With $$|Z|=|X|^r$$ we get

$$E(|X|^r)\le E(|X|^{rp})^\frac{1}{p}.$$

Now, since $$p \ge1$$ we have $$rp \ge r$$ and setting $$rp=s$$ we get

$$E(|X|^r)\le E(|X|^{s})^\frac{r}{s}$$

or

$$E(|X|^r)^\frac{1}{r}\le E(|X|^{s})^\frac{1}{s}.$$

Is there a way to expand this inequality to $$-\infty?

• The original result you state at the beginning is a certain Lyapunov inequality discussed here previously. Oct 14 '18 at 14:17

Yes, assuming the expectations exist (for $$r=0$$, it is $$\lim_{r\to 0}(\mathbb{E}[|X|^r])^{1/r}=\exp(\mathbb{E}[\log |X|])$$). This is just the power mean inequality in the discrete case. So the lazy way to prove it is simply take limit of discrete case. A more honest way to prove this is to establish comparing $$r=0$$ with $$s>0$$ (Jensen's inequality with $$\varphi(x)=-\log x$$ for the random variable $$|X|^s$$)), and then use $$Y=X^{-1}$$ for first the case $$r=-1$$, $$s=0$$, and finally for general $$r,s$$.