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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.

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  • $\begingroup$ Do you know Hölder's inequality? en.wikipedia.org/wiki/H%C3%B6lder%27s_inequality $\endgroup$
    – sharpe
    May 19 '15 at 14:28
  • $\begingroup$ Assuming the inequality I see why it is true. Is there a simple proof of the inequality for this specific case? $\endgroup$ May 19 '15 at 14:58
  • $\begingroup$ On the other hand, I am familiar with Jensen's inequality which can be used to prove that g is monotonic. $\endgroup$ May 19 '15 at 15:32
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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.

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This follows from the general Lyapunov inequality which states that $$\nu_a^{b-c}\nu_c^{a-b}\ge\nu_b^{a-c}\quad\forall a>b>c\ge0,\quad\text{where}\quad \nu_t=\mathbb{E}\left(|X|^t\right)$$

Another form of this inequality is proven here using Hölder's inequality.

Choosing $c=0$ in the generalised inequality we have,

$\nu_a^b>\nu_b^a\quad\forall a>b>0$

$\implies \nu_a^{1/a}>\nu_b^{1/b}\quad\forall a>b>0$

$\implies \nu_t^{1/t}$ is non-decreasing in $t\quad\forall t>0$.


Note that proving the generalised Lyapunov inequality is equivalent to proving $$\alpha g(a)+(1-\alpha)g(c)\ge g\left(\alpha a+(1-\alpha)c\right),$$ where $\alpha=\frac{b-c}{a-c}$ and $g(t)=\ln(\nu_t)$.

Then showing $g$ is convex (by proving $g''\ge0$) proves the inequality.

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