Question

Let $X$be a non-negative random variable. Prove that

$E\left[X\right]\le {\left(E\left[X^2\right]\right)}^{1/2}\le {\left(E\left[X^3\right]\right)}^{1/3}\le \cdots$

Short answer

Apply Lyapunov's inequality (proof is given inside) to a random variable$0<1<2<3<\cdots$.

Step-by-step solution

Given Information.

Let$X$ be a nonnegative random variable.

Explanation.

Suppose that $X$is a nonnegative random variable. The relationship

$E\left[X\right]\le {\left(E\left[X^2\right]\right)}^{1/2}\le {\left(E\left[X^3\right]\right)}^{1/3}\le \cdots$

we need to prove the consequence of Lyapunov's inequality, so let's prove this general result.

Lyapunov's inequality: If $0<s<t\text{, then}{\left(E\left[|X{|}^s\right]\right)}^{1/s}\le {\left(E\left[|X{|}^t\right]\right)}^{1/t}\text{.}$

Proof: Consider the function$f\left(y\right)=|y{|}^r,r>1$. The function $f\left(y\right)$is a convex function. Then, to Jensen's inequality,

$$\begin{gathered} E\left[f\right(Y\left)\right]\ge f\left(E\right[Y\left]\right)\iff E\left[|Y{|}^r\right]\ge \left|E\right[Y]{|}^r \\ r=\frac{t}{s}>1,\text{let}Y=|X{|}^s \\ E\left[|X{|}^t\right]\ge {\left(E\left[|X{|}^s\right]\right)}^{t/s} \end{gathered}$$

| taking the $t$root of each side we get:

${\left(E\left[|X{|}^t\right]\right)}^{1/t}\ge {\left(E\left[|X{|}^s\right]\right)}^{1/s}.$

Since $X\ge 0$we have that$\left|X\right|=X$. Apply Lyapunov's inequality to a random variable $X$=localid="1649893415983" $0<1<2<3<\cdots$.