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Let Xbe a non-negative random variable. Prove that

E[X]≤EX21/2≤EX31/3≤⋯

Short Answer

Expert verified

Apply Lyapunov's inequality (proof is given inside) to a random variable0<1<2<3<⋯.

Step by step solution

01

Given Information.

LetX be a nonnegative random variable.

02

Explanation.

Suppose that Xis a nonnegative random variable. The relationship

E[X]≤EX21/2≤EX31/3≤⋯

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

Lyapunov's inequality: If 0<s<t, thenE|X|s1/s≤E|X|t1/t.

Proof: Consider the functionf(y)=|y|r,r>1. The function f(y)is a convex function. Then, to Jensen's inequality,

E[f(Y)]≥f(E[Y])⇔E|Y|r≥|E[Y]|rr=ts>1,letY=|X|sE|X|t≥E|X|st/s

| taking the troot of each side we get:

E|X|t1/t≥E|X|s1/s.

Since X≥0we have that|X|=X. Apply Lyapunov's inequality to a random variable X=localid="1649893415983" 0<1<2<3<⋯.

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