Thoughts on a Jensen's Inequality Question

Question

Given a non-negative random variable $X$, and positive integers $n$, $m$, show that if $n \le m$, $[E(X^{n})]^{\frac{1}{n}} \le [E(X^{m})]^{\frac{1}{m}}$.

Analysis

Intuitively, the problem may need techniques like Jensen’s Inequality as there is expectation and concave ($a^{\frac{1}{m}}$) & convex functions ($a^{m}$). However, it would be hard to use Jensen directly without proper transformation. Recall that Jensen’s Inequality states:

$$f(EX) \le Ef(X) ~~~~~~~~~ \text{if $f(.)$ is convex}$$ $$f(EX) \ge Ef(X) ~~~~~~~~~ \text{if $f(.)$ is concave.}$$

Attempt 1

We can try to apply Jensen partially to the LHS to see if it gives us a intermediate expression.

We know that $f(a)=a^n$ is convex on $[0,\infty)$ if $n \ge 1$. Using Jensen, we have

$$E(X^n) \ge (EX)^n \\ [E(X^n)]^{\frac{1}{n}} \ge [(EX)^n]^{\frac{1}{n}}=EX$$

This is not we want as Jensen finds a lower bound for the LHS and proving $EX \le [E(X^{m})]^{\frac{1}{m}}$ does not lead to our goal.

Attempt 2

What about applying to the concave function ($a^{\frac{1}{m}}$)?

We know that $f(a)=a^{\frac{1}{n}}$ is concave on $[0,\infty)$ if $n \ge 1$. Using Jensen, we have

$$[E(X^n)]^{\frac{1}{n}} \ge E(X)^{n\frac{1}{n}}=EX$$

Same lower bound. Not helpful, either!

Attempt 3

The failure in finding a good intermediate form motivates me to look back and examine the correctness of the proposition in a basic case.

Recall that variance of a random variable is always non-negative:

$$Var(X) = E(X-EX)^2 = EX^2-2E(XEX)+(EX)^2$$ $$= EX^2-(EX)^2 \ge 0.$$

We have:

$$(EX)^2 \le EX^2 \Leftrightarrow EX \le (EX^2)^{\frac{1}{2}} \Leftrightarrow (EX^1)^{\frac{1}{1}} \le (EX^2)^{\frac{1}{2}},$$

which is true for non-negative random variable.

In this case, $n=1, ~ m = 2$. Can we try to reverse the general case into a well-formatted form like $(EX)^2 \le EX^2$?

$$(EX^n)^{\frac{1}{n}} \le (EX^m)^{\frac{1}{m}}$$ $$\Leftrightarrow (EX^n)^{\frac{m}{n}} \le EX^m$$ $$\Leftrightarrow (EX^n)^{\frac{m}{n}} \le EX^{n\frac{m}{n}}$$

Boom! The reversed roadmap for transforming basic case into our goal worked! Here we have a random variable $Y=X^n$, convex function $g(a)=a^{\frac{m}{n}}$ as $\frac{m}{n} \ge 1$. The final form can be treated as $g(EY) \le E[g(Y)]$, which is true by Jensen’s Inequality! We can finally prove the proposition!

Proof

Let $Y=X^n, g(a) = a^{\frac{m}{n}}$.

Since

$$g''(a) = \frac{m}{n} \cdot (\frac{m}{n}-1) \cdot a^{\frac{m}{n}-2} \ge 0 ~~\forall a \in [0, \infty),$$

$g(a)$ is convex on $[0,\infty)$.

By Jensen’s Inequality:

$$(EX^n)^{\frac{m}{n}} = (EY)^{\frac{m}{n}} \le E(Y^{\frac{m}{n}}) = E([X^n]^{\frac{m}{n}})=EX^m$$ $$\Leftrightarrow [E(X^{n})]^{\frac{1}{n}} \le [E(X^{m})]^{\frac{1}{m}}. ~~~~~~~~~~~\square$$

Conclusion

We solved the problem by tracing back to the basic case and used it as a guide for proving the general case. Sometimes mathematical gymnastics helps us a lot.