So I had a probability test and I couldn't really answer this question. It just asked something like this:

"Considering that $X$ is a random variable, $X$ $\geqslant$ $0$, use the correct inequality to prove what's higher or equal, $E(X^2)^3$ or $E(X^3)^2$.

The only thing I could think of was Jensen's Inequality, but I don't really know how to apply it here.

  • 1
    $\begingroup$ Try Holder's inequality instead. $\endgroup$
    – jbowman
    Commented Nov 25, 2017 at 21:59
  • 1
    $\begingroup$ Please add the self study tag. $\endgroup$ Commented Nov 25, 2017 at 23:13
  • 3
    $\begingroup$ The thread at stats.stackexchange.com/questions/244202/… generalizes this question: just take sixth roots of both sides to apply it. $\endgroup$
    – whuber
    Commented Nov 25, 2017 at 23:15
  • 2
    $\begingroup$ Please see the discussion of homework-style questions in the help center $\endgroup$
    – Glen_b
    Commented Nov 26, 2017 at 0:08

3 Answers 3


This indeed can be proven by Jensen inequality.

Hint: Note that for $\alpha > 1$ the function $x^{\alpha}$ is convex in $\left[0, -\infty\right)$ (That's where you use the assumption $X \ge 0$). Then Jensen inequality gives $$ \mathbb{E}\left[Y\right]^{\alpha} \le \mathbb{E}\left[Y^{\alpha}\right] $$ and for $\alpha < 1$, it is the other way arround.

Now, transform the variables to something comparable, and find the relevant $\alpha$.


Lyapunov's Inequality (See: Casella and Berger, Statistical Inference 4.7.6):

For $1 < r < s < \infty$: $$ \mathbb{E}[|X|^r]^\frac{1}{r} \leq \mathbb{E}[|X|^s]^\frac{1}{s} $$


By Jensens' inequality for convex $\phi(x)$: $\phi(\mathbb{E}X) \leq \mathbb{E}[\phi(x)]$

Consider $\phi(Y) = Y^t$, then $(\mathbb{E}[Y])^t \leq \mathbb{E}[Y^t]$ where $Y = |X|^r$

Substitute $t = \frac{s}{r}$: $(\mathbb{E}[|X|^r])^{\frac{s}{r}} \leq \mathbb{E}[|X|^{r\frac{s}{r}}]$ $\implies \mathbb{E}[|X|^r]^\frac{1}{r} \leq \mathbb{E}[|X|^s]^\frac{1}{s}$

In general for $X >0$ this implies:

$ \mathbb{E}[X] \leq (\mathbb{E}[X^2])^\frac{1}{2} \leq (\mathbb{E}[X^3])^\frac{1}{3} \leq (\mathbb{E}[X^4])^\frac{1}{4} \leq \dots $


Suppose X has a uniform distribution on [0,1] then E(X$^2$)= $\frac{1}{3}$ and so E(X$^2$)$^3$ = $\frac{1}{27}$ and E(X$^3$)=$\frac{1}{4}$ so E(X$^3$)$^2$= $\frac{1}{16}$. So in this case E(X$^3$)$^2$ > E(X$^2$)$^3$. Can you generalize this or find a counterexample?

  • $\begingroup$ Very vague answer. The OP is asked to prove the correct statement. There is no counterexample at all. $\endgroup$
    – Zhanxiong
    Commented Jan 15, 2018 at 5:38

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.