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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 was the Jensen's Inequality, but I don't really know how to apply it here.

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    $\begingroup$ Try Holder's inequality instead. $\endgroup$ – jbowman Nov 25 '17 at 21:59
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    $\begingroup$ Please add the self study tag. $\endgroup$ – Michael Chernick Nov 25 '17 at 23:13
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    $\begingroup$ The thread at stats.stackexchange.com/questions/244202/… generalizes this question: just take sixth roots of both sides to apply it. $\endgroup$ – whuber Nov 25 '17 at 23:15
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    $\begingroup$ Please see the discussion of homework-style questions in the help center $\endgroup$ – Glen_b Nov 26 '17 at 0:08
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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 $\alpha$.

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

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

Proof:

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 $

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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?

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  • $\begingroup$ Very vague answer. The OP is asked to prove the correct statement. There is no counterexample at all. $\endgroup$ – Zhanxiong Jan 15 '18 at 5:38

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