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Say $X$ ~ $Beta(\alpha,\beta)$. I want to prove the following,

$$4E(X)E(X^2)-2E(X^2)^2-4E(X)^2+2E(X^4)-4E(X^3)+4E(X^2)>0$$

Is there a simple way?

3 Answers 3

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  • $4E(X^2) - 4E(X)^2 = 4\cdot\text{var}(X) \geq 0$
  • $2E(X^4) -2E(X^2)^2 = 2\left[E((X^2)^2) - (E(X^2))^2\right] = 2\cdot\text{var}(X^2) \geq 0$
  • $4E(X^3) - 4E(X)E(X^2) = 4\cdot\text{cov}(X,X^2)$

So, writing $Y = X^2$, your expression is $$\begin{align*} 4\cdot\text{var}(X) + 2\cdot\text{var}(Y) - 4\cdot\text{cov}(X,Y) &= 2\cdot\text{var}(X) + 2\cdot\text{var}(X-Y)\\ &= 2\cdot[\text{var}(X) + \text{var}(X^2-X)] \end{align*}$$ as Didier said more succinctly as I was composing my answer.

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The $j$'th moment of the Beta distribution is $$E[X^j] = \dfrac{\alpha(\alpha+1)\ldots(\alpha+j-1)}{(\alpha+\beta)(\alpha+\beta+1)\ldots(\alpha+\beta+j-1)} $$ Expand out $ 4E(X)E(X^2)-2E(X^2)^2-4E(X)^2+2E(X^4)-4E(X^3)+4E(X^2)$ and you get a rather complicated expression in $\alpha$ and $\beta$ whose numerator and denominator have all nonnegative coefficients.

  • 1
    How can you be sure the $numerator$ of the expanded expression will have only nonnegative summands? (note also that I require >0, not >=0)2012-07-13
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    Robert: Did you do what you suggest to do?2012-07-13
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    Yes @did, I did (with help from Maple).2012-07-13
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For every random variable $X$, the expression is twice $\mathrm{Var}(X)+\mathrm{Var}(X^2-X)$.