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?
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?
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.
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.
For every random variable $X$, the expression is twice $\mathrm{Var}(X)+\mathrm{Var}(X^2-X)$.