Prove $\mathbf{E}[X^2] \geq \frac{ \mathbf{E}[X^{2\alpha}]}{ \mathbf{E} [X^{\alpha}]^2}$

Question

I have a square-integrable random variable $X > 0$ with $\mathbf{E}[X] = 1$ and $0 < \alpha < 1$. I would like to prove that: \begin{equation} \mathbf{E}[X^2] \geq \frac{ \mathbf{E}[X^{2\alpha}]}{ \mathbf{E} [X^{\alpha}]^2} \end{equation} It seems to be true numerically, but I can't come up with the analytical proof. Any ideas? Thank you!

Answer 1

The required inequality follows from two applications of Hölder's inequality. Firstly, note that $$\Big(\mathbb{E}\left[X^2\right]\Big)^{\frac{1-\alpha}{2-\alpha}}\,\Big(\mathbb{E}\left[X^\alpha\right]\Big)^{\frac{1}{2-\alpha}}\geq \mathbb{E}\left[X\right]\,.$$ Therefore, $$\frac{\mathbb{E}\left[X^2\right]}{\big(\mathbb{E}\left[X\right]\big)^2}\geq \frac{\Big(\mathbb{E}\left[X^2\right]\Big)^{\frac{\alpha}{2-\alpha}}}{\Big(\mathbb{E}\left[X^\alpha\right]\Big)^{\frac{2}{2-\alpha}}}\,.$$ Secondly, we have $$\Big(\mathbb{E}\left[X^2\right]\Big)^{\frac{\alpha}{2-\alpha}}\,\Big(\mathbb{E}\left[X^\alpha\right]\Big)^{\frac{2-2\alpha}{2-\alpha}}\geq \mathbb{E}\left[X^{2\alpha}\right]\,.$$ This shows that $$\frac{\Big(\mathbb{E}\left[X^2\right]\Big)^{\frac{\alpha}{2-\alpha}}}{\Big(\mathbb{E}\left[X^\alpha\right]\Big)^{\frac{2}{2-\alpha}}}\geq \frac{\mathbb{E}\left[X^{2\alpha}\right]}{\big(\mathbb{E}\left[X^{\alpha}\right]\big)^2}\,.$$ Thus, for any nonnegative random variable $X$ such that $\mathbb{E}\left[X^2\right]$ is a finite positive real number, we have $$\frac{\mathbb{E}\left[X^2\right]}{\big(\mathbb{E}\left[X\right]\big)^2} \geq \frac{\mathbb{E}\left[X^{2\alpha}\right]}{\big(\mathbb{E}\left[X^{\alpha}\right]\big)^2}$$ for any $\alpha\in [0,1]$ (where $0^0$ is defined to be $1$). The inequality becomes an equality if and only if $\alpha=1$ or $X$ is almost surely constant.

As before, $X$ is a nonnegative random variable such that $\mathbb{E}\left[X^2\right]$ is a finite positive real number. Let $\beta\in[1,\infty]$ denote the supremum of the set of positive real numbers $\alpha$ such that $\mathbb{E}\left[X^{2\alpha}\right]$ is finite. The required inequality also shows that the function $f:[0,\beta)\to[0,\infty)$ defined by $$f(\alpha):=\frac{\mathbb{E}\left[X^{2\alpha}\right]}{\big(\mathbb{E}\left[X^{\alpha}\right]\big)^2}$$ for every $\alpha\in[0,\beta)$ is nondecreasing. It is strictly increasing unless $X$ is almost surely constant.

Answer 2

The inequality in question is equivalent to the homogeneous $$ \mathbf{E}[X^2]\cdot\mathbf{E}[X^{\alpha}]^2 \ge \mathbf{E}[X^{2\alpha}]\cdot\mathbf{E}[X]^2, $$ or equivalently, if $\mu$ is the density of $X$, $$ \int_0^\infty x^2\,d\mu(x) \int_0^\infty y^\alpha\,d\mu(y) \int_0^\infty z^\alpha\,d\mu(z) \ge \int_0^\infty x^{2\alpha}\,d\mu(x) \int_0^\infty y\,d\mu(y) \int_0^\infty z\,d\mu(z) . $$ The inequality would follow if it is true that $$ \sum_{sym} x^2y^\alpha z^\alpha \ge \sum_{sym} x^{2\alpha}yz $$ for any positive $x,y,z$, where $\sum_{sym}$ represents the sum of the six terms resulting from applying all six possible permutations to $\{x,y,z\}$.

As user8675309 commented, this last statement is a special case of Muirhead's inequality, since $2\ge\max\{1,2\alpha\}$ and $2+\alpha\ge\max\{2,1+\alpha\}$.