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

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

• What is $\langle X\rangle$ ? Feb 12, 2020 at 23:15
• the expectation value $\langle X \rangle = \text{E}[X]$ Feb 12, 2020 at 23:17

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.

• Thanks a lot of the proof, that's very helpful! When you say "The required inequality also shows that the function f...". How do you go about proving that? Also using Hölder's inequalities for different values of $\alpha$? Mar 5, 2021 at 22:58
• @Chachni Let's say we have $\alpha_1$ and $\alpha_2$ in $[0,\beta)$, with $\alpha_1<\alpha_2$. Consider $Y=X^{\alpha_2}$ and $\alpha:=\dfrac{\alpha_1}{\alpha_2}$. Then, we have $$f(\alpha_2)=\dfrac{\mathbb{E}[X^{2\alpha_2}]}{\big(\mathbb{E}[X^{\alpha_2}]\big)^2}=\dfrac{\mathbb{E}[Y^2]}{\big(\mathbb{E}[Y]\big)^2}$$ and $$f(\alpha_1)=\dfrac{\mathbb{E}[X^{2\alpha_1}]}{\big(\mathbb{E}[X^{\alpha_1}]\big)^2}=\dfrac{\mathbb{E}[Y^\alpha]}{\big(\mathbb{E}[Y^{\alpha}]\big)^2}\,.$$ Mar 9, 2021 at 17:30
• Since $\alpha\in[0,1)$, we get $$f(\alpha_1)=\dfrac{\mathbb{E}[Y^\alpha]}{\big(\mathbb{E}[Y^{\alpha}]\big)^2}\leq \dfrac{\mathbb{E}[Y^2]}{\big(\mathbb{E}[Y]\big)^2}=f(\alpha_2)\,.$$ Mar 9, 2021 at 17:30
• Awesome, thanks! Also wondering if you have thoughts about: math.stackexchange.com/questions/4050762/… Mar 9, 2021 at 18:22

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 $$2+\alpha\ge\max\{2,1+\alpha\}$$.

• isn't your last inequality $\sum_{perm} x^2y^\alpha z^\alpha \ge \sum_{perm} x^{2\alpha}yz$ just Muirhead's inequality? $w_1\begin{bmatrix} 2 \\ a \\ a \end{bmatrix} + w_2\begin{bmatrix} a \\ 2 \\ a \end{bmatrix} + w_3\begin{bmatrix} a \\ a \\ 2 \end{bmatrix} = \begin{bmatrix} 2a \\ 1 \\ 1 \end{bmatrix}$ where $w_i \geq 0$ and $\sum_i w_i =1$ Feb 13, 2020 at 0:49
• @user8675309 Aha, well done! I didn't know this inequality. Feb 13, 2020 at 1:33