Is there a way to prove that $$ S = \left(\,\left\langle\, x^{4}\,\right\rangle - \left\langle\, x^{2}\,\right\rangle^{2}\,\right) \left(\,\left\langle\, x^{2}\,\right\rangle - \left\langle\, x\,\right\rangle^{2}\,\right) - \left(\,\left\langle\, x^{3}\,\right\rangle - \left\langle\, x^{2}\,\right\rangle \left\langle\, x\,\right\rangle\,\right)^{2}\ >\ 0 $$ for a real random variable defined over $-1\ \leq\ x\ \leq\ 1$ with any distribution function $\,\mathrm{f}\left(\, x\, \right)$ whose support is at least three distinct points ?. This inequality seems to hold for all distribution functions I tried. I am even not sure if the $-1\ \leq\ x\ \leq\ 1$ condition is necessary, I encountered this expression in physics research where $x\equiv \cos\left(\,\theta\,\right)$.

Here the expectation value is defined in the usual way

$$\left\langle\, \mathrm{y}\,\right\rangle \equiv \frac{\displaystyle{\int_{-1}^{1}\mathrm{y}\left(\, x\,\right)\,\mathrm{f}\left(\,x\,\right)\, \mathrm{d}x}} {\displaystyle{\int_{-1}^{1}\mathrm{f}\left(\, x\,\right)\,\mathrm{d}x}} $$

for any $\,\mathrm{y}\left(\, x\,\right)$ function where $\,\mathrm{f}\left(\, x\,\right) \geq 0$ for all $-1 \leq x\leq 1$, and $\,\mathrm{f}\left(\, x\,\right) > 0$ for at least three distinct points $x_{1}$, $x_{2}$, and $x_{3}$.

Note that it is straightforward to show that if the distribution function is uniform and restricted to only three points, $x_{1}$, $x_{2}$, and $x_{3}$, then $$ S = \frac{\left(\, x_{1} - x_{2}\,\right)^{\, 2} \,\left(\, x_{2} - x_{3}\right)^{\, 2} \,\left(\, x_{3} - x_{1}\,\right)^{\, 2}}{27} > 0. $$

  • 1
    $\begingroup$ Potentially write $$\langle y\rangle = \frac{\langle y,f\rangle}{\langle 1,f\rangle}$$ where $\langle f,g\rangle=\int_{-1}^1 f(x)g(x)\mathrm{d}x$ is an inner product, and try to apply Cauchy Schwarz. $\endgroup$
    – Mark
    Aug 21, 2016 at 18:06
  • $\begingroup$ If you expand both products you get a cancellation, so you're just trying to show that: $$\newcommand{\E}[1]{\langle #1\rangle} \E{x^4}\E{x^2}-\E{x^4}\E{x}^2-\E{x^2}^3-\E{x^3}^2+2\E{x^2}\E{x}>0$$ $\endgroup$
    – Mark
    Aug 21, 2016 at 22:39
  • $\begingroup$ To prove that the left hand side is positive $$\langle x^2\rangle - \langle x\rangle^2 = \langle x - \langle x\rangle\rangle^2 > 0 $$ and similarly for the other term. $\endgroup$
    – bkocsis
    Aug 23, 2016 at 6:48

1 Answer 1


Assume that $x$ is a real-valued random variable such that $\left\langle x^4\right\rangle <\infty$ (otherwise, the inequality does not make sense). Using the Cauchy-Schwarz Inequality, we have $$\Big\langle \big(x^2-\left\langle x^2\right\rangle\big)^2\Big\rangle\,\Big\langle \big(x-\left\langle x\right\rangle\big)^2\Big\rangle\geq\Big\langle\big(x^2-\left\langle x^2\right\rangle\big)\,\big(x-\left\langle x\right\rangle\big)\Big\rangle^2\,.$$ The left-hand side of the inequality above is precisely $$\Big(\left\langle x^4\right\rangle -\left\langle x^2\right\rangle^2\Big)\,\Big(\left\langle x^2\right\rangle -\left\langle x\right\rangle^2\Big)\,.$$ The right-hand side of the inequality above is the square of $$ \begin{align}\Big\langle x^3-\langle x\rangle\,x^2-x\,\left\langle x^2\right\rangle+\left\langle x\right\rangle\,\left\langle x^2\right\rangle\Big\rangle&=\left\langle x^3\right\rangle-\left\langle x\right\rangle\,\left\langle x^2\right\rangle-\left\langle x\right\rangle\,\left\langle x^2\right\rangle+\left\langle x\right\rangle\,\left\langle x^2\right\rangle \\&=\left\langle x^3\right\rangle -\left\langle x\right\rangle\,\left\langle x^2\right\rangle\,. \end{align}$$

The inequality $$\Big(\left\langle x^4\right\rangle -\left\langle x^2\right\rangle^2\Big)\,\Big(\left\langle x^2\right\rangle -\left\langle x\right\rangle^2\Big)\geq \Big(\left\langle x^3\right\rangle -\left\langle x\right\rangle\,\left\langle x^2\right\rangle\Big)^2$$ becomes an equality if and only if $x=\left\langle x\right\rangle$ almost surely or there exists $\lambda \in\mathbb{R}$ for which $$x^2-\left\langle x^2\right\rangle=\lambda\,\big(x-\langle x\rangle\big)$$ almost surely. Hence, the equality holds iff there exist $u,v\in\mathbb{R}$ such that $x\in\{u,v\}$ almost surely (i.e., the support of $x$ has at most two distinct points). If the support of $x$ has at least three distinct points, then the inequality is strict.

In fact, for all $p,q\in\mathbb{Z}$ such that $\left\langle x^{2p}\right\rangle$ and $\left\langle x^{2q}\right\rangle$ exist and are finite, we have $$\Big(\left\langle x^{2p}\right\rangle -\left\langle x^p\right\rangle^2\Big)\,\Big(\left\langle x^{2q}\right\rangle -\left\langle x^q\right\rangle^2\Big)\geq \Big(\left\langle x^{p+q}\right\rangle-\left\langle x^p\right\rangle\,\left\langle x^q\right\rangle\Big)^2\,.$$ The inequality becomes an equality if and only if $x^q$ takes only one value almost surely, or there exist $\lambda,\mu\in\mathbb{R}$ such that $x^p-\lambda\,x^q=\mu$ almost surely.

It is possible to extend to $p,q\in\mathbb{R}$, but then we have to worry about the possibility that $x<0$, where a non-integral power of $x$ may not make sense. However, the generalized inequality above is certainly true when $x>0$ almost surely, or when $p,q>0$ and $x\geq 0$ almost surely.


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