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Suppose $X$ is a (whatever bounded or not) continuous random variable ($X$ is not constant) with an arbitrary distribution. Is it possible to construct a distribution s.t. $\text{Corr}(X, \text{sign}(X)\cdot X^2)=0$? If not, approximately what's the minimum (in terms of absolute values) of this correlation? Say, is it possible to achieve Pearson correlation below 0.5 for certain continuous distribution?

I read from this post that $\text{Corr}(X, X^2)=0$ if $X$ is symmetric about zero, but I cannot fiture it out when $\text{sign}(X)$ is added. I tried some plots like this, but I felt the correlation is high intuitively.

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    $\begingroup$ Consider any distribution with a finite absolute third moment but infinite fourth moment, such as any Student t distribution having between 3 and 4 degrees of freedom. $\endgroup$
    – whuber
    Oct 19, 2023 at 12:46

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For any random variable (regardless continuous or discrete) that is symmetric about $0$, since $\operatorname{Cov}(X, \operatorname{sign}(X)X^2) = E[|X|^3] \geq 0$, it can be seen that the correlation of your interest is bounded below by $0$. And since $E[|X|^3] = 0$ if and only if $|X|^3 = 0$ almost surely, that is $X = 0$ almost surely, no non-degenerate continuous random variable can attain this $0$ lower bound.


Your second conjecture can be negated by using whuber's comment. Let $X \sim t_\nu$, where $\nu > 4$, it then follows by Theorem 2.2 of MOMENTS OF STUDENT’S T-DISTRIBUTION: A UNIFIED APPROACH by J. LARS KIRKBY, DANG H. NGUYEN, AND DUY NGUYEN that \begin{align*} & \operatorname{Cov}(X, \operatorname{sign}(X)X^2) = E[|X|^3] = \frac{\nu^{3/2}\Gamma(\frac{\nu - 3}{2})}{\sqrt{\pi}\Gamma(\frac{\nu}{2})}, \\ & \operatorname{Var}(X) = E[X^2] = \frac{\Gamma(\frac{3}{2})}{\sqrt{\pi}}\frac{\nu}{\frac{\nu}{2} - 1} = \frac{\nu}{\nu - 2}, \\ & \operatorname{Var}(\operatorname{sign}(X)X^2) = E[X^4] = \frac{\Gamma(\frac{5}{2})}{\sqrt{\pi}}\frac{\nu^2}{(\frac{\nu}{2} - 1)( \frac{\nu}{2} - 2)} = \frac{3\nu^2}{(\nu - 2)(\nu - 4)}. \end{align*} This gives \begin{align*} \operatorname{Corr}(X, \operatorname{sign}(X)X^2) = \frac{\frac{\nu^{3/2}\Gamma(\frac{\nu - 3}{2})}{\sqrt{\pi}\Gamma(\frac{\nu}{2})}}{\sqrt{\frac{\nu}{\nu - 2}\frac{3\nu^2}{(\nu - 2)(\nu - 4)}}} = \frac{\Gamma(\frac{\nu - 3}{2})(\nu - 2)(\nu - 4)^{1/2}}{\sqrt{3\pi}\Gamma(\frac{\nu}{2})}, \end{align*} which converges to $0$ as $\nu \to 4^+$. This implies that there exists a distribution that gives the correlation of your interest arbitrarily close to $0$, implying that a fixed positive lower bound of this correlation does not exist.

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  • $\begingroup$ Compute that the correlation for a Student $t(n)$ distribution is $$\frac{n-2}{(n-1)(n-3)B\left(\frac{n}{2},\frac{1}{2}\right)} \sqrt{\frac{2(n-4)(n-2)}{n-1}} $$ and show the limit as $n\to 4^+$ is zero. $\endgroup$
    – whuber
    Oct 19, 2023 at 19:04
  • $\begingroup$ @whuber I appreciate your hint. The answer was updated. $\endgroup$
    – Zhanxiong
    Oct 19, 2023 at 19:59
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    $\begingroup$ +1 It looks like your calculations are likely correct: I erred in computing the variance of $X^2$ rather than that of $\operatorname{sgn}(X)X^2.$ But it makes little difference: the point is that the variance of $X^2$ diverges as $n\to 4^+$ while the lower moments stay finite and that's enough to achieve correlations arbitrarily close to zero. $\endgroup$
    – whuber
    Oct 19, 2023 at 22:01
  • $\begingroup$ @Zhanxiong You mentioned correlation is non-zero if (non-constant) r.v. is symmetric about 0. What if the r.v. is asymmetric? Intuitively they are positively correlated, but I could not analyze it quantatively. $\endgroup$
    – cat
    Oct 20, 2023 at 2:34
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    $\begingroup$ For the 1-st part of the answer, a r.v. $X$ is positively correlated with any increasing function of itself i.e. with $g(X)$ where $g$ is increasing (provided the variances are finite). We can take $g(x ) = \text{sgn}(x) x^2$ and I believe that the symmetry condition is not required. $\endgroup$
    – Yves
    Oct 20, 2023 at 5:33

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