What is the correlation between a random variable and its probability integral transform? Are there known bounds on the $\operatorname{cor}(X,F(X))$? $X$ is a random variable with CDF $F(X)$. Let $X$ have a fixed variance, for example $\operatorname{var}(X)=1$. What $X$ can maximize or minimize the covariance?


If we assume $\mathbb E^F[X]=0$ then \begin{align} \mathbb
  E^F[XF(X)]&= \frac{1}{2}\int F(x)\{1-F(x)\}\,\text dx
  \end{align}

Indeed, assuming the pdf $f$ is associated with the cdf $F$,
\begin{align}
\mathbb E^F[XF(X)]&= \int x F(x) f(x)\text dx\\
&= \int_{-\infty}^0 x F(x) f(x)\text dx + \int_0^\infty x \{F(x) -1+1\}f(x)\text dx \\
&= \int_{-\infty}^0 x F(x) f(x)\text dx - \int_0^\infty x \{1-F(x) \}f(x)\text dx+ \int_0^\infty x f(x)\text dx\\
&= -\frac{1}{2}\int_{-\infty}^0 F(x)^2\text dx - \frac{1}{2} \int_0^\infty \{1-F(x) \}^2\text dx+ \int_0^\infty \{1-F(x)\}\text dx\\
\end{align}
by integrations by parts. And, since $\mathbb E^F[X]=0$ then
$$\int_0^\infty \{1-F(x)\}\text dx=\int_{ -\infty} ^0 F(x)\text dx$$
Note also that the variance of $X$, $\sigma^2$, does not impact the correlation since
$$\text{corr}(X,F_\sigma(X))=12\dfrac{\mathbb E_\sigma(XF_\sigma(X))}{\text{var}_\sigma(X)}=12\dfrac{\mathbb E_\sigma(\sigma^{-1}XF_1(\sigma^{-1}X))}{\text{var}_\sigma(\sigma^{-1}X)}=12\mathbb E_1(XF_1(X))$$
Another identity of possible interest is
\begin{align} \mathbb  E^F[XF(X)]&= \frac{1}{2}\mathbb E^F[\max\{X_1,X_2\}]  \end{align} when $X_1,X_2$ are iid $F$ with mean $0$
