Proof of negative covariance (for the inverse transformation method) Given a continuous strictly monotone distribution function $F$ and it's quantile function $F^{-1}$, with $U\sim\mathcal{U}(0, 1)$, the inverse transformation method takes $F^{-1}(U)$ and $F^{-1}(1-U)$ as anti-thetic variates (which just means they have negative covariance). But why?
I've seen a proof where $g$ is strict monotone increasing on $[0, 1]$, where Reimann integration is first used to calculate the expectations then the mean valued theorem is applied, but here $F^{-1}$ is not Reimann integrable as it can take on $\pm{\infty}$ values within $[0, 1]$.
 A: I'm not very familiar with measure theory, but for continuous, strictly increasing $F$ I can provide some results:
Note: Covariance is a (signed) value that indicates the degree to which two variables can be linearly related.
Let $X=F^{-1}(U)$ and $Y= F^{-1}(1-U)$
Let's assume that $F(z)$ is differentiable and strictly increasing in $z$, therefore:
$\frac{dX}{dU}> 0$. Also, $\frac{dY}{dU}< 0 \implies \frac{dU}{dY}<0$. From the chain rule, we get:
$$\frac{dX}{dU}\frac{dU}{dY}=\frac{dX}{dY}<0$$
Hence, they will be inversely related to each other and will therefore give a negative covariance.
To generalize this, you'd need to be able to gracefully handle the cases where $F$ has a derivative of zero. For discrete distributions, I think one could use discrete calculus and/or measure theory to show that the resulting $X$ and $Y$ will be inversely related. For pathological distributions (e.g., Cantor Distribution), you'd probably need to get more creative, since its constant almost everywhere.
