If a sequence of distributions converges to a degenerate, does that imply the variance strictly decreases? If $F_i = G(F_{i-1}), F_0 = x$ is a sequence of distributions that converges to a degenerate distribution as $i \to \infty$, does that imply that the variance of $F_i$ decreases with $i$?  
Specifically, I am interested in the inverse Kumaraswamy distribution: $G(x) = (1-(1-x)^a)^b$ when $a=b$... Note that $G$ produces the cdf of the Max of b draws from the distribution of the Min of a draws from its argument (which is going to be a cdf).  
So if it's not true for the general case, but you have insights on how/why it would be true in this case, I would appreciate it.  I can intuit (and graph for specific values of a) that the mean and variance decrease in this case -- taking the Min of $a$ draws from $F$ decreases the mean, then taking the max of $a=b$ draws from the new distribution increases the mean, but not by as much as the first operation lowered it, so the final mean is less than that of the original.   But I am shaky on how to prove it (and my other questions today have basically been trying to get at this point).
PS -- sorry to overload with questions today!  As I mentioned, they've all been in the goal of this question here.
 A: Here is a simple counterexample.
Assume the CDF $F$ is supported on $[0,1]$ and define a new CDF $G[F]$ as follows.  If $\mathbb{E}[F] \gt 1/2$, let
$$G[F](x) = 1 - F(3/2 - 2 x).$$
Otherwise, let
$$G[F](x) = 1 - F(5/6 - 2x/3).$$
The first operation squeezes the distribution by a factor of $2$ towards $1/2$ and flips it around $1/2$, while the second expands the distribution by a factor of $3/2$ away from $1/2$ and also flips it around $1/2$.  The flipping guarantees that if the mean of $F$ is other than $1/2$, then iterating $G$ will alternate between these two operations, because the expectations will alternately be greater than and less than $1/2$.  The net result nevertheless is to compress the support down towards $1/2$, converging to a degenerate distribution.  However, the variance in the first case is multiplied by $1/4$ and in the second case it is multiplied by $9/4$, whence it does not uniformly decrease: it alternately jumps up and down.  Therefore convergence of a sequence of distributions $(G^n[F])$ to an atom does not imply monotonic decrease of the variances.
A: You ask for intuition.  Maybe the following bit of hand-waving will help.
Provided $F$ is continuous, your particular sequence, $G^n[F]$ with $G[F](x) = (1-(1-F(x))^a)^a$, indeed eventually has monotonically decreasing variances because $G$ is a smooth contracting map.  Specifically, $G$ has a fixed point $x_0$ (the unique root of $G[x]=x$ in the interval $(0,1)$) and it contracts all values towards that fixed point.

(The figure shows $F(x)=x$ and the first three iterations of $G$ for $a=2$.)
This alone is not enough for an easy proof (I don't think), but notice in addition that the derivative of $G$ near that fixed point always strictly exceeds $1$.  (It is never smaller at the fixed point than $6 - 2\sqrt{5} \approx 1.528$, when $a=2$.)  After enough iterations, almost all the probability is squeezed into a region near this fixed point, where $G$ acts essentially as a rescaling operation, which will shrink the variance at each step.  (Any probability outside this small neighborhood cannot contribute much to the variance because the support of $G^n[F]$ is bounded independent of $n$.) The shrinking means the variance eventually behaves almost like a geometric series with common ratio $1/G'(x_0)^2$, which will monotonically decrease.  I think a little epsilon-delta analysis could make this argument rigorous.
