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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.

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    $\begingroup$ In general, no, of course not. Let $X_1, X_2, \ldots$ be zero-mean normal random variables with variances $\sigma_i^2$. Choose any sequence of $\sigma_i^2 \geq 0$ such that $\sigma_i^2 \to 0$. (The point being, there is no need for $\sigma_i^2$ to be monotone.) $\endgroup$
    – cardinal
    Oct 11 '11 at 20:51
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    $\begingroup$ If you assume all the $F_i$ are supported in $[0,1]$, then the variance will have to converge to 0. If you only assume the limiting distribution is supported in $[0,1]$, then the variances do not have to converge! (Consider a mixture of a uniform and a Normal distribution, with the location of the Normal distribution diverging faster than the inverse square of its mixture weight.) $\endgroup$
    – whuber
    Oct 11 '11 at 20:51
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    $\begingroup$ After reading my comments and @whuber, it becomes apparent to me that there are at least two ways this question can be read as currently written. :) (My comments have also been colored by the title of this question.) $\endgroup$
    – cardinal
    Oct 11 '11 at 20:54
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    $\begingroup$ @cardinal I think your reading is more accurate (if I have understood it correctly :-)), but it seems to lead to questions of little interest or utility. Because convergence of a sequence of distributions is independent of the phenomenon "decreases with $i$", the two should have no relationship with one another. In fact, swapping every other term in a monotonically convergent sequence won't change its convergence properties at all but it ruins monotonicity... $\endgroup$
    – whuber
    Oct 11 '11 at 20:59
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    $\begingroup$ @JandR: You're not being slow. We're just trying to figure out the real question of interest. If a sequence of random variables converges to a degenerate distribution, then the variance of the limiting distribution will be zero. However, I believe the variances of the random variables in the sequence need not even be defined. Take, for example, a sequence of Cauchy distributions with a scale parameter converging to zero. $\endgroup$
    – cardinal
    Oct 11 '11 at 21:08
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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.

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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.

Figure

(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.

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