Limit of a nested probability function? Let $F_0(x) =x$.
Then, let $F_i(x) = (1-(1-F_{i-1})^2)^2$ where $i>0$.
Is there a way I can calculate $F_{\infty}(x)$? (Terminology-wise, would I say that is a limit of $F_i(x)$ as $i \to \infty$ ?)
And/or, since in this case $F_i(x)$ is a cdf, can I say what any of the central moments (e.g. mean, skewness) of the $X_i$ represented by $F_i(x)$ approach as  $i \to \infty$ ?
 A: Let $u^*=\frac12(3-\sqrt5)$. As already noted by others, for every fixed $x$, $(F_i(x))$ converges to a limit which depends on $F_0(x)$: the limit is $0$ if $F_0(x)< u^*$, $u^*$ if $F_0(x)=u^*$, and $1$ if $F_0(x)>u^*$.
From here, the asymptotics depends on the starting point $F_0$. The OP is interested in the case when $F_0(x)=x$ for every $x$ in $[0,1]$. In this case, $F_i\to G$ pointwise, where $G(x)$ is $0$ if $x<u^*$, $u^*$ if $x=u^*$, and $1$ if $x>u^*$.
This function $G$ is not a CDF but there exists a CDF $H$ such that $F_i(x)\to H(x)$ for every $x$ such that $H$ is continuous at $x$, namely the function $H=\mathbf{1}_{[u^*,+\infty)}$. Since $H$ is the CDF of a random variable $X$, a well known theorem asserts that this is enough to guarantee that $X_i\to X$ in distribution. Since $0\le X_i\le 1$ almost surely for every $i$, $X_i\to X$ in every $L^p$ as well. Of course, $P(X=u^*)=1$ hence $X_i\to u^*$ in distribution and in every $L^p$.
Note that for other starting points $F_0$, the limit can be different. For example, if there exists $w$ such that $F_0$ is the cdf of a random variable $X_0$ such that $P(X_0=w)=1$, then $F_i=F_0$ for every $i$, hence $F_i\to F_0$ for any $w$. 
Likewise, if there exists $v<w$ and $p$ in $(0,1)$ such that $F_0$ is the cdf of a random variable $X_0$ such that $P(X_0=v)=p$ and $P(X_0=w)=1-p$, then one of three asymptotics occurs: 


*
*either $p>u^*$, and then $X_i\to v$, 

*or $p<u^*$, and then $X_i\to w$, 

*or $p=u^*$, and then $F_i=F_0$ for every $i$, hence $F_i\to F_0$.


A pathwise representation of the transformation of interest may help to get some intuition about the asymptotics described above and is as follows. Assume that the random variables $X_i^{(n)}$ are i.i.d. with CDF $F_i$. Then $F_{i+1}$ is the CDF of the random variable $X_{i+1}$ defined by
$$
X_{i+1}=\max\{\min\{X_i^{(1)},X_i^{(2)}\},\min\{X_i^{(3)},X_i^{(4)}\}\}.
$$
A: There are only 3 roots in the unit interval to the steady state equation $F_\infty = (1-(1-F_\infty)^2)^2$, which are $F_\infty = 0$, $F_\infty = 1$ and $F_\infty = \frac{3-\sqrt{5}}{2}$, thus $F_\infty(x)$ can only assume one of these values.
The point $x=\frac{1}{2}( 3 - \sqrt{5})$ is unstable, resulting that for all $x$ you would $F_\infty(x)$ is either 0 or 1. The limiting distribution is therefore degenerate, with all the probability concentrated at $x=\frac{1}{2}( 3 - \sqrt{5})$.

