Let $(S,d)$ be some separable and complete metric space, and let $\mathbb{F}$ be some collection of functions from $S$ to $S$. Endow $\mathbb{F}$ with a suitable sigma algebra such that everything I need to be measurable is. Let $(\mathcal{M}(S),\rho)$ be the space of probability measures on $S$ with the Prokhorov metric $\rho$. This also becomes a separable and complete metric space.
Now let $F$ be some random element of $\mathbb{F}$. We can then define a function $\Phi: \mathcal{M}(S) \to \mathcal{M}(S)$ by that if $X$ is distributed according to $\mu\in \mathcal{M}(S)$, then $F(X)$ is distributed according to $\Phi(\mu)$.
First off, a softball opening question: What conditions on $F$ ensure that $\Phi$ is a continuous function?
Now, the idea is that if we take iid copies $F_i$ of $F$, then for any $x\in S$, the sequence $x, F_1(x), F_2(F_1(x)),\ldots$ becomes a Markov chain. A theorem by Letac gives us conditions on $F$ for this Markov chain to have a stationary distribution, and convergence rates towards it. Specifically, if $K_F$ is the (random) Lipschitz constant of the function $F$, it requires that: 1) $K_F$ have first moment, 2) for some fixed reference point $x_0\in S$, $d(x_0,F(x_0))$ have first moment, and 3) $\mathbb{E}(\log(K_F))<0$. (The moment conditions can be relaxed slightly, but that's besides the point here.) The final condition is called being "contracting on average".
The proof proceeds by coupling trajectories using the random functions, using the contracting on average hypothesis to get that trajectories then converge onto each other, using a trick with iterating in the opposite order to get a Cauchy sequence and establish existence of the stationary distribution, and finally using the following lemma (by Strassen, I believe) to translate from the coupling into a statement about distances between measures:
Lemma: Let $\mu$ and $\nu$ be two probability measures on $S$. Then $\rho(\mu,\nu)<\delta$ if and only if there exists a coupling $(X,Y)$ of $\mu$ and $\nu$ such that $\mathbb{P}(d(X,Y)>\delta)<\delta$.
So, what I was thinking is that this is a complicated proof, with lots of inequality massaging. It would be much nicer if we could "just" prove that $\Phi$ is then a contraction, and the theorem would follow directly from the Banach fixed point theorem.
So, question two: What conditions do we need on $F$ for $\Phi$ to be a contraction on $\mathcal{M}(S)$?
