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We define $T: C[0,1]\to C[0,1]\ni T(f(x))= \sum\limits_{k=1}^{m} p_k (f\circ f_k)(x):=\mathbb E( f(X_{n+1}|X_n=x)$ for a system $X_{n+1}=f_{\omega_n}(X_n), n=0,1,2\dots.$ and $\omega_n$ are i.i.d discrete r.v over $\{1,2,\dots,m\}$, $p_k=\text{ Prob} (\omega_i=k)$, $f_k$ are bounded Lipschitz.

Could anyone explain to me how come $ | T^n(f(x))-T^n(f(y))|=| \mathbb E(f(X_n(x))-f(X_n(y)))|$?

Thanks for the help.

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    $\begingroup$ Please include links to your questions on Math.SE and Stats.SE. Also, please go to those sites and edit your questions there to include a link here, so that every version of the question includes links to the other two. $\endgroup$ – Nate Eldredge Apr 18 at 15:03
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By the tower law, $$ \mathbb{E} [f(X_n) | X_0=x] = \mathbb{E} [\mathbb{E} \{f(X_n) | X_{n-1}\} | X_0=x]. $$ Observe that $\mathbb{E} \{f(X_n) | X_{n-1}\}=Tf(X_{n-1})$ (because $(X_n)_{n\in \mathbb{N}}$ forms a time-homogeneous Markov chain) . By induction it follows that $E[f(X_n) | X_0=x]=T^n f(x)$ and your result follows.

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