Suppose $X_n$ is a Markov chain (w.r.t $\mathcal{F_n}$) with transition probability $P(x,A)$, $x\in E$ and $A\in \varepsilon$. $f$: $E\to\mathbb{R}$ is a bounded measurable function. Define $Pf(x)=\int_{E}P(x,dy)f(y)$ and $$M_{n}=\sum_{k=0}^{n-1} \frac{1}{k+1}\left(P f\left(X_{k}\right)-f\left(X_{k+1}\right)\right).$$ Try to prove:
-
$M_n(n\geq1)$ is a martingale.
-
$\sup_{n\geq1}E(M_n^2)<\infty$
- $\frac{1}{n} \sum_{k=0}^{n-1}\left(P f\left(X_{k}\right)-f\left(X_{k+1}\right)\right) \stackrel{a . s.}{\rightarrow} 0$
The first question is easy for me because $E(f(X_{n+1})|\mathcal{F_n})=\int_EP(X_n,dy)f(y)=Pf(X_n)$, so I can easily get $M_n$ is a martingale. For the second question I try to use Cauchy's Inequality:
\begin{align}
E(M_n^2)&=E\left(\sum_{k=0}^{n-1} \frac{1}{k+1}\left(P f\left(X_{k}\right)-f\left(X_{k+1}\right)\right)\right)^2
\\&\leq \sum_{k=0}^{n-1}\frac{1}{(k+1)^2}E\left(\sum_{k=0}^{n-1}\left(P f\left(X_{k}\right)-f\left(X_{k+1}\right)\right)^2\right)
\end{align}
and then I have no idea to continue. For the third question I think if we can get $\sup_{n\geq1}E(M_n^2)<\infty$ then we can know that $M_n$ is uniformly integrable, thus we can use the convergence of martingale, but I don't know how to improve my idea. So any help is welcome!
Best Answer
Hints:
Part (b):
Part (c):