Thinning lowers observed lag correlation but discards information

Each row of $P$ is nonnegative and sums to one. The uniform law is invariant because each column also sums to one, so $\pi P=\pi$ by [F1] and [F2]. By [F5], the transition rule is time homogeneous. Repeated use of the Markov property gives, for times $n_1<\cdots <n_r$ and states $x_1,\dots ,x_r$, the path probability $\pi (x_1){\prod }_{j=1}^{r-1}{P}^{n_{j+1}-n_j}(x_j,x_{j+1})$: the first marginal is $\pi$ because $\pi P=\pi$, and each later gap uses the corresponding matrix iterate. This expression depends only on the gaps, so a common time shift leaves every finite-dimensional law unchanged, which is strict stationarity by [F3].

Under $\pi$, $\mathbb{E}[Y_n]=0$ and $\mathrm{Var}(Y_n)=1$ by [F6]. Given $Y_n=1$, the next value is $1$ with probability $.9$ and $-1$ with probability $.1$; given $Y_n=-1$, the signs reverse. Thus $\mathbb{E}[Y_{n+1}\mid Y_n]=.8Y_n$. Iteration gives $\mathbb{E}[Y_{n+k}\mid Y_n]={.8}^k{Y}_n$. Since $\mathbb{E}{Y}_n=0$ and $\mathrm{Var}(Y_n)=1$, [F6] gives $\rho (k)=\mathbb{E}(Y_n{Y}_{n+k})={.8}^k$.

The sum is geometric, so $\tau =1+2{\sum }_{k\ge 1}{.8}^k=1+2(.8\mathrm{/}.2)=9$. By [F4], the ESS for $n$ unthinned transitions is $n\mathrm{/}9$.

Keeping every tenth state gives retained lag-$j$ correlation ${.8}^{10j}$. Therefore ${\tau }_{10}=1+2{\sum }_{j\ge 1}{.8}^{10j}=(1+{.8}^{10})\mathrm{/}(1-{.8}^{10})\approx 1.240$, and there are $n\mathrm{/}10$ retained states. Its ESS is approximately $(n\mathrm{/}10)\mathrm{/}1.240=.0806n$, less than $n\mathrm{/}9=.1111n$. The displayed lag-one correlation is lower, but the total information under a fixed transition budget is also lower.