Let

  • $(\Omega,\mathcal A,\operatorname P)$ be a probability space
  • $\pi$ be a probability measure on $(\mathbb R,\mathcal B(\mathbb R))$
  • $(X_n)_{n\in\mathbb N}$ be a real-valued stationary stochastic process on $(\Omega,\mathcal A,\operatorname P)$ with $X_n\sim\pi$ with respect to $\operatorname P$ for all $n\in\mathbb N$
  • $f\in\mathcal L^2(\pi)$, $$Y_n:=f(X_n)\;\;\;\text{for }n\in\mathbb N$$ and $$\mu_N:=\frac1N\sum_{n=1}^NY_n\;\;\;\text{for }N\in\mathbb N$$

Let's define the autocorrelation function $$\gamma(m,n):=\operatorname{Cov}[Y_m,Y_n]\;\;\;\text{for }m,n\in\mathbb N$$ and its normalized version $$\rho(m,n):=\frac{\gamma(m,n)}{\sigma_m\sigma_n}\;\;\;\text{for }m,n\in\mathbb N,$$ where $\sigma_n:=\sqrt{\operatorname{Var}[Y_n]}$ for $n\in\mathbb N$. By stationarity, $$\gamma(m,n)=\gamma_{|m-n|}:=\operatorname{Cov}[Y_1,Y_{1+|m-n|}]\;\;\;\text{for all }m,n\in\mathbb N\tag1$$ and $$\rho(m,n)=\rho_{|m-n|}:=\frac{\gamma_{|m-n|}}{\gamma_0}\;\;\;\text{for all }m,n\in\mathbb N\tag2$$

With the notation above, we obtain $$\operatorname{Var}[\mu_N]=\frac{\sigma_1^2}N2\tau_N\;\;\;\text{for all }N\in\mathbb N,\tag3$$ where $$\tau_N:=\frac12+\sum_{k=1}^N\left(1-\frac kN\right)\rho_k\;\;\;\text{for }N\in\mathbb N.$$ Now, I've read that $$\tau_N\xrightarrow{N\to\infty}\frac12+\sum_{k=1}^\infty\rho_k\tag4.$$ Is this really true? It seems to me that we need further assumptions for $(4)$ to hold. In any case, how can we prove $(4)$ (imposing additional assumptions, when necessary)?

  • this answer gives a sufficient condition (absolute summability): stats.stackexchange.com/questions/154070/… – Taylor Oct 11 at 12:31
  • @Taylor You mean $\sum_{k\in\mathbb N_0}\left|\gamma_k\right|<\infty$? If so, this condition is not very handy to use, is it? Is there a (nontrivial) condition on $X$ or $f$ which ensures $\sum_{k\in\mathbb N_0}\left|\gamma_k\right|<\infty$? – 0xbadf00d Oct 11 at 17:02
  • If you want a handy condition, you have to tell us more about your application. Is it MCMC? Just a guess based on some of the notation. The linked answer is quite handy if you're working with time series models. – Taylor Oct 11 at 17:57
  • @Taylor Yes, it's MCMC. I'm trying to understand the paper by Roberts and Rosenthal. – 0xbadf00d 17 hours ago

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