Proper way to estimate integrated autocorrelation time of a chain

Question

Let's say that I have some quantity $\mathcal{O}$ that I wish to estimate by a sequence $(m_1,m_2,...,m_N)$ of data points (which are used to construct an estimate of $\mathcal{O}$ at each $i$, called $\mathcal{O}_i$) generated by Monte Carlo, such that the estimate of $\mathcal{O}$ is $\hat{\mathcal{O}}=\langle\mathcal{O}_i\rangle$ (the average over each $\mathcal{O}_i$). I wish to find the integrated autocorrelation time of $\mathcal{O}$, $\tau_{\mathcal{O}}$, so I use an estimator defined by

$$\hat{\tau}_{\mathcal{O}}=1+2\sum_{i=1}^{N}\rho_{\mathcal{O_i}},$$

where

$$\rho_{\mathcal{O}_t}=c_{\mathcal{O}}(t)/c_{\mathcal{O}}(0); \\c_{\mathcal{O}}(t)=\frac{1}{N-t}\sum_{i=1}^{N-t}\big(\mathcal{O}_{i+t}-\langle\mathcal{O}\rangle\big)\big(\mathcal{O}_i-\langle\mathcal{O}\rangle\big).$$

The Monte Carlo will accept some proposals for $m_i$ and reject others. When estimating autocorrelation time this way, should one only use the parts of the chain that correspond to moves that were accepted, or should the entirety of the chain (which includes elements of the chain that were accepted and elements of the chain that stayed the same because the proposed move was rejected) be used?

Answer 1

• Use the entire chain, with all duplicates. If you skip the duplicates, the chain you generated would not even have correct distribution anymore.

• But you can (and maybe should) do a "thinning out" of the markov chain. That means only keeping every second element of the chain (or every 10'th or even every 1000th. really depends on your problem), throwing away the rest. But in that procedure, duplicate elements (due to non-accepted updates) count as multiple elements. In practice this usually means that in the thinned out chain, no duplicates are actually left.

• Don't use the the formula explicitly as written, but instead with a cutoff like $$ \hat\tau_\mathcal{O} \approx 1 + 2\sum_{i=0}^{t_\mathrm{max}} \rho_{\mathcal{O}_i}$$ with $t_\mathrm{max}=5\cdot \hat\tau_\mathcal{O}$ or similar. Reason is that the estimated $\rho_{\mathcal{O}_i}$ becomes very noisy for large $i$.