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Basically, the title of the question. I'm wondering if it is reasonable to use the mESS (defined here) to estimate the autocorrelation time $\tau$ as:

$$\tau = \frac{N}{mESS}$$

where $N$ is the size of my MCMC chain. I'm asking because I have not seen this used/recommended anywhere and I want to be sure I'm not using it wrong.

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This would require defining a multivariate interpretation of integrated autocorrelation time.

Let $Y_1, Y_2, \dots, Y_n$ be a $p-$dimensional Markov chain with invariant distribution $\pi$; $Y_1 = (Y_{11}, Y_{12}, \dots, Y_{1p})^T$. Suppose the effective sample size and autocorrelation time of the average is of interest, the average being: $$\bar{Y} = \dfrac{1}{n} \sum_{t=1}^{n} Y_{t}\,. $$

Let us first describe the univariate autocorrelation time, $\tau_1$ for the first component only. Suppose the asymptotic variance of $\bar{Y}_{1} is \sigma^2_{1}$, then \begin{align*} \sigma^2_1 & = \text{Var}_{\pi}(Y_{11}) + 2 \sum_{k=1}^{\infty} \text{Cov}\left(Y_{11}, Y_{1(1+k)}\right) \\ \Rightarrow \dfrac{\sigma^2_1}{\text{Var}_{\pi}(Y_{11})} & = 1 + 2 \Rightarrow\sum_{k=1}^{\infty} \text{Corr}\left(Y_{11}, Y_{1(1+k)}\right) \\ \dfrac{N}{ESS_1} &= \tau_1\,, \end{align*} where $ESS_1$ is the univariate effective sample size for the first component. A similar construction is done for each component. However, such univariate constructions ignore the cross-covariances in the Markov chain, and the cross covariances structure in $\pi$.

Now let's see what the multivariate case looks like. Suppose $\Sigma$ is the asymptotic variance-covariance matrix of $\bar{Y}$, then $\Sigma$ has diagonals $\sigma^2_1, \sigma^2_2, \dots, \sigma^2_p$, and the off diagonal elements are cross-covariances from the Markov chain. That is, $$\Sigma = \underbrace{\text{Var}(Y_1)}_{\text{a matrix}} + \sum_{k=1}^{\infty} \left[\text{Cov}(Y_1, Y_{1+k}) + \text{Cov}(Y_1, Y_{1+k})^T \right]\,, $$ where note that the cross-covariance matrix $\text{Cov}(Y_1, Y_{1+k})$ need not be symmetric. Let $\text{Var}(Y_1) = \Lambda$ \begin{align*} \Sigma &= \text{Var}(Y_1) + \sum_{k=1}^{\infty} \left[\text{Cov}(Y_1, Y_{1+k}) + \text{Cov}(Y_1, Y_{1+k})^T \right] \\ \Rightarrow \Lambda^{-1/2} \Sigma \Lambda^{-1/2} & = I_p + \sum_{k=1}^{\infty} \Lambda^{-1/2}\left[\text{Cov}(Y_1, Y_{1+k}) + \text{Cov}(Y_1, Y_{1+k})^T \right]\Lambda^{-1/2}\\ \Rightarrow \det(\Lambda^{-1/2} \Sigma \Lambda^{-1/2})^{1/p} &= \det\left( I_p + \sum_{k=1}^{\infty} \Lambda^{-1/2}\left[\text{Cov}(Y_1, Y_{1+k}) + \text{Cov}(Y_1, Y_{1+k})^T \right]\Lambda^{-1/2} \right)^{(1/p)}\\ \Rightarrow \dfrac{n}{mESS}& = \det\left( I_p + \sum_{k=1}^{\infty} \Lambda^{-1/2}\left[\text{Cov}(Y_1, Y_{1+k}) + \text{Cov}(Y_1, Y_{1+k})^T \right]\Lambda^{-1/2} \right)^{(1/p)}\,. \end{align*}

Thus the qualtity $n/mESS$ corresponds to that particular determinant on the right. The reason this is not a straightforward generalization of the integrated autocorrelation time is because $\Lambda^{-1/2}\left[\text{Cov}(Y_1, Y_{1+k}) + \text{Cov}(Y_1, Y_{1+k})^T \right]\Lambda^{-1/2}$ is not the cross-correlation matrix (at least I don't think it is). It would be the cross-correlation matrix if the $\Lambda$ matrix was replaced by the diagonals of the matrix.

However, I do suspect that $n/mESS$ gives a multivariate interpretation of the integrated autocorrelation time; it is just a bit unclear to me right now, how different this is to the univaariate interpretation. However in the above if $p =1$, you get back the univariate quantity.

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  • $\begingroup$ Thank you very much for this amazing answer Greenparker! Given that I can obtain the mESS independently of the acor time, would you say that obtaining the acor is then not really important? $\endgroup$
    – Gabriel
    Commented Jul 18, 2018 at 15:17
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    $\begingroup$ @Gabriel Happy to help. I think effective sample size and acor time provide almost entirely the same information, except for the fact that acor time is not dependent on the size of the sample. Usually in the MCMC community, effective sample size is reported more often than acor time. If you report mESS/N then you're reporting a standardized effective sample size, irrespective of N. So in short, yes, acor is not really important then. $\endgroup$ Commented Jul 18, 2018 at 15:27

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