# Are the mean of samples taken from Metropolis-Hastings MCMC normally distributed?

I've come across the following theorem while studying MCMC. It seems to suggest that the sample mean taken from the MCMC – the posterior marginal expectation – should be normally distributed, using just $\phi(x) = x$:

Theorem 4.4. (A Central Limit Theorem). For a Harris recurrent, $\pi$-invariant Markov chain, and a function $\phi : \mathbb X \to \mathbb R$ satisfying enough regularity conditions, $$\sqrt{t} \left[ \frac1t \sum_{i=1}^t \phi(X_t) - \int_{\mathbb X} \, \phi(x) \pi(x) \, dx \right] \mathop{\longrightarrow}\limits_{t \to \infty}^D \mathcal N(0, \sigma^2(\phi))$$ where $$\sigma^2(\phi) = \mathbb V[\phi(X_1)] + 2 \sum_{k=2}^\infty \mathbb{Cov}[\phi(X_1), \phi(X_k)] .$$ The variance and covariance in the expression above are with respect to the distribution $\pi$ of the Markov chain in its stationary regime.

I am wondering is there some extension of this result when dimension of the parameter space is larger than 1? So in this case $\phi : \mathbb X \to \mathbb R^n$.

• I've transcribed your screenshot into text, since the big page gap made reading difficult. It would also be helpful if you gave a source; I searched briefly and didn't find it. Oct 16, 2015 at 22:18
• I'd expect that a similar convergence holds: taking $\phi_i(x) = x^T e_i$ for each vector in a basis $\{e_i\}$, for example, shows that all one-dimensional projections become normal, I believe uniformly across the different $\phi_i$, which strongly suggests it should converge to multivariate normal. In a few minutes of looking I didn't find such a theorem, though. Oct 16, 2015 at 22:21
• Hi, thank you for the input. I found this lecture notes here, the chapter 4 link. sites.google.com/site/pierrejacob/advanced-simulation Oct 17, 2015 at 5:22
• The ergodic average is asymptotically normal, this is simply the central limit theorem. Which also works in larger dimensions. Oct 17, 2015 at 12:55

Indeed when the univariate Markov chain central limit theorem holds, the multivariate CLT also holds. In fact, the the univariate CLT is a consequence of the multivariate CLT via the Cramer-Wold Theorem. Look at Page 7-9 here.

However, the Theorem 4.4 you indicated has the hidden line of "satisfying enough regularity conditions". These regularity conditions are difficult to verify but do hold in general. These conditions are that of the rate of convergence of the Markov chain to its stationary distribution. This paper summarizes most of the research done in this field. For the Metropolis-Hasting MCMC (which is reversible), you need that the chain be geometrically ergodic with a finite 2nd moment.

If these conditions hold true, the the multivariate CLT takes the following form: If $\phi: \mathbb{X} \to \mathbb{R}^n$, and $E_{\pi} \|\phi(X_1) \|^2 <\infty$ and $\{X_i\}$ is geometrically ergodic, then there exists a $n \times n$ positive definite matrix, $\Sigma$ such that, $$\sqrt{t} \left[\dfrac{1}{t}\sum_{i=1}^t{ \phi(X_i)} - \int_\mathbb{X} \phi(x) \pi(x) \right] \to N_n(0, \Sigma),$$

where if $\phi(X_t) = Y_t = (Y^{(1)}_t, Y^{(2)}_t, \dots, Y^{(n)}_t)$, then the $(i,j)th$ entry of $\Sigma$ is

$$\Sigma_{ij} = Cov_{\pi}(Y^{(i)}_1, Y^{(j)}_1) + 2 \sum_{k=1}^{\infty} Cov_{\pi}(Y^{(i)}_1, Y^{(j)}_{1+k}).$$

• Thanks for the answer. I have two question. 1. In practice we wait for the chain to run for some time and then start to take samples. Say the first sample I take is from $X_k$ where k = 50. Can I replace the X_1 with X_50 in the variance function? Oct 18, 2015 at 17:38
• Second question. Could you specify what the dimension of $Var_\pi \phi(X_1)$ and $Cov_\pi(\phi(X_1),\phi(X_{1+k})$ Oct 18, 2015 at 17:39
• Yes. The statement of the CLT says that this holds true regardless of the starting value. This is the reason that there is also a school of thought that says there is no need to throw away initial samples. Second question: The matrics $Var_{\pi}\phi(X_1)$ and $Cov_{\pi}(\phi(X_1), \phi(X_{1+k}))$ $n\times n$ matrices. The Var matrix is positive definite, but the infinite sum of the covariances might not be positive definite. Oct 18, 2015 at 17:40
• Huh interesting. So even if I throw away, say first 50 initial samples, the infinit sum of $cov(\phi(X_{50}),\phi(X_{50+k})$ will not be much smaller than $cov(\phi(X_1),\phi(X_{1+k})$? I understand that there is the infinite sum, but I thought if the chain converges, say after 50 iterations, then $cov(\phi(X_{50}),\phi(X_{51}))$ will definitely be smaller than $cov(\phi(X_1),\phi(X_2))$, or there is no such result? Oct 18, 2015 at 17:46
• Wald should probably be Wold.
– KOE
Aug 12, 2016 at 13:30