I have the following question regarding the Gibbs sampler, although it might be considered a simple question on conditional probability.

For sake of simplicity, let us say we are trying to sample from a bivariate normal distribution: $$ [X,Y]^T \sim \left([0,0], \begin{bmatrix}1 & \rho \\ \rho & 1\end{bmatrix}\right) $$ To perform Gibbs sampling, we generate samples: $$ x_{n+1} \sim X|Y^{(n)} $$ $$ y_{n+1} \sim Y|X^{(n+1)} $$ and output samples $(x_n,y_n)$. Here the notation $X|Y$means in words "the conditional distribution of $X$ given $Y$.

Given this information, is it possible to calculate the "marginalized" conditional distribution: $$ X_{n+1}|X_n $$ directly? How would one do it? I assume this is a basic conditional probability fact unrelated to Gibbs sampling.


1 Answer 1


Here is an excerpt (p.340) from our book Monte Carlo Statistical Methods that gives an answer to this question:

Example 9.1. Normal bivariate Gibbs. For the special case of the bivariate normal density, \begin{equation} \tag{9.3} (X, Y) \sim {\cal N}_2 \left( 0, \left( \begin{array}{cc} 1 & \ \ \rho \\ \rho & \ \ 1 \end{array} \right) \right) \;, \end{equation} the Gibbs sampler is

Given $y_t$, generate \begin{eqnarray}\tag{9.4} X_{t+1} \mid y_t & \sim &{\cal N}(\rho y_t,\ 1 - \rho^2) \;,\\ Y_{t+1} \mid x_{t+1} & \sim &{\cal N}(\rho x_{t+1},\ 1- \rho^2). \nonumber \end{eqnarray}

The Gibbs sampler is obviously not necessary in this particular case, as iid copies of $(X,Y)$ can be easily generated using the Box-Muller algorithm (see Example 2.8). Note that the corresponding marginal Markov chain in $X$ is defined by the AR$(1)$ relation $$ X_{t+1} = \rho^2 X_t + \sigma\epsilon_t,\qquad \epsilon_t\sim\mathcal{N}(0,1)\,, $$ with $\sigma^2 = 1-\rho^2+\rho^2(1-\rho^2) = 1 - \rho^4$. As shown in Example 6.43,the stationary distribution of this chain is indeed the normal distribution $$\mathcal{N}\left(0,\frac{1-\rho^4}{1 - \rho^4}\right)$$

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    $\begingroup$ Thank you, I'll do the calculations in some detail and then accept the answer. $\endgroup$ May 30, 2020 at 16:29
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    $\begingroup$ Yes, I realize now that one must use the tower property: $\mathbb{E}[[X_{n+1}|Y_n]|x_n] = \mathbb{E}[X_{n+1}|X_n]$ $\endgroup$ May 30, 2020 at 17:16

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