Computing a "Double Conditional Distribution"

Question

This question comes in the context of Gibbs sampling, and I have posted it on the Stats Stack Exchange.

Let us say we are considering random vectors in $\mathbb{R}^2$ of the form $(X,Y)^T$, such that: $$ (X,Y) \sim \mathcal{N} \left([0,0], \begin{bmatrix}1 & \rho \\\rho & 1\end{bmatrix}\right) $$ We sample from these distributions as follows:

Assume we have sampled $X^{(n)},Y^{(n-1)}$, then we recursively sample:

1. $Y^{(n)} \sim f_{Y|X = X^{(n)}}$ (We sample $Y^{(n)}$ from the conditional distribution given the previous sample $X^{(n)}$)
2. $X^{(n+1)} \sim f_{X|Y = Y^{(n)}}$ (We sample $X^{(n+1)}$ from the conditional distribution given the previous sample $Y^{(n)}$)

Given this sampling procedure, how can we compute the distribution of: $$ X^{(n+1)} | X^{(n)} \quad ? $$

That is, if we know $X^{(n)}$, how can we compute the conditional distribution of the next draw for $X^{(n+1)}$? It is easy to compute the distribution of the next draw of $X$ given the current draw of $Y$, but this draw of $Y$ depends on the previous draw of $X$.

Answer 1

I will provide details of the calculation. Essentially, we need to compute: $$ \mathbb{E}[X_{n+1} | X_n] $$ and $$ \mathrm{Var}[X_{n+1}|X_n] $$ We know that $X_{n+1}$ conditional on $X_n$ will be normally distributed, because $X_{n+1} | Y_n$ is normal and $Y_n | X_n$ is normal (not sure if this sort of transitivity is true in general, if someone could make me aware of any details I am glossing over I would appreciate it).

Recall the very general tower property of conditional expectation, if $(\Omega, \mathcal{F}, \mathbb{P})$ is a probability space and $G_1 \subset G_2 \subset \mathcal{F}$ are nested subalgebras, then we have that: $$ \mathbb{E}[\mathbb{E}[X|G_2]|G_1] = \mathbb{E}[X|G_1] $$ for any $X$ that is in $L^1(\Omega)$. Let $G_1 = \sigma(X_n)$ and $G_2 = \sigma(Y_n)$. This satisfies the above because we generate $X_n$ first, and then generate $Y_n$. In particular, it follows: $$ \mathbb{E}[X_{n+1}|X_n] = \mathbb{E}[\mathbb{E}[X_{n+1}|Y_n] |X_n] = \mathbb{E}[\rho Y_n |X_n] = \rho \cdot (\rho X_n) = \rho^2X_n $$

For the case of the variance, we have an identity (cf. Wikipedia) that states: $$ \mathrm{Var}(\mathrm{Var}(X_{n+1}|Y_n)|X_n) = \mathbb{E}[\mathrm{Var}(X_{n+1}|Y_n)|X_n] + \mathrm{Var}(\mathbb{E}[X_{n+1}|Y_n]|X_n) $$ The conditional variance of $X|Y$ is $(1-\rho^2)$. Thus, we can finish the calculation above to get: $$ 1 - \rho^2 + \rho^2(1-\rho^2) = 1-\rho^4 $$

Thus, we arrive at: $$ X_{n+1}|X_n \sim (\rho^2X_n, 1-\rho^4) $$