Gibbs Sampler transition kernel

Let $\pi$ be the target distribution on $(\mathbb{R}^d,\mathcal{B}(\mathbb{R^d}))$ which is absolutely continuously wrt to the $d$-dimensional Lebesgue measure, i.e :

$\pi$ admits a density $\pi(x_1,…,x_d)$ wrt to $\lambda^d$ with
$$\lambda^d(dx_1,…,dx_d) = \lambda(dx_1) \cdot \cdot \cdot \lambda (dx_d)$$

Let us assume that the full conditionals $\pi_i(x_i|x_{-i})$ from $\pi$ are known.
So the transition kernel of the Gibbs-Sampler is clearly the product of the full conditionals from $\pi$.

Is the transition kernel absolutely continuously wrt to the $d$-dimensional Lebesgue measure too ?


If you write down the transition of the systematic Gibbs sampler kernel, you get
$$\mathbb{P}(X’\in A_1\times\cdots\times A_d|X=x)=\int_{A_1} \pi_1(x’_1|x_{-1})\Big\{\int_{A_2} \pi_2(x’_2|x’_1,x_{-1:2})\cdots \left\{\int_{A_d} \pi_d(x’_1|x_{-d}’)\lambda(\text{d}x_d’)\right\} \cdots\lambda(\text{d}x_2′)\Big\}\lambda(\text{d}x_1′)$$
for any product set $A_1\times\cdots\times A_d\in\mathcal{B}(\mathbb{R^d})$ and therefore
is the density of a probability measure that is absolutely continuous against the Lebesgue measure on $(\mathbb{R}^d,\mathcal{B}(\mathbb{R^d}))$.

Source : Link , Question Author : user2016445 , Answer Author : Xi’an

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