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 ?

• i am so confused about the chapter of convergence properties of the gibbs sampler written by casella and Robert. sry for this question it is rather obvious but i need to be sure because it is for my master thesis – user2016445 Feb 6 '15 at 16:06
• sorry about our chapter confusing you...! – Xi'an Feb 6 '15 at 17:11
• You're fortunate to have one of the authors answering your question. – Glen_b Feb 6 '15 at 17:28

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 $$K(x,x')=\pi_1(x'_1|x_{-1})\times\pi_2(x'_2|x'_1,x_{-1:2})\times\cdots\times\pi_d(x'_d|x_{-d}')$$ is the density of a probability measure that is absolutely continuous against the Lebesgue measure on $(\mathbb{R}^d,\mathcal{B}(\mathbb{R^d}))$.