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Suppose we want to obtain samples of the density $f(\mathbf{x})$ where $\mathbf{x}$ is a $d$-dimensional vector, i.e. $\mathbf{x} = (x_1, x_2, \dots, x_d)$. To that end, we choose the Metropolis-Hastings algorithm with a proposal such that:

  • At iteration $i+1$ only one element of the proposed sample $\mathbf{x}^*$ will be different from the current sample $\mathbf{x}^i$. We denote that element with $j$ so that $\mathbf{x}^*_{\backslash j} = \mathbf{x}^i_{\backslash j}$, where $\mathbf{x}_{\backslash j}$ is used to denote the vector $\mathbf{x}$ with its $j^\text{th}$ component removed. That is,

$$ \mathbf{x}_{\backslash j}=(x_1, x_2, ..., x_{j-1}, x_{j+1}, ..., x_d). $$

  • The index $j$ is randomly chosen from an integer uniform distribution in the interval $[1,d]$.

With this description, the proposal can be written as:

$$ q(\mathbf{x}^*|\mathbf{x}^i) = \sum_{j=1}^d \frac{1}{d} \ q_j(x^*_j|\mathbf{x}^i,\mathbf{x}_{\backslash j}^*) \ \mathbb{1}(\mathbf{x}^*_{\backslash j}-\mathbf{x}^i_{\backslash j}), $$

where $\mathbb{1}(\cdot)$ denotes an indicator function that is zero for every argument except for zero when it is equal to 1.

This proposal distribution is singular. Its integral is equal to zero because it is a mixture of lower-dimensional proposals (in other words, $q$ is concentrated on a set whose Lebesgue measure is zero).

Is it correct to use such a proposal in the Metropolis-Hastings algorithm or does it violate any necessary condition?

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    $\begingroup$ This is similar to a component-wise Metropolis Hastings, where one sweep of the algorithm updates each component independently. Why not just do that? $\endgroup$
    – Greenparker
    Mar 7 '16 at 12:02
  • $\begingroup$ Indeed, I agree with you. However, I am analysing an algorithm already existing in the literature which follows the description in the question. $\endgroup$
    – iglesias
    Mar 8 '16 at 11:17
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    $\begingroup$ This is a random-scan Metropolis-within-Gibbs algorithm, which is only singular when you look one step at a time. Nonetheless, providing the proposals have a large enough support, the resulting Markov kernel is irreducible and ergodic over the whole space. $\endgroup$
    – Xi'an
    Apr 6 '16 at 12:07
  • $\begingroup$ I think the interpretation of the proposal is incorrect. The proposal as written down seems correct as a density, but is not the joint proposal in this case. I believe you cannot write it down as a join proposal. Each $q_j$ updates the $j$th component only, and thus $q_j$ is non singular for all $j$. Each $q_j$ leads to a Markov transition kernel $P_j$ and $\sum \frac{1}{d}P_j$ is the joint Markov kernel. Thus there is no problem with singularity. $\endgroup$
    – Greenparker
    Apr 11 '16 at 19:15
  • $\begingroup$ @Greenparker, why do you believe the the proposals $q_j$ cannot be written as a linear combination in order to provide an expression for the joint proposal? $\endgroup$
    – iglesias
    Apr 19 '16 at 19:16

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