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?
