Q10. Metropolis-Hastings with an asymmetric-proposal extension

Topic: Markov chains, Gibbs sampling, and Metropolis–Hastings · more questions on this topic
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A posterior density is known up to a normalising constant: p(\theta\mid{\boldsymbol{x}})\propto\theta^2e^{-\theta}, \qquad \theta>0. From the current state \theta^{(t)}>0, propose \theta^*=\theta^{(t)}+\varepsilon, \qquad \varepsilon\sim\mathcal{N}(0,s^2), where s>0 is fixed.

(a) Derive the acceptance probability, including the case \theta^*\leq0. Explain why the unknown normalising constant is not needed. [3 marks]

(b) Give pseudocode for one iteration, including the acceptance decision and the next state after rejection. [3 marks]

(c) One run uses a very small s and accepts almost every proposal. Another uses a very large s and accepts very few. Explain why either run may be inefficient. [2 marks]

(d) If the proposal density were not symmetric, what additional factor would be required in the acceptance ratio? Define its numerator and denominator. [2 marks]

Worked answer (click to show)

Definitions. The target is proportional to \theta^2e^{-\theta} on \theta>0 and is zero otherwise. Let q(\theta^*\mid\theta) be the proposal density, \alpha(\theta,\theta^*) the acceptance probability, and (t) the iteration index. The proposal N(\theta,s^2) has variance s^2>0, fixed across states and iterations; the current state must lie in the target support.

The target density has kernel \pi(\theta)\propto \theta^2e^{-\theta}, \qquad \theta>0.

(a) Acceptance probability

For a random-walk proposal \theta^*\mid\theta^{(t)}\sim N\left(\theta^{(t)},s^2\right), the proposal is symmetric: q(\theta^*\mid\theta^{(t)}) =q(\theta^{(t)}\mid\theta^*). Consequently, the proposal-density terms cancel from the Metropolis–Hastings ratio. If \theta^*>0, \begin{aligned} \alpha\left(\theta^{(t)},\theta^*\right) &= \min\left\{ \frac{\pi(\theta^*)}{\pi(\theta^{(t)})},1 \right\}\\ &= \boxed{ \min\left\{ \left(\frac{\theta^*}{\theta^{(t)}}\right)^2 \exp\left[-\left(\theta^*-\theta^{(t)}\right)\right], 1 \right\}. } \end{aligned} If \theta^*\leq 0, then \pi(\theta^*)=0, so the proposal is rejected and \alpha=0.

(b) Normalising constant

The unknown normalising constant appears in both the numerator and denominator of the target-density ratio and therefore cancels. Only the target kernel is needed.

(c) One sampler iteration

One iteration of the algorithm is:

1. Generate a proposal \theta^*\sim N\left(\theta^{(t)},s^2\right).

2. Calculate \alpha(\theta^{(t)},\theta^*) as above.

3. Generate U\sim\operatorname{Uniform}(0,1).

4. Set \theta^{(t+1)}= \begin{cases} \theta^*, & U\leq\alpha(\theta^{(t)},\theta^*),\\ \theta^{(t)}, & U>\alpha(\theta^{(t)},\theta^*). \end{cases}

(d) Proposal scale

If s is extremely small, proposals remain close to the current state and the acceptance rate will usually be high. However, the chain moves only in small steps, producing high autocorrelation and slow exploration.

If s is extremely large, many proposals will be negative or will fall in regions where the target density is very small. Most will be rejected, so the chain will repeatedly remain at the same state and again mix slowly.

(e) Asymmetric proposals

For an asymmetric proposal, the proposal terms do not cancel. The acceptance probability becomes \boxed{ \alpha(\theta,\theta^*) = \min\left\{ \frac{\pi(\theta^*)q(\theta\mid\theta^*)} {\pi(\theta)q(\theta^*\mid\theta)}, 1 \right\}. } The additional factor is therefore \boxed{ \frac{q(\theta\mid\theta^*)}{q(\theta^*\mid\theta)}, } that is, the reverse-proposal density divided by the forward-proposal density.