Random walk metropolis-hasting algorithm

Estimation ML/Bayesian estimation
#1

Hi.

I have two questions about the random walk metropolis-hasting algorithm.

We have a proposal distribution in this algorithm such as q(x^{\star} / x^{0}). q is a proposal distribution. My first question is that how this proposal distribution for example Normal distribution has relation to the previous sample or x^{0}.
For example we can select a new sample from a normal distribution as a proposal distribution but why in the literature this new sample or x^{\star} is subject to the previous sample x^{0}?

The acceptance ratio is :

α(x^{\star}/x^{0})=min{\frac{π(x^{\star})\times q(x^{\star}/x^{0})}{π(x^{0})\times q(x^{0}/x^{\star})} ,1}

The numerator is posterior density in x^{\star} and denominator is posterior density in x^{0} ??

#2

If you draw a candidate, \theta^*, from the proposal distribution and you run the RWMH, then you draw from \mathcal{N}(\theta^{(s-1)},\Sigma). So the previous draw/state of the chain, \theta^{(s-1)}, enters as the (conditional) mean. The reason is that you are drawing from a random walk in order to explore the parameter space.

Since for the random walk with normal increments the proposal density is symmetric (due to symmetry of the normal w.r.t x and \mu) it holds that q(\theta^*\vert\theta^{(s-1)})=q(\theta^{(s-1)}\vert\theta^*).

Hence the acceptance probability simplifies to \alpha(\theta^*,\theta^{(s-1)})= min\{\frac{\pi(\theta^*)}{\pi(\theta^{(s-1)})},1\} and the numerator is posterior density in \theta^* and denominator is posterior density in \theta^{(s-1)}.

#3

You may also want to have a look at https://drive.google.com/file/d/1iLQ7QjI-HRGgN1NMrRjWQbQOtzxQ-vWP/view
The underlying issue is to get the Reversibility Condition to hold, i.e. moving from \theta^{j-1} to \theta^{j} with the same likelihood as moving back. For that, the likelihood of moving between points in the proposal obviously matters.

#4

I think that you mean q(x^{\star}/x^{0}) means that for example for a normal distribution N(x^{0},\Sigma) and for the second iteration our normal proposal distribution is N(x^{\star},\Sigma). And for the third iteration we have N(x^{\star \star} , \Sigma) as normal proposal distribution. Therefore each new sample is subject to the previous sample because mean of our proposal normal distribution changes in each iteration.

Or other intuition is that this issue or dependence of new sample is only for the first iteration, in other words in the first iteration we draw from a normal proposal distribution such as N(x^{0}/\Sigma) and new sample or x^{\star} depends to this distribution but other iterations or other new samples are from this distribution only and mean of proposal distribution do not change after the first step.

Can you show symmetry of the normal distribution with normal distribution graph for better understanding ??

#5
  1. With the random walk MH, the proposal is always centered at the previous value, i.e. N(x^i,\Sigma).
  2. The normal distribution is symmetric around its mean. See e.g. https://my.ilstu.edu/~gjin/hsc204-hed/Module-5-Summary-Measure-2/Figure-3-9.png
#6

Thanks so much professor.:+1::+1: