confused about proposal distribution in MCMC

Question

This is a question from notes

I have some questions regarding the proposal distribution which is $N(x,1)$

1. Is the proposal distribution symmetric i.e. $g(x_p|x)=g(x|x_p)$?

2. I'm not sure whether it is correct: let's x=0 first, and rand() in excel gives 0.95 as culumative probability so $x_p=1.64$ and $g(1.64|0)=0.95$, but how come $g(0|1.64)$=$g(1.64|0)=0.95$?

This is a binomial example

Answer 1

1. Yes: the normal distribution cares only about the distance between the mean and the point. Note that in $g(x \mid x_p)$, the mean is $x_p$ now, not $x$, so the distance between the mean in the two cases are the same and the variance in the two cases are the same. (In this symmetric situation, the algorithm is commonly called just Metropolis, rather than Metropolis-Hastings.)
2. The $f$ in the formula is the PDF of the desired sampling distribution, not the CDF. (In a discrete problem it would be the PMF, not the CDF.) The same is true of $g$, although because of the symmetry, $g$ will simply cancel out anyway.

By the way, because the shape of your distribution is much like a normal distribution with variance $1/2$, not $1$, you will probably get better performance if your sampling uses a variance $1/2$ normal r.v. rather than $1$. (That said, in general choosing the candidate distribution is more art than science: it needs to be "adventurous" enough to efficiently explore the configuration space, "conservative" enough that most jumps are accepted, and computationally tractable to draw many samples from it.)