# Why accept Metropolis-Hastings sample if more probable than previous sample?

One step in the Metropolis-Hastings sampling algorithm is to determine whether a sample $x_i$ is to be accepted based on the previous sample $x_{i-1}$. My understanding is that $x_i$ is accepted with probability $P(x_i)/P(x_{i-1})$ (treated as certain if greater than 1). I essentially understand the reasoning behind this scheme, but why would we want to compare the probability of a sample to the previous sample? In other words, what is wrong with the following naive sampling algorithm?

1. Choose an arbitrary $x_0$.
2. Accept $x_0$ with probability $P(x_0)$.
3. Repeat for $x_i$.

What is the need for a specialized sampling distribution?

• I have changed $P(x{i-1})$ to $P(x_{i-1})$, as I believe was intended. (Close inspection of the latex reveals {} around the "i-1" but it was missing the "_" in front.) In the event my interpretation is incorrect then please revert my edit. Commented Oct 13, 2015 at 17:56

You typically cannot compute the probability $P$, it is only known up to a multiplicative constant. To see why, consider Bayes theorem:
$$P(\theta|x) = \frac{P(x|\theta)P(\theta)}{\displaystyle{\int_{\theta'} P(x|\theta')P(\theta')~\mathrm{d}\theta'}}$$