# In the Metropolis algorithm, if the ratio of probabilities $r$ is less than $1$, why not directly reject instead of accepting with probability $r$? [duplicate]

Suppose that we want to sample from a posterior distribution $p(\theta|y)$ but we do not know how to directly sample. Suppose instead that we have a working set of values $\{\theta^{(1)}, \ldots, \theta^{(S)}\}$ and that we have a new draw $\theta^{*}$ that we want to know if we want to set equal to $\theta^{(S+1)}$, the next value of the set.

Then, the algorithm states to look at:

$$r = \frac{p(\theta^{*}|y)}{p(\theta^{S}|y)}$$

If $r>1$, then $\theta^{*}$ is in a sense more likely than $\theta^{(S)}$, so we set $\theta^{*} = \theta^{(S+1)}$.

Now, if $r<1$, then $\theta^{*}$ is in a sense less likely than $\theta^{(S)}$. Then, we set $\theta^{*} = \theta^{(S+1)}$ with probability $r$ and $\theta^{*} = \theta^{(S)}$ with probability $1-r$.

This is where I am confused, if $p(\theta^{*}|y) < p(\theta^{S}|y)$, shouldn't we just directly reject $\theta^{*}$ since it is less likely to occur? Why do we even have this scheme where we will accept it with probability $r$? Why not directly remove it? Thanks.

• Because you want to sample from the distribution, not just find the (local) maximum. Also, your $r$ equation is only for a symmetric proposal. – jaradniemi Mar 24 '17 at 0:51