# Bayesian Posterior: SD = SE?

#### hlsmith

##### Less is more. Stay pure. Stay poor.
I was curious if the standard deviation in a Bayesian posterior is equal to the standard error?

#### Dason

A standard error is literally just the standard deviation of the sampling distribution for whatever estimator you're using.

So I'd say it's analogous but isn't exactly the same.

#### Jake

With that said, we might still ask if there are cases where these two quantities numerically coincide, i.e., $$SD_{posterior} = SD_{sampling}$$. To focus the discussion, consider the case of the regression coefficients in a regression model (i.e., the $$\beta$$s). So the answer here is, I believe, "in some special cases the two coincide, but not in general or even in the most common real-world cases."
For one thing, if the Bayesian model in question involves non-flat priors, then the two will not coincide, except by accident. Now, if one uses a flat prior on $$\beta$$ AND we consider the residual variance $$\sigma^2$$ to be fixed/known for both the Bayesian and classical models, then I think that the two do coincide, since the posterior distribution $$\beta | y$$ is then Normal and I think numerically identical to the Normal sampling distribution of $$\hat{\beta}$$. But of course we usually consider $$\sigma^2$$ to be unknown and something we need to estimate. In that case, the posterior $$\beta | y$$ is no longer Normal, since it's marginalized over whatever the distribution for $$\sigma^2$$ is. But the sampling distribution of $$\hat{\beta}$$ is still Normal. So the SDs of these two distributions shouldn't match except by accident. However I think it's true that $$\beta | y, \sigma^2=\hat{\sigma^2}$$ -- the posterior that conditions on $$\sigma^2$$ being the point estimate $$\hat{\sigma^2}$$ from the classical model -- is still Normal and matching the sampling distribution as before. Whatever that buys you. Anyway, some actual statistician type person should check over all this.