# Interpreting the sampling distribution of $\bar{X}$ as the posterior distribution of $\mu$

Suppose we wanted to do inference on the mean $\mu$ of some population for which the variance is known.

The conventional frequentist approach is as follows. Given a random sample $X_1, \dots, X_n$ with $\text{E}(X_i) = \mu$ and $\text{Var}(X_i) = \sigma^2$, the CLT tells us that

$$\frac{\bar{X} - \mu}{\sigma / \sqrt{n}}$$

converges in distribution to a standard normal. We can then use this fact to construct confidence intervals or compute p-values using actual data $x_1, \dots, x_n$ collected from the population, provided $n$ is large enough.

Here's my question: Is there a Bayesian reformulation of this problem that allows us to essentially treat the above sampling distribution of $\bar{X}$ as a posterior distribution on $\mu$? In other words, is there a case in which

$$\mu \sim N(\bar{x}, \sigma^2/n)$$

is actually a valid conclusion?

• The question is: why would you like to have a Bayesian estimate to obtasin frequentist confidence intervals and p-values?
– Tim
Oct 24, 2017 at 7:58
• I think my question was motivated by the fact that non-statisticians are often tempted to use frequentist methods to derive a sampling distribution and then use that distribution to make probabilistic statements that treat the parameter of interest as a random variable, e.g. computing $P(\mu > 1)$. Oct 25, 2017 at 6:48

Yes. If $$X_1,\ldots,X_n \stackrel{ind}{\sim} N(\mu,\sigma^2)$$ with $\sigma^2$ known and your prior on $\mu$ is an improper prior over the whole real line, i.e. $p(\mu)\propto 1$, then $$\mu|x_1,\ldots,x_n \sim N(\bar{x},\sigma^2/n).$$