# The role of variance of the distribution plays in Bayesian inference

Given prior $$\mu \sim \mathcal{N}(\mu_0, \tau^2)$$, likelihood $$X_i | \mu \sim \mathcal{N}(\mu, \sigma^2)$$, we know the closed-form solution of posterior is $$\mu | X_1, X_2, \ldots, X_n \sim \mathcal{N}(\mu_n, \tau_n^2)$$ with $$\frac{1}{\tau_n^2} = \frac{1}{\tau^2} + \frac{n}{\sigma^2}$$ and $$\mu_n = \frac{\frac{1}{\tau^2} \mu_0 + \frac{n}{\sigma^2} \bar{X}}{\frac{1}{\tau^2} + \frac{n}{\sigma^2}}$$. It can be observed that given the same number of observations $$n$$,the larger the $$\sigma$$ is, the larger the variance in the posterior is. I find it hard to get an intuition of this, because I feel fitting the same number of observations (i.e., MLE) should have the same level of certainty regardless of how spread out the observations are. In other words, I feel it is easy to understand as the number of observations gets large, the uncertainty in posterior diminishes, but the uncertainty should be independent of the sample variance (obviously I am wrong but I want to have the intuition why that is the case).

Think about a simpler problem: estimating the mean $$\mu$$ in a frequentist setting.
It is sensible to use the MLE, $$\bar X$$, as our estimator. How accurate is this on average? We know that $$\bar X \sim \mathcal{N}(\mu, \sigma^2/n)$$ so the mean squared error is $$\sigma^2/n$$, i.e proportional to the population variance.
This is quite a fundamental point. It is easier to estimate a population mean when the population variance is low, because (on average) your observations will be closer to this mean. For example, if the effect of a medical intervention is very similar for all patients (low $$\sigma$$), you don't need a large sample to estimate the mean effect accurately.