# What is the correct formula for Bayesian update for normal distribution with known variance [duplicate]

As question title states, I'm interesting in Bayesian update of normally distributed data with known variance. I compared three sources and they seems to contradict each other. I use some kind of unified notation:

$$\mu_{post} = \frac{n\times\sigma_{prior}^2}{\sigma_{known}^2+n \times\sigma_{prior}^2}\times \bar x+\frac{\sigma_{known}^2}{\sigma_{known}^2+n\times\sigma_{prior}^2}\times \mu_{prior}$$

$$\sigma_{post}^2=(\frac{1}{\sigma_{prior}^2}+\frac{n}{\sigma_{known}^2})^{-1}$$

$$\mu_{post} = \frac{\sigma_{prior}^2}{\frac{\sigma_{known}^2}{n}+\sigma_{prior}^2}\times \bar x+\frac{\sigma_{known}^2}{\frac{\sigma_{known}^2}{n}+\sigma_{prior}^2}\times \mu_{prior}$$

$$\sigma_{post}^2=(\frac{1}{\sigma_{prior}^2}+\frac{n}{\sigma_{known}^2})^{-1}$$

$$\mu_{post} = \frac{\frac{n}{\sigma_{known}^2}}{\frac{n}{\sigma_{known}^2}+\sigma_{prior}^2}\times \bar x+\frac{\sigma_{prior}^2}{\frac{n}{\sigma_{known}^2}+\sigma_{prior}^2}\times \mu_{prior}$$

$$\sigma_{post}^2=(\sigma_{prior}^2+\frac{n}{\sigma_{known}^2})^{-1}$$

• The most basic difference is that Murply uses normal distribution parametrized by variance $\sigma^2$, while Gelman uses precission $\tau^2$ (you incorrectly used the same symbol in both cases).
• Yes, that correct, Gelman uses precission, but I converted his precission $\tau^2$ to $\frac{1}{\sigma^2}$, am I? Dec 10, 2018 at 13:44