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:

1) Murphy K. P. (2007)

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


2) Jordan M. I. (2010)

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


3) Gelman A. et. al (2014): 42

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


I'm asking for two things:

1) verification of my notation changes from original sources

2) if they are correct, explanation of the difference between them

  • 1
    $\begingroup$ 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). $\endgroup$
    – Tim
    Dec 10, 2018 at 13:33
  • $\begingroup$ Yes, that correct, Gelman uses precission, but I converted his precission $\tau^2$ to $\frac{1}{\sigma^2}$, am I? $\endgroup$
    – aGricolaMZ
    Dec 10, 2018 at 13:44


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