Likelihood of n independent samples from a Normal distribution

[@willhitchcock5139]

if Mu_l = y.bar, then mu_l ^2 does not equal (1/n) sum(y_i^2) and it doesnt match up with what you had before

[@BROWNKEY]

You have made my life easier . Thank you girl. My notes are very poor at explaining

[@namitaa]

queen thank you

[@TuNguyen-ox5lt]

I think the product of two gaussian pdf is just "proportional" to another gaussion pdf .

[@deetoher]

Tú Nguyễn this is a short recording done in a class where I emphasise that once we can show the functional form of a posterior then if it is a known form of a distribution, then we can easily find the constant of integration. Thus in all these cases (looking at conjugacy), we are only interested in proportionality.

[@alexdiaz2066]

ms. patrickfmt

[@joacorapela]

This result proved false: Counter example: n=2, y_1=0, y_2=0, \mu=0, \sigma^2=1. Then f(y_1, y_2|\mu, \sigma^2)=N(0|0, 1)*N(0|0, 1)=0.3989*0.3989=0.1592 which is different from N(\mu|\bar{y}, \sigma^2/2)=N(0|0,0.5)=0.5642. Can anybody guess what is wrong in the proof?