This is a part of a larger proof, which is irrelevant here. What I'm stuck on is showing that this:
k*(u0-u)^2+n*(y-u)^2
is the same as
(k*n/(k+n))*(y-u0)^2+(k+n)*((k*u0+n*y)/(k+n)-u)^2
Note that u0 and u are two different variables.
I've been factoring and pushing things around for more than 2 hours. I'm embarrassed to be stuck
princeps- Thank you so much!!!
You're right I probably should have. Although it's a bit easier for me to use notation similar to my original variables (selfish, I know). 'u' refers to the "true mean" and 'u0' refers to the mean of the prior. It's from the process of updating the prior of a normal-inverse chi squared distribution.