Didn't see an easy bayes so decided to try bhulmann since it's a conjugate prior but i'm getting the wrong answer....
mu = .9*5*.01 + .1*4*5*.01 = 0.065
V1 = 5(0.01)^2 = 0.0005
V2 = (4^2)*5(0.01)^2 = 0.008
v = .9(0.0005) + .1(0.008) = 0.00125
a = var(0.9E(g) + 0.1*4*E(g))
= var(1.3*E(g))
= (1.3^2)var(E(g))
= (1.3^2)(5*(0.01^2)) = 0.000845
Z = 1/(1+k) = 0.403
from to the solution the Z should be very small around 0.01-0.02
Pretty sure I did something wrong with a or v.
I can't figure out these mixture questions! Ahh!
Mahler PE18 Q26