Question 8-17
I get the same as you but get 46/21. Check your numerator [6*1 + 5*4*2].
I get 6 + 40 = 46, not 28.
Ashwin
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Original Text
From: "Sheri De La Boursodiere" <Sheri_De_La_Boursodiere@hsbc.ca>, on
3/22/99 4:20 PM:
Question 8-17 assumes:
(1) X1, X2,..., Xn are indentically distributed normal rv with mean theta
and variance 1.
(2) X1, X2, ..., Xn are conditionally independent, given theta.
(3) The prior density of tehat is a normal density, with mean 6 and
variance 5.
Given values n=4, X1=-1, X2=0, X3=7, X4=4, determine the mean of the
posterior normal distribution of theta.
I solve this problem by using the formula for the mean of a posterior
normal distribution that has a Normal distribution for the conditional and
prior distributions.
U= [U2 * variance1] + [variance2* n *(average of X's)]
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variance1 + (n * variance2)
where U2 and variance2 are the mean and variance of the conditional
distribution, and variance2 is the variance of the prior distribution.
Substituting the values into the formula I get:
U= [6*1 + 5*4*2]/[1 + 4*5] = 28/21
The book and solutions manual answers are U=26/21
Cna someone help me figure out what I am doing wrong here?