91Ó°ÊÓ

The prior distribution ofθis a normal distribution for which the mean is 68 inches and the standard deviation is 1 inch

In the notation of Theorem 7.3.3,

Suppose thatX₁,...,X_nform a random sample from a normal distribution for which

the value of the meanθis unknown and the value of the variance \({\sigma ^2} > 0\) is known.

Suppose also that the prior distribution ofθis the normal distribution with mean \({\mu _0}\)

and variance \({v_0}^2\). Then the posterior distribution ofθgiven thatX_i=xi(i=1,...,n)

is the normal distribution with mean \({\mu _1}\) and variance \({v_1}^2\) where

\({\mu _1} = \frac{{{\sigma ^2}\mu + n{v^2}{{\bar x}_n}}}{{{\sigma ^2} + n{v^2}}}\)

\({v_1}^2 = \frac{{{\sigma ^2}{v^2}}}{{{\sigma ^2} + n{v^2}}}\)

we have\({\sigma ^2} = 4,\mu = 68,{v^2} = 1,n = 10,\,and\,{\bar x_n} = 69.5\)

The posterior distribution of\(\theta \)is given by,

\(\begin{aligned}{\mu _1} = \frac{{{\sigma ^2}\mu + n{v^2}{{\bar x}_n}}}{{{\sigma ^2} + n{v^2}}}\\ = \frac{{4 \times 68 + 10 \times 1 \times 69.5}}{{4 + 10 \times 1}}\\ = 69.07143\\ \simeq 69.07\end{aligned}\)

The mean is 69.07

\(\begin{aligned}{v_1}^2 = \frac{{{\sigma ^2}{v^2}}}{{{\sigma ^2} + n{v^2}}}\\ = \frac{{4 \times 1}}{{4 + 10 \times 1}}\\ = 0.285714\\ \simeq 0.2857\end{aligned}\)

The variance is 0.2857