91Ó°ÊÓ

The 10 data points are: 0.86, 1.53, 1.57, 1.81, 0.99, 1.09, 1.29, 1.78, 1.29, 1.58.

The additional 20 data points are: 1.68, 1.9, 1.06, 1.3, 1.52, 1.74, 1.16, 1.49, 1.63, 1.99, 1.15, 1.33, 1.44, 2.01, 1.31, 1.46, 1.72, 1.25, 1.08, 1.25.

The prior in example 8.6.2 is \({\mu _0} = 1,{\lambda _0} = 1,{\alpha _0} = 0.5,{\beta _0} = 0.5\)

Let \(\mu \,\,{\rm{and}}\,\,\tau \) be random variables. Suppose that the conditional distribution of \(\mu \,\,{\rm{given}}\,\,\tau \) is the normal distribution with mean \({\mu _0}\) and precision \({\lambda _0}\tau \) .

Suppose also that the marginal distribution of \(\,\tau \) is the gamma distribution with parameters \({\alpha _0}\,\,{\rm{and}}\,\,\,{\beta _0}\).

Then we say that the joint distribution of\(\mu \,\,{\rm{and}}\,\,\,\tau \) is the normal-gamma distribution with hyperparameters \({\mu _0},{\lambda _0},{\alpha _0},{\beta _0}\).

We know that,

\(\begin{align}{\mu _1} &= \frac{{{\lambda _0}{\mu _0} + n\overline {{x_n}} }}{{{\lambda _0} + n}}\\{\lambda _1} &= {\lambda _0} + n\\{\alpha _1} &= {\alpha _0} + \frac{n}{2}\\{\beta _1} &= {\beta _0} + \frac{{{s_n}^2}}{2} + \frac{{n{\lambda _0}{{\left( {\overline {{x_n}} - {\mu _0}} \right)}^2}}}{{2\left( {{\lambda _0} + n} \right)}}\end{align}\)

Also, the values of:

\(\begin{align}\overline {{x_n}} &= \frac{{0.86 + 1.53 + 1.57 + {\rm{ }} \ldots + 1.25 + {\rm{ }}1.08 + {\rm{ }}1.25}}{{30}}\\ &= 1.442\end{align}\)

\(\begin{align}{s_n}^2 &= \frac{{\sum\limits_{i = 1}^n {\left( {{x_i} - \overline {{x_n}} } \right)} }}{{n - 1}}\\ &= \frac{{2.67108}}{{29}}\\ &= 0.0921\end{align}\)

\(\begin{align}{\mu _1} &= \frac{{1 \times 1 + 30 \times 1.442}}{{1 + 30}}\\ &= 1.427\\{\lambda _1} &= 1 + 30\\ &= 31\\{\alpha _1} &= 0.5 + \frac{{30}}{2}\\ &= 15.5\\{\beta _1} &= 1 + \frac{{0.0921}}{2} + \frac{{30 \times 1 \times {{\left( {1.442 - 1} \right)}^2}}}{{2\left( {1 + 30} \right)}}\\ &= 1.1405\end{align}\)

Therefore, the answer is: \({\mu _1} = 1.427,{\lambda _1} = 31,{\alpha _1} = 15.5,{\beta _1} = 1.1405\)

Now,\({\mu _0} = 1.345,{\lambda _0} = 11,{\alpha _0} = 5.5,{\beta _0} = 1.0484\), \(\overline {{x_n}} = 1.442\), \({s_n}^2 = 0.0921\).

\(\begin{align}{\mu _1} &= \frac{{11 \times 1.345 + 30 \times 1.442}}{{1 + 30}}\\ &= 1.872\\{\lambda _1} &= 11 + 30\\ = 41\end{align}\)

\(\begin{align}{\alpha _1} &= 5.5 + \frac{{30}}{2}\\ &= 82.5\\{\beta _1} &= 1 + \frac{{0.0921}}{2} + \frac{{30 \times 11 \times {{\left( {1.442 - 1.345} \right)}^2}}}{{2\left( {11 + 30} \right)}}\\ &= 1.0839\end{align}\)

Therefore, the answer is: \({\mu _1} = 1.872,{\lambda _1} = 41,{\alpha _1} = 82.5,{\beta _1} = 1.0839\)