91Ó°ÊÓ

The posterior distribution conditional on the first 10 observations that are to be treated as prior are: \({\mu _0} = 1.379,{\lambda _0} = 10,{\alpha _0} = 4.5,{\beta _0} = 0.4831\)

The 20 observations 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.

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\(\mu \,\,and\,\,\tau \) is the normal-gamma distribution with hyperparameters \({\mu _0},{\lambda _0},{\alpha _0},{\beta _0}\).

We know that,

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

Also, the values of:

\(\begin{aligned}{c}\overline {{x_n}} &= \frac{{1.68 + 1.9 + 1.06 + {\rm{ }} \ldots + 1.25 + {\rm{ }}1.08 + {\rm{ }}1.25}}{{20}}\\ &= 1.4735\end{aligned}\)

\(\begin{aligned}{c}{s_n}^2 &= \frac{{\sum\limits_{i = 1}^n {{{\left( {{x_i} - \overline {{x_n}} } \right)}^2}} }}{{n - 1}}\\ &= \frac{{1.64525}}{{19}}\\ &= 0.086592\end{aligned}\)

\(\begin{aligned}{c}{\mu _1} &= \frac{{10 \times 1.379 + 20 \times 1.4735}}{{1 + 20}}\\ &= 2.06\\{\lambda _1} &= 10 + 20\\ &= 30\end{aligned}\)

\(\begin{aligned}{c}{\alpha _1} &= 4.5 + \frac{{20}}{2}\\ &= 14.5\\{\beta _1} &= 1 + \frac{{0.0865}}{2} + \frac{{20 \times 10 \times {{\left( {1.4735 - 1.379} \right)}^2}}}{{2\left( {10 + 20} \right)}}\\ &= 4.0730\end{aligned}\)

Therefore, the answer is: \({\mu _1} = 2.06,{\lambda _1} = 30,{\alpha _1} = 14.5,{\beta _1} = 4.0730\)