91Ó°ÊÓ

One can get the desired product by using the conclusion in the mentioned example and the probability density functions of normal distribution and gamma distribution. The required product is a product of the following two functions

\(exp - \frac{{{u_0}{{\left( {\psi - {\psi _0}} \right)}^2}}}{2} - \int_{i = 1}^p {{\tau _i}} \beta + \frac{{{n_i}{{\left( {{\mu _i} - {{\bar y}_i}} \right)}^2} + {w_i} + {\lambda _0}{{\left( {\mu - \psi } \right)}^2}}}{2}\)

and the following expression

and\({w_i}\)is given in the same way as in the example

\({w_i} = _{j = 1}^{{n_i}}{\left( {{y_{ij}} - {{\bar y}_i}} \right)^2},\;\;\;{\kern 1pt} i = 1,2, \ldots ,p\)

The second term corresponds to the prior distribution.

As mentioned in the hint, the conditional distributions stay practically the same as in the example.

If they look at a product as a function,\({\tau _i},\)then you get a gamma distribution with parameters

\({\alpha _0} + \frac{{{n_i} + 1}}{2} and \beta + \frac{{{n_i}{{\left( {{\mu _i} - {{\bar y}_i}} \right)}^2} + {w_i} + {\lambda _0}{{\left( {\mu - \psi } \right)}^2}}}{2}\)

Next, as a function of\(\psi ,\)by writing it in a form widely known, one gets a normal distribution with parameters

Before commenting on the distribution of\(\beta .\)given all others, notice that the product above looks like a probability density function of a normal distribution when it is a function of\({\mu _i}\)for the same reason. The mean and precision are given with

\(\frac{{{n_i}{{\bar y}_i} + {\lambda _0}\psi }}{{{n_i} + {\lambda _0}}}\;\;\;{\kern 1pt} and \;\;\;{\kern 1pt} {u_0}{\tau _i}\left( {{n_i} + {\lambda _0}} \right)\)

Finally, if the product is seen as a function of\(\beta .\), it approximately probability density function of gamma distribution with parameters

The parameters that should be used are given in the exercise, and the others are:

\(\begin{aligned}{l}{n_1} = 20 for beef \\{n_2} = 17 for meat \\{n_3} = 17 for poultry \\{n_4} = 9 for specialty. \end{aligned}\)

Means\({\mu _i},i = 1,2,3,4\)correspond to the indices of\({n_i}, i = 1,2,3,4.\)

The code used for this simulation uses\(N = 20000\)Markov chains with\(I = 100000\)iterations or steps. It means that there is a total of\(I = 100000\)parameter vectors from which the result is obtained.

Note that the given code should be changed as the initial parameters are different. The estimated posterior means of\({\mu _i},i = 1,2,3,4\)are, respectively, 156.8, 158.4, 120.3, and 160.1. Similarly, the estimated posterior means for the\(1/{\tau _i},i = 1,2,3,4\)are, respectively,\(494.9,609.4,545.6,\)and\(570.5\)