Question

Consider the analysis performed in Example 8.6.2. This time, use the usual improper before computing the parameters' posterior distribution.

Short answer

\({\mu _1} = 1.379,{\lambda _1} = 10,{\alpha _1} = 4.5,{\beta _1} = 0.4831\)

Step-by-step solution

Given information

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 prior hyperparameters are: \({\mu _0} = 1,{\lambda _0} = 1,{\alpha _0} = 0.5,{\beta _0} = 0.5\)

Define Normal-Gamma distribution

Let \(\mu \,\,and\,\,\tau \) be random variables. Suppose that the conditional distribution of \(\mu \,\,given\,\,\tau \) 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 parameter \({\alpha _0}\,\,and\,\,{\beta _0}\). Then we say that the joint distribution of\(\mu \,\,and\,\,\tau \) is the normal-gamma distribution with hyperparameters \({\mu _0},{\lambda _0},{\alpha _0},{\beta _0}\).

Calculate the mean and sample variance

\(\begin{align}\overline {{x_n}} &= \frac{{0.86 + 1.53 + 1.57 + {\rm{ }} \ldots + 1.78 + {\rm{ }}1.29 + {\rm{ }}1.58}}{{10}}\\ &= 1.379\end{align}\)

\(\begin{align}{s_n}^2 &= \frac{{\sum\limits_{i = 1}^n {\left( {{x_i} - \overline {{x_n}} } \right)} }}{{n - 1}}\\ &= 0.96629\end{align}\)

The posterior distribution is:

The posterior hypermeters by the improper prior method are:

\(\begin{align}{\mu _1} &= \overline {{x_n}} \\ &= 1.379\\{\lambda _1} &= n\\ &= 10\end{align}\)

\(\begin{align}{\alpha _1} &= \frac{{n - 1}}{2}\\ &= 4.5\\{\beta _1} &= \frac{{{s_n}^2}}{2}\\ &= 0.4831\end{align}\)

Therefore, the answer is \({\mu _1} = 1.379,{\lambda _1} = 10,{\alpha _1} = 4.5,{\beta _1} = 0.4831\)