Question

Consider again the conditions of Exercise 12. Suppose also that in a random sample of size n = 8, it is found that \(\sum\limits_{{\bf{i = 1}}}^{\bf{n}} {{{\bf{x}}_{\bf{i}}}{\bf{ = 16}}} \,\,{\bf{and}}\,\,\sum\limits_{{\bf{i = 1}}}^{\bf{n}} {{{\bf{x}}_{\bf{i}}}^{\bf{2}}{\bf{ = 48}}} \) . Find the shortest possible interval such that the posterior probability that \({\bf{\mu }}\) lies inthe interval is 0.99.

Short answer

(0.724,3.336)

Step-by-step solution

Given information

The conditions in exercise 12 state that \(E(\tau ) = 1,\,\,Var(\tau ) = \frac{1}{3},\,\Pr \left( {\mu > 3} \right) = 0.5\,\,and\,\,\Pr \left( {\mu > 0.12} \right) = 0.9\,\) .

It is also given that \(\sum\limits_{i = 1}^n {{x_i} = 16} \,\,and\,\,\sum\limits_{i = 1}^n {{x_i}^2 = 48} \).

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

Find sample mean and variance and define a new variable

Sample mean is,

\(\begin{align}\overline {{x_n}} &= \frac{{\sum\limits_{i = 1}^n {{x_i}} }}{n}\\ &= \frac{{16}}{8}\\ &= 2\end{align}\)

Sample variance is,

\(\begin{align}{s_n}^2 &= \frac{{\sum\limits_{i = 1}^n {{x_i}^2 - \left( {\frac{{{{\left( {\sum\limits_{i = 1}^n {{x_i}} } \right)}^2}}}{n}} \right)} }}{{n - 1}}\\ &= \frac{{48}}{7} - \frac{{{{16}^2}}}{{7 \times 8}}\\ &= 2.2857\end{align}\)

Let,

\(\begin{align}U &= {\left( {\frac{{n\left( {n - 1} \right)}}{{{s_n}^2}}} \right)^{\frac{1}{2}}}\left( {\mu - \overline {{x_n}} } \right)\\ &= {\left( {\frac{{8\left( {8 - 1} \right)}}{{2.2857}}} \right)^{\frac{1}{2}}}\left( {\mu - 2} \right)\\ &= \left( {4.9497} \right)\left( {\mu - 2} \right)\end{align}\)

Here U follows t distribution with \(n - 1\) degrees of freedom, that is 7.

Calculating the posterior probability that \({\bf{\mu }}\)  lies in the interval.

Let the confidence interval be of the form \(P\left( {a < \mu < b} \right) = 0.99\).

To find a and b, convert the parameter \(\mu \) to U by using the above-mentioned transformation.

\(\begin{align} \Rightarrow P\left( {\left( {4.9497} \right)\left( {a - 2} \right) < \left( {4.9497} \right)\left( {\mu - 2} \right) < \left( {4.9497} \right)\left( {b - 2} \right)} \right) &= 0.99\\ \Rightarrow P\left( {\left( {4.9497} \right)\left( {a - 2} \right) < U < \left( {4.9497} \right)\left( {b - 2} \right)} \right) &= 0.99\end{align}\)

Now U follows t distribution with 7 degrees of freedom. For 99% confidence interval the lower and upper limit of distribution is -6.3158and 6.6127

Substituting this, we get,

\(\begin{align}\left( {4.9497} \right)\left( {a - 2} \right) &= - 6.3158\\ \Rightarrow a &= 0.724\end{align}\)

And

\(\begin{align}\left( {4.9497} \right)\left( {b - 2} \right) &= 6.6127\\ \Rightarrow b &= 3.336\end{align}\)

Therefore, the interval is (0.724,3.336)