Question

Consider the conditions of Exercise 10 again. Suppose also that it is found in a random sample of size n = 10 \(\overline {{{\bf{x}}_{\bf{n}}}} {\bf{ = 1}}\,\,{\bf{and}}\,\,{{\bf{s}}_{\bf{n}}}^{\bf{2}}{\bf{ = 8}}\) . Find the shortest possible interval so that the posterior probability \({\bf{\mu }}\) lies in the interval is 0.95.

Short answer

(0.446,1.530)

Step-by-step solution

Given information

It is given that two variables \(\mu \,\,and\,\,\tau \)have the joint normal-gamma distribution such that \(E\left( \mu \right) = 0\,\,\,,E\left( \tau \right) = 2\,,E\left( {{\tau ^2}} \right) = 2\,\,\,and\,\,\Pr \left( {\left| \mu \right| < 1.412} \right) = 0.5\).

Also, it is given that a random sample of size n = 10, it is found that \(\overline {{x_n}} = 1\,\,and\,\,{s_n}^2 = 8\) .

Define Normal-Gamma distribution

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

Define a new variable

Let,

\(\begin{align}U &= {\left( {\frac{{n\left( {n - 1} \right)}}{{{s_n}^2}}} \right)^{\frac{1}{2}}}\left( {\mu - \overline {{x_n}} } \right)\\ &= {\left( {\frac{{10\left( {10 - 1} \right)}}{8}} \right)^{\frac{1}{2}}}\left( {\mu - 1} \right)\\ &= \left( {3.3541} \right)\left( {\mu - 1} \right)\end{align}\)

Here U follows t distribution with \(n - 1\) degrees of freedom, that is 9.

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

\(\begin{align}P\left( {a < \mu < b} \right) = 0.95\\ \Rightarrow P\left( {\left( {3.3541} \right)\left( {a - 1} \right) < \left( {3.3541} \right)\left( {\mu - 1} \right) < \left( {3.3541} \right)\left( {b - 1} \right)} \right) = 0.95\\ \Rightarrow P\left( {\left( {3.3541} \right)\left( {a - 1} \right) < U < \left( {3.3541} \right)\left( {b - 1} \right)} \right) = 0.95\end{align}\)

The confidence intervals derived for U are -2.26216 and 2.26216

Therefore, the value of a and b is,

\(\begin{align}\left( {3.3541} \right)\left( {a - 1} \right) = - 2.26216\\ \Rightarrow a = 0.446\end{align}\)

And,

\(\begin{align}\left( {3.3541} \right)\left( {b - 1} \right) = 2.26216\\ \Rightarrow b = 1.530\end{align}\)

Therefore, the interval is (0.446,1.530)