Question

Suppose that two random variables\({\bf{\mu }}\,\,{\bf{and}}\,\,{\bf{\tau }}\)have the joint normal-gamma distribution with hyperparameters\({{\bf{\mu }}_{\bf{0}}}{\bf{ = 4,}}{{\bf{\lambda }}_{\bf{0}}}{\bf{ = 0}}{\bf{.5,}}{{\bf{\alpha }}_{\bf{0}}}{\bf{ = 1,}}{{\bf{\beta }}_{\bf{0}}}{\bf{ = 8}}\)Find the values of (a)\({\bf{Pr}}\left( {{\bf{\mu > 0}}} \right)\)and (b)\({\bf{Pr}}\left( {{\bf{0}}{\bf{.736 < \mu < 15}}{\bf{.680}}} \right)\).

Short answer

(a) the answer is 0.78868

(b) the answer is 0.25005

Step-by-step solution

Given information

It is given that two variables \(\mu \,\,and\,\,\tau \)have the joint normal-gamma distribution such that \({\mu _0} = 4,{\lambda _0} = 0.5,{\alpha _0} = 1,{\beta _0} = 8\)

Define Normal-Gamma distribution

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 \({\mu _0},{\lambda _0},{\alpha _0},{\beta _0}\).

Define a new variable

Let,

\(\begin{align}U &= {\left( {\frac{{{\lambda _0}{\alpha _0}}}{{{\beta _0}}}} \right)^{\frac{1}{2}}}\left( {\mu - {\mu _0}} \right)\\ &= {\left( {\frac{{0.5 \times 1}}{8}} \right)^{\frac{1}{2}}}\left( {\mu - 4} \right)\\ &= 0.25\left( {\mu - 4} \right)\end{align}\)

Here U follows t distribution with \(2{\alpha _0}\) degrees of freedom, that is 2.

(a) Find the confidence interval

\(\begin{align}\Pr \left( {\mu > 0} \right)\\ &= \Pr \left( {\left( {\mu - 4} \right) \times 0.25 > - 4 \times 0.25} \right)\\ &= \Pr \left( {U > - 1} \right)\\ &= {\rm{0}}{\rm{.78868}}\end{align}\)

Therefore, the answer is 0.78868

(b) Find the confidence interval

\(\begin{align}\Pr \left( {0.736 < \mu < 15.680} \right)\\ &= \Pr \left( {\left( {0.736 - 4} \right) \times 0.25 < \left( {\mu - 4} \right) \times 0.25 < \left( {15.680 - 4} \right) \times 0.25} \right)\\ &= \Pr \left( { - 0.816 < U < 2.92} \right)\\ &= 0.99994 - 0.74989\\ &= 0.25005\end{align}\)

Therefore, the answer is 0.25005