Question

Suppose that \({{\bf{X}}_{\bf{1}}}{\bf{, }}{\bf{. }}{\bf{. }}{\bf{. , }}{{\bf{X}}_{\bf{n}}}\) form a random sample from the normal distribution with unknown mean \({\bf{\mu }}\,\,{\bf{and}}\,\,{\bf{\tau }}\), and also that the joint prior distribution of \({\bf{\mu }}\,\,{\bf{and}}\,\,{\bf{\tau }}\) is the normal-gamma distribution satisfying the following conditions: \({\bf{E}}\left( {\bf{\tau }} \right){\bf{ = 1,Var}}\left( {\bf{\tau }} \right){\bf{ = }}\frac{{\bf{1}}}{{\bf{3}}}\,\,\,{\bf{and}}\,\,{\bf{Pr}}\left( {{\bf{\mu > 3}}} \right){\bf{ = 0}}{\bf{.5}}\,\,{\bf{and}}\,\,{\bf{Pr}}\left( {{\bf{\mu > 0}}{\bf{.12}}} \right){\bf{ = 0}}{\bf{.9}}\,\)

Determine the prior hyper parameters \({{\bf{\mu }}_{\bf{0}}}{\bf{,}}{{\bf{\lambda }}_{\bf{0}}}{\bf{,}}{{\bf{\alpha }}_{\bf{0}}}{\bf{,}}{{\bf{\beta }}_{\bf{0}}}\)

Short answer

\({\mu _0} = - 9.8,{\lambda _0} = 0.016,{\alpha _0} = 3,{\beta _0} = 3\)

Step-by-step solution

Given information

It is given that two variables \(\mu \,\,and\,\,\tau \)have the joint normal-gamma distribution such that \(E\left( \tau \right) = 1,Var\left( \tau \right) = \frac{1}{3}\,\,\,and\,\,\Pr \left( {\mu > 3} \right) = 0.5\,\,and\,\,\Pr \left( {\mu > 0.12} \right) = 0.9\,\)

 Step 2: 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 \({\mu _0},{\lambda _0},{\alpha _0},{\beta _0}\).

Solve for \({{\bf{\alpha }}_{\bf{0}}}{\bf{,}}{{\bf{\beta }}_{\bf{0}}}\) 

Since in the definition it is given that the marginal distribution of \(\,\tau \) is the gamma distribution with parameters \({\alpha _0}\,\,and\,\,{\beta _0}\).

Therefore, by properties of gamma distribution,

\(E\left( {\,\tau } \right) = \frac{{{\alpha _0}\,}}{{{\beta _0}}},Var\left( {\,\tau } \right) = \frac{{{\alpha _0}\,}}{{{\beta _0}^2}}\)

But it is given that,

\(\begin{align}E\left( \tau \right) &= 1\\\,\,Var\left( \tau \right) &= \frac{1}{3}\end{align}\)

So equating equations,

\(\begin{align}\frac{{{\alpha _0}\,}}{{{\beta _0}}} &= 1 \ldots \left( 1 \right)\\\frac{{{\alpha _0}\,}}{{{\beta _0}^2}} &= \frac{1}{3} \ldots \left( 2 \right)\end{align}\)

Solving (1) and (2)

\({\alpha _0} = 3,{\beta _0} = 3\)

Solve for \({{\bf{\mu }}_{\bf{0}}}{\bf{,}}{{\bf{\lambda }}_{\bf{0}}}\) 

It is given that \(\,\Pr \left( {\mu > 3} \right) = 0.5\,\,and\,\,\Pr \left( {\mu > 0.12} \right) = 0.9\,\).

By transforming to Z variables, and solving for the mean and variance,

\(\begin{align}\,\Pr \left( {Z > \frac{{3 - \mu }}{\sigma }} \right) &= 0.5\,\,and\,\,\Pr \left( {Z > \frac{{0.12 - \mu }}{\sigma }} \right) &= 0.9\,\\ \Rightarrow \frac{{3 - \mu }}{\sigma } &= 1.65\,\,\,and\,\,\frac{{0.12 - \mu }}{\sigma } &= 1.28\\ \Rightarrow \mu &= - 9.8\,\,and\,\,\sigma &= 7.78\end{align}\)

Therefore,

\(E\left( {\mu \,\,|\,\,\tau } \right) = {\mu _0} \ldots \left( 3 \right)\)

It is also given,

\(E\left( \mu \right) = - 9.8\,\, \ldots \left( 4 \right)\)

Therefore, by taking expectation of (3) and equating it with (4)

\(\begin{align}E\left( {E\left( {\mu \,\,|\,\,\tau } \right)} \right) &= E\left( \mu \right)\\ &= - 9.8\end{align}\)

Therefore, \({\mu _0} = - 9.8\) .

Since the precision is inverse of variance, therefore,

\(Var\left( {\mu \,\,|\,\,\tau } \right) = \frac{1}{{{\lambda _0}\tau }}\),

The inverse of variance is:

\(\begin{align}{\lambda _0} &= \frac{1}{{{{7.78}^2}}}\\ &= 0.016\end{align}\)

Hence, \({\mu _0} = - 9.8,{\lambda _0} = 0.016,{\alpha _0} = 3,{\beta _0} = 3\)