91Ó°ÊÓ

It is given that two variables \(\mu \,\,and\,\,\tau \)have the joint normal-gamma distribution such that \(E\left( \mu \right) = 0\,\,\,\,,E\left( \tau \right) = 2,E\left( {{\tau ^2}} \right) = 5\,\,\,and\,\,\Pr \left( {\left| \mu \right| < 1.412} \right) = 0.5\)

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

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,

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

Also,

But it is given that,

\(\begin{align}E\left( \tau \right) &= 2\,\\\,\,Var\left( \tau \right) &= 5 - {2^2}\\ &= 5 - 4\\ &= 1\end{align}\)

So equating equations,

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

Solving (1) and (2)

\({\alpha _0} = 4,{\beta _0} = 2\)

Since in the definition it is given that the conditional distribution of \(\mu \,\,given\,\,\tau \) is the normal distribution with mean \({\mu _0}\) and precision \({\lambda _0}\tau \)

Which means,

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

It is also given,

\(E\left( \mu \right) = 0\,\, \ldots \left( 4 \right)\)

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

\(\begin{align}E\left( {E\left( {\mu \,\,|\,\,\tau } \right)} \right) &= E\left( \mu \right)\\ &= 0\end{align}\)

Therefore,\({\mu _0} = 0\).

Since the precision is inverse of variance, therefore,

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

To find variance,

We have to covert the probability\(\Pr \left( {\left| \mu \right| < 1.412} \right) = 0.5\)with Z transformation.

\(\Pr \left( {\left| Z \right| < \frac{{1.412}}{\sigma }} \right) = 0.5\)

The Z value is 1.96, which means,

\(\begin{align} \Rightarrow \frac{{1.412}}{\sigma } &= \pm 1.96\\ \Rightarrow \sigma &= 1.388\end{align}\)

Therefore, the inverse of variance is:

\(\begin{align}{\lambda _0} &= \frac{1}{{{{1.388}^2}}}\\ &= 0.51\end{align}\)

Hence,\({\mu _0} = 0,{\lambda _0} = 0.51,{\alpha _0} = 4,{\beta _0} = 2\)