91Ó°ÊÓ

It is given that two variables \(\mu \,\,and\,\,\tau \)have the joint normal-gamma distribution such that \(E\left( \mu \right) = - 5\,\,,Var\left( \mu \right) = 1\,\,\,,E\left( \tau \right) = \frac{1}{2}\,\,and\,\,Var\left( \tau \right) = \frac{1}{8}\).

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

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 the 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,

\(E\left( \tau \right) = \frac{1}{2}\,\,and\,\,Var\left( \tau \right) = \frac{1}{8}\)

So equating equations,

\(\begin{align}\frac{{{\alpha _0}\,}}{{{\beta _0}}} &= \frac{1}{2} \ldots \left( 1 \right)\\\frac{{{\alpha _0}\,}}{{{\beta _0}^2}} &= \frac{1}{8} \ldots \left( 2 \right)\end{align}\)

Solving (1) and (2)

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

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,

\(E\left( \mu \right) = - 5\,\, \ldots \left( 4 \right)\)

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

\(\begin{align}E\left( {E\left( {\mu \,\,|\,\,\tau } \right)} \right) &= E\left( \mu \right)\\ &= - 5\end{align}\)

Therefore, \({\mu _0} = - 5\) .

Since the precision is the inverse of variance, therefore,

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

Now

\(\begin{align}Var\left( \mu \right) &= E\left( {Var\left( {\mu \,\,|\,\,\tau } \right)} \right) + Var\left( {E\left( {\mu \,\,|\,\,\tau } \right)} \right)\\1 &= E\left( {\frac{1}{{{\lambda _0}\tau }}} \right) + Var\left( { - 5} \right)\\{\lambda _0} &= E\left( {\frac{1}{\tau }} \right) + 0\\{\lambda _0} &= 4\end{align}\)

Note that since \(\,\tau \) is the gamma distribution with parameters \({\alpha _0} = 2,{\beta _0} = 4\)

By its properties, \(E\left( {\frac{1}{\tau }} \right) = \frac{2}{{E\left( \tau \right)}}\) .

Hence, \({\mu _0} = - 5,{\lambda _0} = 4,{\alpha _0} = 2,{\beta _0} = 4\)