Question

Show that two random variables \({\bf{\mu }}\,\,{\bf{and}}\,\,{\bf{\tau }}\)cannot have the joint normal-gamma distribution such that

\({\bf{E}}\left( {\bf{\mu }} \right){\bf{ = 0}}\,\,{\bf{,Var}}\left( {\bf{\mu }} \right){\bf{ = 1,E}}\left( {\bf{\tau }} \right){\bf{ = }}\frac{{\bf{1}}}{{\bf{2}}}\,\,{\bf{and}}\,\,{\bf{Var}}\left( {\bf{\tau }} \right){\bf{ = }}\frac{{\bf{1}}}{{\bf{4}}}\)

Short answer

The conditions imply that \({\alpha _0} = 1\)and \(E\left( \mu \right)\) exist only for \({\alpha _0} > \frac{1}{2}\)

Step-by-step solution

Given information

It is given that two variables \(\mu \,\,and\,\,\tau \)have the joint normal-gamma distribution such that \(E\left( \mu \right) = 0\,\,,Var\left( \mu \right) = 1,E\left( \tau \right) = \frac{1}{2}\,\,{\rm{and}}\,\,Var\left( \tau \right) = \frac{1}{4}\)

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

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,

\(E\left( \tau \right) = \frac{1}{2}\,\,{\rm{and}}\,\,Var\left( \tau \right) = \frac{1}{4}\)

So equating equations,

\(\begin{align}\frac{{{\alpha _0}\,}}{{{\beta _0}}} &= \frac{1}{2} \ldots \left( 1 \right)\\\frac{{{\alpha _0}\,}}{{{\beta _0}^2}} &= \frac{1}{4} \ldots \left( 2 \right)\end{align}\)

Solving (1) and (2)

\(\begin{align}{\alpha _0} &= 1\\{\beta _0} &= 2\end{align}\)

The conditions imply that \({\alpha _0} = 1\)and \(E\left( \mu \right)\) which is equal to 0 exists only for \({\alpha _0} > \frac{1}{2}\)