Question

Show that two random variables \({\bf{\mu }}\,\,{\bf{and}}\,\,{\bf{\tau }}\)cannot have the joint normal-gamma distribution such that \({\bf{E}}\left( {\bf{\mu }} \right){\bf{ = 0}}\,\,{\bf{,E}}\left( {\bf{\tau }} \right){\bf{ = 1}}\,\,{\bf{and}}\,\,{\bf{Var}}\left( {\bf{\tau }} \right){\bf{ = 4}}\)

Short answer

The conditions imply that \({\alpha _0} = \frac{1}{4}\)and \(E\left( \mu \right)\) exist only for \({\alpha _0} > \frac{1}{2}\)\({\alpha _0}\,\,{\rm{and}}\,\,{\beta _0}\)

Step-by-step solution

Given information

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

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 .

Then we say that the joint distribution of\(\mu \,\,{\rm{and}}\,\,\tau \) 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,

\(E\left( \tau \right) = 1\,\,and\,\,Var\left( \tau \right) = 4\)

So equating equations,

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

Solving (1) and (2)

\({\alpha _0} = {\beta _0} = \frac{1}{4}\)

The conditions imply that \({\alpha _0} = \frac{1}{4}\)and \(E\left( \mu \right)\) which is equal to 0 exists only for \({\alpha _0} > \frac{1}{2}\)