Question

We will draw a sample of size n = 11 from the normal distribution with the mean \({\bf{\mu }}\,\,{\bf{and}}\,\,{\bf{\tau }}\). We will use a natural conjugate prior for the parameters \({\bf{\mu }}\,\,{\bf{and}}\,\,{\bf{\tau }}\) from the normal-gamma family with hyperparameters \({{\bf{\alpha }}_{\bf{0}}}{\bf{ = 2,}}{{\bf{\beta }}_{\bf{0}}}{\bf{ = 1,}}{{\bf{\mu }}_{\bf{0}}}{\bf{ = 3}}{\bf{.5}}\,\,{\bf{and}}\,\,{{\bf{\lambda }}_{\bf{0}}}{\bf{ = 2}}\)

The sample yields an average of \(\overline {{{\bf{x}}_{\bf{n}}}} {\bf{ = 7}}{\bf{.2}}\,\,{\bf{and}}\,\,{{\bf{s}}_{\bf{n}}}^{\bf{2}}{\bf{ = 20}}{\bf{.3}}\)

a. Find the posterior hyperparameters.

b. Find an interval that contains 95% of the posterior distribution of \({\bf{\mu }}\).

Short answer

1. \({\mu _1} = 6.631\,\,,{\lambda _1} = 13,\,\,{\alpha _1} = 7.5,\,\,{\beta _1} = 22.73\)
2. (5.602,7.660).

Step-by-step solution

Given information

It is given that two variables \(\mu \,\,and\,\,\tau \)have the joint normal-gamma distribution such that \({\alpha _0} = 2,{\beta _0} = 1,{\mu _0} = 3.5\,\,\,{\rm{and}}{\lambda _0} = 2\).

It is also given that n=11 \(\overline {{x_n}} = 7.2\,{\rm{and}}\,\,{s_n}^2 = 20.3\).

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 \({\alpha _0}\,\,and\,\,{\beta _0}\).

Then we say that the joint distribution of\(\mu \,\,\,{\rm{and}}\,\tau \) is the normal-gamma distribution with hyperparameters \({\mu _0},{\lambda _0},{\alpha _0},{\beta _0}\).

(a) Define the new variables

We know that,

\(\begin{align}{\mu _1} &= \frac{{{\lambda _0}{\mu _0} + n\overline {{x_n}} }}{{{\lambda _0} + n}}\\{\lambda _1} &= {\lambda _0} + n\\{\alpha _1} &= {\alpha _0} + \frac{n}{2}\\{\beta _1} &= {\beta _0} + \frac{{{s_n}^2}}{2} + \frac{{n{\lambda _0}{{\left( {\overline {{x_n}} - {\mu _0}} \right)}^2}}}{{2\left( {{\lambda _0} + n} \right)}}\end{align}\)

(b) Substitute the values

\(\begin{align}{\mu _1} &= \frac{{2 \times 3.5 + 11 \times 7.2}}{{2 + 11}}\\ &= 6.631\\{\lambda _1} &= 2 + 11\\ &= 13\end{align}\)

\(\begin{align}{\alpha _1} &= 2 + \frac{{11}}{2}\\ &= 7.5\\{\beta _1} &= 1 + \frac{{20.3}}{2} + \frac{{11 \times 2 \times {{\left( {7.2 - 3.5} \right)}^2}}}{{2\left( {2 + 11} \right)}}\\ &= 22.73\end{align}\)

Find the interval that contains 95% of the posterior distribution of \({\bf{\mu }}\)

Define a new variable

Let,

\(\begin{align}U &= {\left( {\frac{{{\lambda _1}{\alpha _1}}}{{{\beta _1}}}} \right)^{\frac{1}{2}}}\left( {\mu - {\mu _1}} \right)\\ &= 2.071\left( {\mu - 6.631} \right)\end{align}\)

Here U follows t distribution with \(2{\alpha _1}\) degrees of freedom, that is 15.

Now,

\(\begin{align}P\left( {a < \mu < b} \right) &= 0.95\\ \Rightarrow P\left( {2.071\left( {a - 6.631} \right) < 2.071\left( {\mu - 6.631} \right) < 2.071\left( {b - 6.631} \right)} \right) &= 0.95\\ \Rightarrow P\left( {2.071\left( {a - 6.631} \right) < U < 2.071\left( {b - 6.631} \right)} \right) &= 0.95\end{align}\)

The confidence intervals derived for U are -2.13145and 2.13145

Therefore, the value of a and b is,

\(\begin{align}2.071\left( {a - 6.631} \right) &= - {\rm{2}}{\rm{.13145}}\\ \Rightarrow a &= 5.602\end{align}\)

And,

\(\begin{align}2.071\left( {b - 6.631} \right) &= {\rm{2}}{\rm{.13145}}\\ \Rightarrow b &= 7.660\end{align}\)

Therefore, the interval is (5.602,7.660).