Question

Question: Consider the calorie count data described in Example7.3.10 on page 400. Now assume that each observation has the normal distribution with unknown mean \({\bf{\mu }}\,\,{\bf{and}}\,\,{\bf{\tau }}\) given the parameter \({\bf{\mu }}\,\,{\bf{and}}\,\,{\bf{\tau }}\). Use the normal-gamma conjugate prior distribution with prior hyper parameters

\({{\bf{\alpha }}_{\bf{0}}}{\bf{ = 1,}}{{\bf{\beta }}_{\bf{0}}}{\bf{ = 60,}}{{\bf{\mu }}_{\bf{0}}}{\bf{ = 0}}\,\,{\bf{and}}\,\,{{\bf{\lambda }}_{\bf{0}}}{\bf{ = 1}}\)The value of \({{\bf{s}}_{\bf{n}}}^{\bf{2}}\) is 2102.9.

a. Find the posterior distribution of \({\bf{\mu }}\,\,{\bf{and}}\,\,{\bf{\tau }}\)

b. Compute \({\bf{Pr(\mu > 1|x)}}{\bf{.}}\)

Short answer

(a) \({\mu _1} = 0.109,{\lambda _1} = 21,{\alpha _1} = 11,{\beta _1} = 1111.45\)

(b) \(0.3403\)

Step-by-step solution

Given information

It is given that two variables \(\mu \,\,and\,\,\tau \)have the joint normal-gamma distribution such that\({\alpha _0} = 1,{\beta _0} = 60,{\mu _0} = 0\,\,and\,\,{\lambda _0} = 1\).The value of \({s_n}^2\) is 2102.9 and \(\overline {{x_n}} = 0.1154,\,\,{\sigma ^2} = 4.62\). n=20 in the question.

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

Define the new variables

We know that,

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

Substitute the values

a)

\(\begin{aligned}{}{\mu _1} &= \frac{{1 \times 0 + 20 \times 0.1154}}{{1 + 20}}\\ &= 0.109\\{\lambda _1} &= 1 + 20\\ &= 21\\{\alpha _1} &= 1 + \frac{{20}}{2}\\ &= 11\\{\beta _1} &= 60 + \frac{{2102.9}}{2} + \frac{{20 \times 1 \times {{\left( {0.1154 - 0} \right)}^2}}}{{2\left( {1 + 20} \right)}}\\ &= 1111.45\end{aligned}\)

Calculate the probability

b.

\(\begin{aligned}{}\Pr (\mu > {\bf{ }}1|x) &= 1 - \Phi \left( {\frac{{1 - 0.1154}}{{\sqrt {4.62} }}} \right)\\ &= 1 - \Phi \left( {1.12} \right)\\ &= 0.3403\end{aligned}\)