Question

Continue the analysis in Example 8.6.2 on page 498. Compute an interval (a, b) such that the posterior probability is 0.9 that a <μ<b. Compare this interval with the 90% confidence interval from Example 8.5.4 on page 487.

Short answer

(-1.9199, 3.9199)

Step-by-step solution

Given information

It is given that two variables \(\mu \,\,and\,\,\tau \)have the joint normal-gamma distribution such that \({\mu _0} = 1,{\lambda _0} = 1,{\alpha _0} = 0.5,{\beta _0} = 0.5\)

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 a new variable

Let,

\(\begin{align}U &= {\left( {\frac{{{\lambda _0}{\alpha _0}}}{{{\beta _0}}}} \right)^{\frac{1}{2}}}\left( {\mu - {\mu _0}} \right)\\ &= \left( {\mu - 1} \right)\end{align}\)

Here U follows t distribution with \(2{\alpha _0}\) degrees of freedom, that is 2.

Calculating the posterior probability that \({\bf{\mu }}\)  lies in the interval.

\(\begin{align}P\left( {a < \mu < b} \right) &= 0.90\\ \Rightarrow P\left( {a - 1 < U < b - 1} \right) &= 0.90\end{align}\)

Since it is derived in step 3 that U follows a t distribution with 2 degrees of freedom, for 90%b confidence the lower and upper limits are: -2.91999 and +2.91999

Therefore, the value of a and b is,

\(\begin{align}a - 1 &= - {\rm{2}}{\rm{.91999}}\\ \Rightarrow a &= - 1.9199\end{align}\)

And,

\(\begin{align}b - 1 &= {\rm{2}}{\rm{.91999}}\\ \Rightarrow b &= 3.{\rm{91999}}\end{align}\)

Therefore, the interval is (-1.9199,3.9199)