Question

Using the prior and data in the numerical example on nursing homes in New Mexico in this section, find (a) the shortest possible interval such that the posterior probability that \({\bf{\mu }}\) lies in the interval is 0.90, and (b) the shortest possible confidence interval for \({\bf{\mu }}\) which the confidence coefficient is 0.90.

Short answer

1. (157.83, 210.07)
2. (152.55, 211.79)

Step-by-step solution

Given information

It is given that two variables \(\mu \,\,and\,\,\tau \)have the joint normal-gamma distribution such that \({\mu _0} = 183.95,{\lambda _0} = 20,{\alpha _0} = 11,{\beta _0} = 50925.37\)

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

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

Define a new variable

Let,

\(\begin{align}U &= {\left( {\frac{{{\lambda _0}{\alpha _0}}}{{{\beta _0}}}} \right)^{\frac{1}{2}}}\left( {\mu - {\mu _0}} \right)\\ &= 0.0657\left( {\mu - 183.95} \right)\end{align}\)

Here U follows t distribution with \(2{\alpha _0}\) degrees of freedom, that is 22.

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

\(\begin{align}P\left( {a < \mu < b} \right) &= 0.90\\ \Rightarrow P\left( {0.0657\left( {a - 183.95} \right) < 0.0657\left( {\mu - 183.95} \right) < 0.0657\left( {b - 183.95} \right)} \right) &= 0.90\\ \Rightarrow P\left( {0.0657\left( {a - 183.95} \right) < U < 0.0657\left( {b - 183.95} \right)} \right) &= 0.90\end{align}\)

The confidence intervals derived for U are -1.717 and +1.717

Therefore, the value of a and b is,

\(\begin{align}0.0657\left( {a - 183.95} \right) &= - 1.717\\ \Rightarrow a &= 157.83\end{align}\)

And,

\(\begin{align}0.0657\left( {b - 183.95} \right) &= 1.717\\ \Rightarrow b &= 210.07\end{align}\)

Therefore, the interval is (157.83, 210.07)

(b) Determine the confidence interval for \({\bf{\mu }}\)

To find this, we need to know the expectation and variance of \(\mu \) .

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

Therefore,

\(\begin{align}E\left( \mu \right) &= 183.95\\Var\left( \mu \right) &= \frac{{50925.37}}{{20\left( {11 - 1} \right)}} &= 254.62\end{align}\)

Hence,

\(\begin{align}P\left( {a < \mu < b} \right) &= 0.90\\ \Rightarrow P\left( {\frac{{a - 183.95}}{{\sqrt {254.62} }} < Z < \frac{{b - 183.95}}{{\sqrt {254.62} }}} \right) &= 0.90\end{align}\)

The value for 0.90 confidence is \({Z_{\frac{\alpha }{2}}} = 1.645\)

Solving the inequality, we get a=152.55, b=211.79

Therefore, the interval is (152.55, 211.79)