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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

Expert verified
  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

01

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 \(\overline {{x_n}} = 7.2\,{\rm{and}}\,\,{s_n}^2 = 20.3\).

02

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

03

(a) Define the new variables

We know that,

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

04

(b) Substitute the values

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

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

05

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).

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Most popular questions from this chapter

Question:Consider an infinite sequence of Bernoulli trials for which the parameter p is unknown (0 <p< 1), and suppose that sampling is continued until exactly k successes have been obtained, where k is a fixed integer (k ≥ 2). Let N denote the total number of trials that are needed to obtain the k successes. Show that the estimator (k − 1)/(N − 1) is an unbiased estimator of p.

Consider the conditions of Exercise 10 again. Suppose also that it is found in a random sample of size n = 10 \(\overline {{{\bf{x}}_{\bf{n}}}} {\bf{ = 1}}\,\,{\bf{and}}\,\,{{\bf{s}}_{\bf{n}}}^{\bf{2}}{\bf{ = 8}}\) . Find the shortest possible interval so that the posterior probability \({\bf{\mu }}\) lies in the interval is 0.95.

Suppose that a point(X, Y )is to be chosen at random in thexy-plane, whereXandYare independent random variables, and each has the standard normal distribution. If a circle is drawn in thexy-plane with its center at the origin, what is the radius of the smallest circle that can be chosen for there to be a probability of 0.99 that the point(X, Y )will lie inside the circle?

Show that two random variables \({\bf{\mu }}\,\,{\bf{and}}\,\,{\bf{\tau }}\)cannot have the joint normal-gamma distribution such that

\({\bf{E}}\left( {\bf{\mu }} \right){\bf{ = 0}}\,\,{\bf{,Var}}\left( {\bf{\mu }} \right){\bf{ = 1,E}}\left( {\bf{\tau }} \right){\bf{ = }}\frac{{\bf{1}}}{{\bf{2}}}\,\,{\bf{and}}\,\,{\bf{Var}}\left( {\bf{\tau }} \right){\bf{ = }}\frac{{\bf{1}}}{{\bf{4}}}\)

Consider again the situation described in Example 8.2.3. How small wouldσ2 need to be in order for Pr(Y≤0.09)≥0.9?

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