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Suppose that in Exercise 15 the parameter is taken asthe standard deviation of the normal distribution, ratherthan the variance. Determine a conjugate family of priordistributions for samples from a normal distribution witha known value of the meanμand an unknown value ofthe standard deviationσ.

Short Answer

Expert verified

\(\xi \left( \sigma \right) = \frac{{2{\beta ^\alpha }}}{{\Gamma \left( \alpha \right)}}{\sigma ^{ - \left( {2\alpha + 1} \right)}}\exp \left( { - \beta /{\sigma ^2}} \right)\,\,\,\,for\,\,\sigma > 0\)

The family of distributions for which the p.d.f has this form, for all values of\(\alpha > 0\,\,and\,\,\beta > 0\), will be a conjugate family of the prior distribution.

Step by step solution

01

Given information

ξ(θ)be a p.d.f. that is defined as follows for constants

α >0 andβ >0:

\(\xi \left( \theta \right) = \left\{ \begin{aligned}{l}\frac{{{\beta ^\alpha }}}{{\Gamma \left( \alpha \right)}}{\theta ^{ - \left( {\alpha + 1} \right)}}{e^{ - \beta /\theta }}\,\,\,for\,\,\theta > 0\\0\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,for\,\,\theta \le 0\end{aligned} \right.\)

A distribution with this p.d.f. is called an inverse gamma

distribution.

02

Showing conjugate family of distribution

If X has the normal distribution with a known value of the mean\(\mu \)and an unknown value of the standard deviation\(\sigma \), then the p.d.f of X has the form

\(f\left( {x|\sigma } \right) \propto \frac{1}{\sigma }\exp \left( { - \frac{{{{\left( {x - \mu } \right)}^2}}}{{2{\sigma ^2}}}} \right)\)

Therefore, if the prior p.d.f\(\xi \left( \theta \right)\)has the form

\(\xi \left( \theta \right) \propto {\sigma ^{ - a}}\exp \left( { - b/{\sigma ^2}} \right),\)

Then the posterior p.d.f of\(\sigma \)will also have the same form, with a replaced by a+1 and b replaced by\(b + {\left( {x - \mu } \right)^2}/2\)

Therefore,

\(\int_0^\infty {{\sigma ^{ - a}}\exp \left( { - b/{\sigma ^2}} \right)} d\sigma = \frac{1}{2}\int_0^\infty {{y^{\left( {a - 3} \right)/2}}\exp \left( { - by} \right)dy} \)

The integral will be finite if a>1 and b>0, and its value will be

\(\frac{{\Gamma \left( {\frac{1}{2}\left( {a - 1} \right)} \right)}}{{{b^{\left( {a - 1} \right)/2}}}}\)

Hence, for a>1 and b>0, the following function will be a p.d.f for\(\sigma > 0\)

\(\xi \left( \sigma \right) = \frac{{2{b^{\left( {a - 1} \right)/2}}}}{{\Gamma \left( {\frac{1}{2}\left( {a - 1} \right)} \right)}}{\sigma ^{ - a}}\exp \left( { - b/{\sigma ^2}} \right)\)

Finally, a more standard form for this p.d.f, by replacing a and b by\(\alpha = \left( {a - 1} \right)/2\,\,{\rm{and}}\,\,\beta = b\)

\(\xi \left( \sigma \right) = \frac{{2{\beta ^\alpha }}}{{\Gamma \left( \alpha \right)}}{\sigma ^{ - \left( {2\alpha + 1} \right)}}\exp \left( { - \beta /{\sigma ^2}} \right)\,\,\,\,for\,\,\sigma > 0\)

The family of distributions for which the p.d.f has this form, for all values of \(\alpha > 0\,\,and\,\,\beta > 0\) , will be a conjugate family of the prior distribution.

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Most popular questions from this chapter

Consider again the conditions of Exercise 6, and suppose that the value of θ must be estimated by using the squared error loss function. Show that the Bayes estimators, for n = 1, 2,..., form a consistent sequence of estimators of θ.

Question: Suppose that \({{\bf{X}}_{\bf{1}}}{\bf{,}}...{\bf{,}}{{\bf{X}}_{\bf{n}}}\) form a random sample from a distribution for which the pdf. f (x|θ ) is as follows:

\(\begin{array}{c}{\bf{f}}\left( {{\bf{x|\theta }}} \right){\bf{ = }}{{\bf{e}}^{{\bf{\theta - x}}}}\,\,{\bf{,\theta > 0}}\\{\bf{ = 0}}\,\,\,{\bf{otherwise}}\end{array}\)

Also, suppose that the value of θ is unknown (−∞ <θ< ∞).

a. Show that the M.L.E. of θ does not exist.

b. Determine another version of the pdf. of this same distribution for which the M.L.E. of θ will exist, and find this estimator.

Question: Suppose that a certain large population contains k different types of individuals (k ≥ 2), and let \({{\bf{\theta }}_{\bf{i}}}\)denote the proportion of individuals of type i, for i = 1,...,k. Here, 0 ≤ \({\theta _i}\)≤ 1 and\({{\bf{\theta }}_{\bf{1}}}{\bf{ + }}...{\bf{ + }}{{\bf{\theta }}_{\bf{k}}}{\bf{ = 1}}\). Suppose also that in a random sample of n individuals from this population, exactly ni individuals are of type i, where\({{\bf{n}}_{\bf{1}}}{\bf{ + }}...{\bf{ + }}{{\bf{n}}_{\bf{k}}}{\bf{ = n}}\). Find the M.L.E.’s of \({{\bf{\theta }}_{\bf{1}}}{\bf{,}}...{\bf{,}}{{\bf{\theta }}_{\bf{n}}}\)

Identify two statistical inferences mentioned in Example 7.1.3.

Suppose that the number of defects in a 1200-foot roll of magnetic recording tape has a Poisson distribution for which the value of the mean θ is unknown, and the prior distribution of θ is the gamma distribution with parameters \(\alpha = 3\) and \(\beta = 1\). When five rolls of this tape are selected at random and inspected, the numbers of defects found on the rolls are 2, 2, 6, 0, and 3. If the squared error loss function is used, what is the Bayes estimate of θ?

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