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Let 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.

a. Verify thatξ(θ)is actually a p.d.f. by verifying that

\(\int_0^\infty {\xi \left( \theta \right)} d\theta = 1\)

b. Consider the family of probability distributions that can be represented by a p.d.f.ξ(θ)having the given form for all possible pairs of constantsα >0 andβ >

0. Show that this family is a conjugate family of prior distributions for samples from a normal distribution with a known value of the meanμand an unknown

value of the varianceθ.

Short Answer

Expert verified

a. This is an actual p.d.f because integration over the whole range is 1

b. This is a conjugate family of prior distributions.

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

Verifying the actual p.d.f

a.

\(Let,\,y = 1/\theta .\,Then\,\theta = 1/y\,and\,d\theta = - dy/{y^2}\)

Hence,

\(\begin{aligned}\int_0^\infty {\xi \left( \theta \right)} d\theta = \int_0^\infty {\frac{{{\beta ^\alpha }}}{{\Gamma \left( \alpha \right)}}} {y^{\alpha - 1}}\exp \left( { - \beta y} \right)dy\\ = \frac{{{\beta ^\alpha }}}{{\Gamma \left( \alpha \right)}}\int_0^\infty {{y^{\alpha - 1}}\exp \left( { - \beta y} \right)dy} \\ = \frac{{{\beta ^\alpha }}}{{\Gamma \left( \alpha \right)}} \times \frac{{\Gamma \left( \alpha \right)}}{{{\beta ^\alpha }}}\\ = 1\end{aligned}\)

Here the integration over whole range is 1.

So this is an actual p.d.f.

03

Step 3:Showing conjugate pair

b.

If an observation X has a normal distribution with a known value of the mean\(\mu \)and an unknown value of the variance\(\theta \), then p.d.f of X has the form:

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

Also, the prior p.d.f of\(\theta \)has the form:

\(\xi \left( \theta \right) \propto {\theta ^{ - \left( {\alpha + 1} \right)}}\exp \left( { - \beta /\theta } \right)\).

Therefore, the posterior p.d.f\(\xi \left( {\theta |x} \right)\)has the form:

\(\begin{aligned}\xi \left( {\theta |x} \right) \propto \xi \left( \theta \right)f\left( {x|\theta } \right)\\ \propto {\theta ^{ - \left( {\alpha + 3/2} \right)}}\exp \left\{ { - \left( {\beta + \frac{1}{2}{{\left( {x - \mu } \right)}^2}} \right) \cdot \frac{1}{\theta }} \right\}\end{aligned}\).

Hence, the posterior p.d.f. of \(\theta \) has the same form as \(\xi \left( \theta \right)\) with \(\alpha \) replaced by \(\alpha + 1/2\) and \(\beta \) replaced by \(\beta + 1/2{\left( {x - \mu } \right)^2}\), Since this distribution will be the prior distribution of \(\theta \) for future observations, it follows that the posterior distribution after any number of observations will also belong to the same family of distribution.

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