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Consider again the problem described in Exercise 6, but suppose now that the prior p.d.f. of \(\theta \) is as follows:\(\xi \left( \theta \right) = \left\{ \begin{aligned}{l}2\left( {1 - \theta } \right)\;\;\;\;\;\;for\;0 < \theta < 1\\0\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;otherwise\end{aligned} \right.\)

As in Exercise 6, suppose that in a random sample of eight items, exactly three are found to be defective. Determine the posterior distribution of \(\theta \)

Short Answer

Expert verified

The posterior pdf of \(\theta \) is given by \(\xi \left( {\theta |{X_1}...{X_n}} \right) = \frac{{\left| \!{\overline {\, {\left( {4 + 6} \right)} \,}} \right. }}{{\left| \!{\overline {\, 4 \,}} \right. \left| \!{\overline {\, 6 \,}} \right. }}{\theta ^{4 - 1}}{\left( {1 - \theta } \right)^{6 - 1}}\)

Step by step solution

01

Explaining posterior distribution.

An unknown parameter\(\theta \)needs to be estimated using a random sample\({X_1}...{X_n}\)of size n from the distribution for which\(\theta \)is the parameter.

Let\(\xi \left( \theta \right)\)be the prior distribution\(\theta \) and\(f\left( {x|\theta } \right)\)be the joint pdf of\({X_1}...{X_n}\).

The conditional pdf of \(\theta \) given \({X_1} = {x_1},...{X_n} = {x_n}\) is called the posterior distribution of \(\theta \) and is denoted by \(\xi \left( {\theta |{X_1}...{X_n}} \right)\). Then \(\xi \left( {\theta |{X_1}...{X_n}} \right) \propto {f_n}\left( {X|\theta } \right)\xi \left( \theta \right)\)

02

Computing the posterior distribution of \(\theta \)

Let\(\theta \)be the proportion of defective items in a large manufactured lot.

A random sample of\(n = 8\)times yielded exactly 3 defectives.

Let\(\xi \left( \theta \right)\)be the prior distribution\(\theta \)

\(\xi \left( \theta \right) = \left\{ \begin{aligned}{l}2\left( {1 - \theta } \right)\;\;\;\;for\;0 < \theta < 1\\0\;\;\;\;\;\;\;\;\;\;\;\;\;otherwise\end{aligned} \right.\)

Let\({X_i} = 0\)if the inspected item is defective and\({X_i} = 1\)is the item is not defective.

Comparing the situation with Bernoulli distribution with parameter\(\theta \). Further, as each of the inspected items is independent of the other.

Hence joint pdf of \({X_1}...{X_n}\) is given by:

\(\begin{aligned}{f_n}\left( {X|\theta } \right) = \prod\limits_{i = 1}^8 {{\theta ^{\left( {{x_i}} \right)}}{{\left( {1 - \theta } \right)}^{\left( {1 - {x_i}} \right)}}} \\ = {\theta ^3}{\left( {1 - \theta } \right)^5}\end{aligned}\)

Consider the product of the joint pdf of\({X_1}...{X_n}\)and the prior pdf of\(\theta \)

\(\begin{aligned}{f_n}\left( {X|\theta } \right)\xi \left( \theta \right) = {\theta ^3}{\left( {1 - \theta } \right)^5}2\left( {1 - \theta } \right)\\ = 2{\theta ^{4 - 1}}{\left( {1 - \theta } \right)^{7 - 1}}\end{aligned}\)

We know that:

\(\begin{aligned}\xi \left( {\theta |{X_1}...{X_n}} \right) \propto {f_n}\left( {X|\theta } \right)\xi \left( \theta \right)\\ \propto 2{\theta ^{4 - 1}}{\left( {1 - \theta } \right)^{7 - 1}}\\{\theta ^{4 - 1}}{\left( {1 - \theta } \right)^{7 - 1}}\end{aligned}\)

Comparing the above expression with beta pdf with parameters\(\alpha = 4\)\(\beta = 7\), it’s clear that the conditional distribution of\(\theta \)is proportional to the beta distribution. That

is \(\frac{{\left| \!{\overline {\, {\left( {\alpha + \beta } \right)} \,}} \right. }}{{\left| \!{\overline {\, \alpha \,}} \right. \left| \! {\overline {\, \beta \,}} \right. }}\)

If multiplied with the constant term, the posterior pdf \(\theta \)will be beta with\(\alpha = 4\)and\(\beta = 7\). Thus, the posterior pdf of\(\theta \)is given by

\(\xi \left( {\theta |{X_1}...{X_n}} \right) = \frac{{\left| \!{\overline {\, {\left( {4 + 7} \right)} \,}} \right. }}{{\left| \!{\overline {\, 4 \,}} \right. \left| \!{\overline {\, 7 \,}} \right. }}{\theta ^{4 - 1}}{\left( {1 - \theta } \right)^{7 - 1}}\)

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Most popular questions from this chapter

Suppose that\({{\bf{X}}_{\bf{1}}}{\bf{,}}...{\bf{,}}{{\bf{X}}_{\bf{n}}}\)form a random sample from a distribution with the following p.d.f.:

\({\bf{f}}\left( {{\bf{x}}\left| {{\bf{\beta ,\theta }}} \right.} \right){\bf{ = }}\left\{ {\begin{align}{}{{\bf{\beta }}{{\bf{e}}^{{\bf{ - \beta }}\left( {{\bf{x - \theta }}} \right)}}}&{{\bf{for}}\,\,{\bf{x}} \ge {\bf{\theta }}}\\{\bf{0}}&{{\bf{otherwise,}}}\end{align}} \right.\)

where\({\bf{\beta }}\,\,{\bf{and}}\,\,{\bf{\theta }}\,\,{\bf{are}}\,\,{\bf{unknown}}\,\,\left( {{\bf{\beta > 0,}} - \infty {\bf{ < \theta < }}\infty } \right)\). Determine a pair of jointly sufficient statistics.

Suppose that \({X_1},...,{X_n}\) form a random sample from an exponential distribution for which the value of the parameter β is unknown (β > 0). Is the M.L.E. of β a minimal sufficient statistic.

Question: Suppose that the lifetime of a certain type of lamp has an exponential distribution for which the value of the parameter \({\bf{\beta }}\) is unknown. A random sample of n lamps of this type are tested for a period of T hours and the number X of lamps that fail during this period is observed, but the times at which the failures occurred are not noted. Determine the M.L.E. of \({\bf{\beta }}\) based on the observed value of X.

Suppose that a single observation X is to be taken from the uniform distribution on the interval \(\left[ {{\bf{\theta - }}\frac{{\bf{1}}}{{\bf{2}}}{\bf{,\theta + }}\frac{{\bf{1}}}{{\bf{2}}}} \right]\), the value of θ is unknown, and the prior distribution of θ is the uniform distribution on the interval [10, 20]. If the observed value of X is 12, what is the posterior distribution of θ?

Suppose that \({X_1}...{X_n}\) form a random sample from a distribution for which the p.d.f. is \(f\left( {x|\theta } \right)\), the value of \(\theta \) is unknown, and the prior p.d.f. of\(\theta \) is. Show that the posterior p.d.f. \(\xi \left( \theta \right)\) is the same regardless of whether it is calculated directly by using Eq. (7.2.7) or sequentially by using Eqs. (7.2.14), (7.2.15), and (7.2.16).

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