Question

Suppose that a machine produces parts that are defective with probability P, but P is unknown. Suppose that P has a continuous distribution with pdf.

\({\bf{f}}\left( {\bf{p}} \right){\bf{ = }}\left\{ \begin{array}{l}{\bf{10}}{\left( {{\bf{1 - p}}} \right)^{\bf{9}}}\;\;{\bf{if}}\,{\bf{0 < p < 1,}}\\{\bf{0}}\;\;{\bf{otherwise}}{\bf{.}}\end{array} \right.\).

Conditional on\(P = p\), assume that all parts are independent of each other. Let X be the number of non defective parts observed until the first defective part. If we observe X = 12, compute the conditional pdf. of P given X = 12.

Short answer

Conditional pdf of p given X=12 is \(g\left( {p|12} \right) = 506p{\left( {1 - p} \right)^{21}}\)

Step-by-step solution

Given information

P be the probability of defected parts which is unknown..

P follows a continues distribution with pdf

\(f\left( p \right) = \left\{ \begin{array}{l}10{\left( {1 - p} \right)^9}\;\;if\,0 < p < 1,\\0\;\;otherwise.\end{array} \right.\)

Computing the Joint pdf 

The joint pf of X and P is\(f\left( p \right)\)times the geometric pf with [parameter\(p\)is :

\(p{\left( {1 - p} \right)^x}10{\left( {1 - p} \right)^9} = p{\left( {1 - p} \right)^{x + 9}},for\,x = 0,1..\) and\(0 < p < 1\). .....(1)

Computing the marginal pdf 

The marginal pdf of X at x=12 is the integral of (1) over p with x=12 is :

\(\begin{array}{c}{\int\limits_1^0 {p\left( {1 - p} \right)} ^{21}}dp = \int\limits_1^0 {{p^{21}}\left( {1 - p} \right)} dp\\ = \frac{1}{{22}} - \frac{1}{{23}}\\ = \frac{1}{{506}}\end{array}\)

Computing the conditional pdf 

The conditional pdf of\(P\)given X=12 is:

\(\begin{array}{c}g\left( {p|12} \right) = \frac{{p{{\left( {1 - p} \right)}^{x + 9}},}}{{\frac{1}{{506}}}}\\ = \frac{{p{{\left( {1 - p} \right)}^{21}}}}{{\frac{1}{{506}}}}\end{array}\)

\(g\left( {p|12} \right) = 506p{\left( {1 - p} \right)^{21}}\)for\(o < p < 1\)

Hence Conditional pdf of p given X=12 is \(g\left( {p|12} \right) = 506p{\left( {1 - p} \right)^{21}}\).