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Suppose that the time in minutes required to serve a customer at a certain facility has an exponential distribution for which the value of the parameter θis unknown and that the prior distribution ofθis a gamma distributionfor which the mean is 0.2 and the standard deviation is 1. If the average time required to serve a random sample of 20 customers is observed to be 3.8 minutes, what is the posterior distribution ofθ?

Short Answer

Expert verified

The posterior distribution is gamma with parameters are 20.04 and 76.2

Step by step solution

01

Given information

The time in minutes required to serve a customer at a certain facility has an exponential distribution for which the value of the parameterθis unknown and the prior distribution ofθis a gamma distribution for which the mean is 0.2 and the standard deviation is 1.

02

Finding the posterior distribution of \({\bf{\theta }}\)

Let\(\alpha \,and\,\beta \)denote the parameters of the prior gamma distribution of\(\theta \).

Then,\(\alpha /\beta = 0.2\,\,{\rm{and}}\,\,{\alpha ^2}/\beta = 1\)

Therefore,\(\alpha = 0.04\,\,{\rm{and}}\,\,\beta = 0.2\)

Furthermore, the total time period required to serve the sample of 20 customers is\(y = 20 \times 3.8 = 76\)

Therefore, by Theorem 7.3.4, the posterior distribution of\(\theta \)is the gamma distribution for which the parameters are\(\alpha + n\,\,{\rm{and}}\,\,\beta + \sum\limits_{i = 1}^n {{x_i}} \)

Suppose thatX1, . . . ,Xnform a random sample from the exponential distributionwith parameterθ >0 that is unknown. Suppose also that the prior distribution ofθis the gamma distribution with parametersα >0 andβ >0. Then the posteriordistribution ofθgiven thatXi=xi(i=1, . . . , n) is the gamma distribution with parameters.

\(\alpha + n\,\,{\rm{and}}\,\,\beta + \sum\limits_{i = 1}^n {{x_i}} \)

\({\rm{i}}{\rm{.e}},\alpha + n\, = 0.04 + 20 = 20.04\)

\({\rm{i}}{\rm{.e}},\,\beta + \sum\limits_{i = 1}^n {{x_i}} = 0.2 + 76 = 76.2\)

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