Question

Silver-Coated Nylon Fiber. Silver-coated nylon fiber is used in hospitals for its anti-static electricity properties, as well as for antibacterial and antimycotic effects. In the production of silver-coated nylon fibers, the extrusion process is interrupted from time to time by blockages occurring in the extrusion dyes. The time in hours between blockages, T, has an exponential E(1/10) distribution, where 1/10 is the rate parameter. Find the probabilities that (a) a run continues for at least 10 hours, (b) a run lasts less than 15 hours, and (c) a run continues for at least 20 hours, given that it has lasted 10 hours. If you use software, be careful about the parametrization of exponentials.

Also, the times (in hours) between blockages of the extrusion process, T, had an exponential E(λ) distribution. Suppose that the rate parameter λ is unknown, but there are three measurements of interblockage times, T1 = 3, T2 = 13, and T3 = 8.

(a) How would classical statistician estimate λ?

(b) What is the Bayes estimator of λ if the prior is π(λ) = √ 1 λ ,λ > 0.

Hint: In (b) the prior is not a proper distribution, but the posterior is. Identify the posterior from the product of the likelihood from (a) and the prior, no need to integrate.

Answer 1

Given T follows exponential distribution with rate parameter 1/10

A run T continues for at least 10 hours is P((T>=10) =   =

                                                                            =       =   

                                                                            = = .

P(a run lasts less than 15 hours) = P((T<15) =     =   

                                                                =   =

                                                                = =    .

P( a run continues for at least 20 hours, given that it has lasted 10 hours) = P(T>20/T>10) =

                                                                                                          =   =

                                                                                                          =     = .

Average value of the variable is inverse of the parameter .

Hence, here = 1/ [(3+13+8)/3]= 1/8.