Maximum For instance the maximum can be used to model The lifetime of a system

Maximum for instance the maximum can be used to model

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There are cases when we are interested in the maximum or minimum of a random sample. Maximum: For instance, the maximum can be used to model: The lifetime of a system of n independent components connected in parallel, The completion time of a project of n independent subprojects, which can be completed simultaneously. Setup of problem: You are given a pdf of n independent random variables X 1 , X 2 , ..., X n . Question: Find the pdf of the maximum of X 1 , X 2 , ..., X n Steps: Let Y = max( X 1 , X 2 , ..., X n ). To find the pdf of Y we start by finding its cdf. F Y ( y ) = P ( Y y ) = P ( X 1 y, X 2 y, ..., X n y ) = P ( X 1 y ) P ( X 2 y ) ...P ( X n y ) (because X i ’s are independent) = F X 1 ( y ) F X 2 ( y ) ...F X n ( y ) Furthermore, if all the X i s have the same pdf, F Y ( y ) = [ F X ( y )] n 11
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We found the cdf of Y , now how do we get the pdf of Y ? f Y ( y ) = F 0 Y ( y ) = n [ F X ( y )] n - 1 f X ( y ) We will work on ex 4.8 and 4.9 from the course text. Example 8 . A system consists of five components connected in parallel. The lifetime (in thousands of hours) of each component is an exponential random variable with mean μ = 3. (a) Calculate the median and standard deviation for each component (b) Calculate the probability that the system fails before 3500 hours. Compare this with the probability that a component fails before 3500 hours. (c) Calculate the median life, mean life and standard deviation for the system. 12
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Minimum: The minimum can be used to model: The lifetime of a system of n independent components connected in series, The completion time of a project pursued by n independent competing teams Setup of problem: You are given the pdf of n independent random variables X 1 , X 2 , ..., X n . Question: Find the pdf of the minimum of X 1 , X 2 , ..., X n Steps: Let Y = min( X 1 , X 2 , ..., X n ). Again, to find the pdf of Y we start by finding its cdf. F Y ( y ) = P ( Y y ) = 1 - P ( Y > y ) = 1 - P ( X 1 > y, X 2 > y, ..., X n > y ) = 1 - [ P ( X 1 > y ) P ( X 2 > y ) ...P ( X n > y )] (because X i ’s are independent) = 1 - [1 - F X 1 ( y )][1 - F X 2 ( y )] ... [1 - F X n ( y )] Furthermore, if all the X i s have the same pdf, F Y ( y ) = 1 - [1 - F X ( y )] n How do we get the pdf of Y ? f Y ( y ) = F 0 Y ( y ) = - n [1 - F X ( y )] n - 1 ( - f X ( y )) = n [1 - F X ( y )] n - 1 f X ( y ) Example 9 . A system consists of five components connected in series. The lifetime (in thousands of hours) of each component is an exponential random variable with mean μ = 3. (a) Calculate the probability that the system fails before 3500 hours. Compare this with the probability that a component fails before 3500 hours.
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  • Summer '11
  • Yew-Wei
  • Statistics, Probability theory, σ, λ, 3500 hours

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