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Math Help - CDF of the Weibull Distribution

  1. #1
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    CDF of the Weibull Distribution

    I'm trying to find the CDF of the weibull distribution but have no idea how to integrate it from 0 to positive infinity. Can someone please show me how I can integrate this? k, b, and a are all constants.

    integrate from 0 to positive infinity
    kx^(B-1)*e^(-ax^B)dx
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  2. #2
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    Quote Originally Posted by xuyuan View Post
    I'm trying to find the CDF of the weibull distribution but have no idea how to integrate it from 0 to positive infinity. Can someone please show me how I can integrate this? k, b, and a are all constants.

    integrate from 0 to positive infinity
    kx^(B-1)*e^(-ax^B)dx
    Substitute t = ax^B.

    If you need more help, please show all your working and say where you get stuck.
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  3. #3
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    Take a look and notice how the Weibull is related to the Gamma function.
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  4. #4
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    hm so I subsituted

    u=-ax^b
    du=-aBX^(B-1)dx
    [1/(-aB)]du=x^(B-1)dx

    Integration from 0 to t gives

    [K/(-aB)](e^(-atB)-1)

    Is this correct?

    How would I subsitute it back to f(x)/(1-F(x)?
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  5. #5
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    Quote Originally Posted by xuyuan View Post
    hm so I subsituted

    u=-ax^b
    du=-aBX^(B-1)dx
    [1/(-aB)]du=x^(B-1)dx

    Integration from 0 to t gives

    [K/(-aB)](e^(-atB)-1)

    Is this correct?

    How would I subsitute it back to f(x)/(1-F(x)?
    Is it b or B? More detail is needed in the integral you got after the substitution and how you solved it. Did you substitute for the new integral terminals too?

    Your answer can be checked against the result given here: Weibull distribution - Wikipedia, the free encyclopedia
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  6. #6
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    Ah I figured it out! I need to first solve k in terms of alpha and beta. That simplies the integration and eventually I end up with the result I need. Thanks!
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  7. #7
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    Re: CDF of the Weibull Distribution

    Consider a Weibull distribution with the following pdf:
    f(x)= (Ɵ2/ Ɵ1)*(x/ Ɵ1)^( Ɵ2-1)*exp{-( x/ Ɵ1)^ Ɵ2} For x>0, Ɵ1>0, Ɵ2>0
    (a) if now Y=( x/ Ɵ1)^ Ɵ2. Show that Y follows an exponential distribution with a mean of 1.
    Explain how to
    (b) Explain how to generate a random variate from this Weibull distribution based on Y.
    (c) Use part (a) to generate 500 random variates from this Weibull distribution when (Ɵ1, Ɵ2)=(3,3) and draw a histogram.
    (d) Compare the histogram in part (c) with the actual pdf of Weibull (3,3).
    (e) Now assume that this Weibull distribution is truncated at points (a,b) = (1,10). Generate 500 random variates from this trunacated Weibull distribution and compute the average.
    (f) Compute the (theoretical expected value of this truncated population. Is the average in part(e) close to this expected value?
    I am New in statistics and got this question as my homework. I tried my best but not able to solve this. Please help me both in Theoretical as well as Matlab coding for the above question
    I am inter
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