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Math Help - CDF & Gamma function

  1. #1
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    CDF & Gamma function

    f(x)=a/b(x/b)^(a-1)exp(-(x/b)^a) for x>0.
    i). find the cumulative distribution function F(x).I believe that i should start by antidifferentiation of f(x), but I was stuck there.
    ii). calculate the n-th moment E(X^n) for all n=1,2,..Express it in terms of the Gamma function.
    Thanx
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  2. #2
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    Hello,
    Quote Originally Posted by Jojo123 View Post
    f(x)=a/b(x/b)^(a-1)exp(-(x/b)^a) for x>0.
    i). find the cumulative distribution function F(x).I believe that i should start by antidifferentiation of f(x), but I was stuck there.
    ii). calculate the n-th moment E(X^n) for all n=1,2,..Express it in terms of the Gamma function.
    Thanx
    i) Yes, you have to find an antiderivative for f.

    F(x) is 0 if x<0.
    Now, if x>0 :

    F(x)=\int_0^x \frac ab \left(\frac tb\right)^{a-1} \exp\left(-\left(\frac tb\right)^a\right) ~dt

    Just substitute y=\left(\frac tb\right)^a
    and the dy will simplify with \left(\frac tb\right)^{a-1}


    ii)
    E(X^n)=\frac ab \int_0^\infty t^n \cdot \left(\frac tb\right)^{a-1} \exp\left(-\left(\frac tb\right)^a\right) ~dt


    The Gamma function is \Gamma(z)=\int_0^\infty y^{z-1} e^{-y} ~dy

    Again, let y=\left(\frac tb\right)^a, to get -y in the exponential
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  3. #3
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    For the second part ..

    I got ..

     n \int^{\infty}_0 t^{n-1}e^{-y} dy

    How do I proceed any further from here?
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  4. #4
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    Quote Originally Posted by panda* View Post
    For the second part ..

    I got ..

     n \int^{\infty}_0 t^{n-1}e^{-y} dy

    How do I proceed any further from here?
    Hmm.. I don't see why you get an n at the beginning of the integral

    Making the substitution, we get :

    \int_0^\infty t^n e^{-y} ~dy

    But now, you gotta transform t^n with respect to y.


    y=\left(\frac tb\right)^a \Rightarrow y^{1/a}=\frac tb \Rightarrow by^{1/a}=t \Rightarrow t^n=b^n y^{n/a}

    Hence the integral is :

    \mathbb{E}(X^n)=b^n \int_0^\infty y^{n/a} e^{-y} ~dy=b^n \Gamma\left(\tfrac na+1\right)

    Does it look clear ? Don't hesitate to tell if it is not.

    I'm not that sure about the result, but I don't see where there would be a mistake ><
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  5. #5
    Moo
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    Yay ! \o/
    It's correct, according to this : Weibull distribution - Wikipedia, the free encyclopedia
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