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Math Help - Exponential to Weibull

  1. #1
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    Question Exponential to Weibull

    Can you please help?

    Problem. An exponential distribution is a member of the Gamma Family; that is, when \alpha = 1, the pdf reduces to the exponential distribution (with \lambda = 1/\beta). Let the r.v. X ~ Exponential (\beta), then Y = X^\frac{1}{\gamma}, \gamma>0, has a Weibull (\gamma, \beta) distribution. Show the density function of Y is

    f_Y(y|\gamma, \beta) = \frac{\gamma}{\beta}y^{\gamma-1} e^\frac{-y^\gamma}{\beta}.

    I have in my notes that for X~Exponential, f_X(x) = \lambda e^{-\lambda x}.

    Thanks!
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  2. #2
    MHF Contributor matheagle's Avatar
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    pasting....

    Let f_X(x) = \lambda e^{-\lambda x}I(x>0) with \lambda = 1/\beta.

    So your density is f_X(x) = {1\over \beta}e^{-x/\beta}I(x>0)

    which is how I write my exponential density, but you do need the indicator function.
    Otherwise you need to say x>0.

    If Y = X^\frac{1}{\gamma}, \gamma>0, then

    f_Y(y) = \frac{\gamma}{\beta}y^{\gamma-1} e^\frac{-y^\gamma}{\beta}I(y>0).

    NOW I can prove that

    f_Y(y) = f_X(x)\biggl|{dx\over dy}\biggr| but that follows from the derivative of the cdfs.

    So f_Y(y) = {1\over \beta}e^{-y^{\gamma}/\beta}|\gamma y^{\gamma -1}|I(y>0)

    = {\gamma\over \beta} y^{\gamma -1}e^{-y^{\gamma}/\beta} I(y>0)

    since y = x^{1/\gamma} hence x = y^{\gamma} and {dx\over dy} =\gamma y^{\gamma -1}
    Last edited by matheagle; December 13th 2009 at 07:01 PM.
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  3. #3
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    Arrow my incorrect solution

    This is what I did earlier in the year. Incorporating some of your notation, I think it corresponds to what you showed me above. However, I didn't get the right answer. Can you tell me what I did wrong?

    Thanks.

    My incorrect "solution".
    Let f_X(x) = \lambda e^{-\lambda x}, \forall x>0, and \lambda = \frac{1}{\beta}.

    By substituting into the Gamma density function, Gamma (\alpha=1, \beta=1/\lambda),
    we get *f_X(x) = \frac{1}{\Gamma(\alpha)\beta^\alpha} x^{\alpha-1} e^{-x/\beta} = \frac{1}{\beta} e^{-x/\beta},  \forall x>0.

    So F_Y(y) = P(Y \leq y), since y = x^{1/\gamma}, this means
    =P(X^{1/\gamma} \leq y)
    =P(X \leq y^\gamma)
    =F_X(y^\gamma).

    Thus F_Y'(y) = f_X(y^\gamma)\gamma y^{\gamma-1}.

    By substituting y^\gamma for x in *f_X(x),
    we get  f_Y(y) = \frac{1}{\beta} e^{-y\gamma/\beta}, \forall y>0.

    The end of my incorrect solution....
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  4. #4
    MHF Contributor matheagle's Avatar
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    It's perfect except the last line.
    you don't need the   \forall y>0 by the way, y>0 is fine.
    But you left out the chain rule which you computed.............

    Thus f_Y(y)=F_Y'(y) = f_X(y^\gamma)\gamma y^{\gamma-1} .

    By substituting y^\gamma for x in *f_X(x),

    AND multiplying by the \gamma y^{\gamma-1}

    we get  f_Y(y) = \frac{1}{\beta} e^{-y^\gamma/\beta}\gamma y^{\gamma-1} , y>0.

    which is the same as mine...

    This.... f_Y(y)=F_Y'(y) = f_X(y^\gamma)\gamma y^{\gamma-1} .

    is f_Y(y)=f_X(y^\gamma){dx\over dy}.
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  5. #5
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    last line

    Would you show me how my last line should read? And why doesn't my last line look like this
    f_Y(y|\gamma, \beta) = \frac{\gamma}{\beta}y^{\gamma-1} e^\frac{-y^\gamma}{\beta}?

    Instead it reads
    f_Y(y) = \frac{1}{\beta} e^{-y\gamma/\beta}, y>0.

    (LaTex ?: how do I add a space after the "," and "y" in my last line above?)

    Thanks for everything!!!
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  6. #6
    MHF Contributor matheagle's Avatar
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    GN poisson ivy
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