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Math Help - Integrating exponential distribution

  1. #1
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    Integrating exponential distribution

     Prob(X \leq x | \theta) = 1 - e^{- \theta x}<br />

    <br />
\theta is gamma distribution so that the PDF of  \theta is g( \theta )

     g( \theta ) = \frac{\beta^{\alpha}}{\Gamma(\alpha)} \theta^{\alpha -1}e^{ - \beta \theta}

    Find the marginal distribution of X.

    Stuck on trying to integrate with respect to \theta, please could someone give me a hand.

    Thanks!
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  2. #2
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    Quote Originally Posted by peterpan View Post
     Prob(X \leq x | \theta) = 1 - e^{- \theta x}<br />

    <br />
\theta is gamma distribution so that the PDF of  \theta is g( \theta )

     g( \theta ) = \frac{\beta^{\alpha}}{\Gamma(\alpha)} \theta^{\alpha -1}e^{ - \beta \theta}

    Find the marginal distribution of X.

    Stuck on trying to integrate with respect to \theta, please could someone give me a hand.

    Thanks!
    \Pr(X \leq x | \theta) = F(x| \theta) = 1 - e^{- \theta x}<br />

    f(x|\theta) = \frac{\partial}{\partial x}F(x|\theta) = \theta\ e^{-\theta x}


    Therefore, the marginal pdf f(x) is:


    f(x) = \int_0^\infty f(x|\theta) \, g(\theta) \, d\theta =<br />
\int_0^\infty g(\theta) \, \theta \, e^{-\theta x} d\theta =<br />
\underbrace{ \frac{\alpha}{\beta}(1 + x/\beta)^{-\alpha - 1}}_{\text{Laplace transform of} ~ g(\theta) \, \theta }
    for x > 0 and zero otherwise


    Furthermore (in anticipation), now that you've got the marginal pdf of X you can get the cdf by integrating this pdf:

    F(x) = \int_{0}^{x} \frac{\alpha}{\beta} \left(1 + \frac{t}{\beta} \right)^{-\alpha - 1} dt = ........
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    Hi mr fantastic,

    Thank for your help, much appreciated. Would you mind explaining the first two steps as im not sure why we do these, is the change from F -> f a typo, or does it have meaning? Why do we differentiate with respect to x?

    \Pr(X \leq x | \theta) = F(x| \theta) = 1 - e^{- \theta x}<br />

    f(x|\theta) = \frac{\partial}{\partial x}F(x|\theta) = \theta\ e^{-\theta x}

    Peter
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  4. #4
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    Quote Originally Posted by peterpan View Post
    Hi mr fantastic,

    Thank for your help, much appreciated. Would you mind explaining the first two steps as im not sure why we do these, is the change from F -> f a typo, or does it have meaning? Why do we differentiate with respect to x?

    \Pr(X \leq x | \theta) = F(x| \theta) = 1 - e^{- \theta x}<br />

    f(x|\theta) = \frac{\partial}{\partial x}F(x|\theta) = \theta\ e^{-\theta x}

    Peter
    F is the (conditional) cumulative distribution and f the (conditional) density.

    RonL
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  5. #5
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    Thanks for the reply - now on the second part:

    \int_0^\infty g(\theta) \, \theta \, e^{-\theta x} d\theta =<br />
\underbrace{ \frac{\alpha}{\beta}(1 + x/\beta)^{-\alpha - 1}}_{\text{Laplace transform of} ~ g(\theta) \, \theta }

    how did you make this step? Do you have to use a Laplace transformation (i am not familiar with these at all) ? can you not integrate by parts with:

     u = \theta e^{-\theta x}

     dv = g (\theta )
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  6. #6
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    Quote Originally Posted by peterpan View Post
    Thanks for the reply - now on the second part:

    \int_0^\infty g(\theta) \, \theta \, e^{-\theta x} d\theta =<br />
\underbrace{ \frac{\alpha}{\beta}(1 + x/\beta)^{-\alpha - 1}}_{\text{Laplace transform of} ~ g(\theta) \, \theta }

    how did you make this step? Do you have to use a Laplace transformation (i am not familiar with these at all) ? can you not integrate by parts with:

     u = \theta e^{-\theta x}

     dv = g (\theta )
    Being lazy, I went for the easy option.

    It can be done by parts. If you need a hand, don't hesitate to say so.
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  7. #7
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    thanks, here is my attempt so far, not sure where to go next...

     <br />
\int_0^\infty g(\theta) \theta e^{-\theta x} d\theta<br />

     <br />
\int_0^\infty \frac{\beta^{\alpha}}{\Gamma(\alpha)} \theta^{\alpha-1}e^{-\beta \theta}\theta e^{-\theta x} d\theta<br />

     <br />
\frac{\beta^{\alpha}}{\Gamma(\alpha)} <br />
\int_0^\infty \theta^{\alpha-1} \theta e^{-\beta \theta} e^{-\theta x} d\theta<br />

     <br />
\frac{\beta^{\alpha}}{\Gamma(\alpha)} <br />
\int_0^\infty \theta^{\alpha} e^{- \theta(\beta + x)} d\theta<br />

    let t = \theta(\beta +x)
    \theta = \frac{t}{\beta + x}
    d\theta = \frac{dt}{\beta + x}

     <br />
\frac{\beta^{\alpha}}{\Gamma(\alpha)(\beta + x)} <br />
\int_0^\infty \left(\frac{t}{\beta+x}\right)^{\alpha} e^{- t} dt<br />
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  8. #8
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    Quote Originally Posted by peterpan View Post
    thanks, here is my attempt so far, not sure where to go next...

     <br />
\int_0^\infty g(\theta) \theta e^{-\theta x} d\theta<br />

     <br />
\int_0^\infty \frac{\beta^{\alpha}}{\Gamma(\alpha)} \theta^{\alpha-1}e^{-\beta \theta}\theta e^{-\theta x} d\theta<br />

     <br />
\frac{\beta^{\alpha}}{\Gamma(\alpha)} <br />
\int_0^\infty \theta^{\alpha-1} \theta e^{-\beta \theta} e^{-\theta x} d\theta<br />

     <br />
\frac{\beta^{\alpha}}{\Gamma(\alpha)} <br />
\int_0^\infty \theta^{\alpha} e^{- \theta(\beta + x)} d\theta<br />

    let t = \theta(\beta +x)
    \theta = \frac{t}{\beta + x}
    d\theta = \frac{dt}{\beta + x}

     <br />
\frac{\beta^{\alpha}}{\Gamma(\alpha)(\beta + x)} <br />
\int_0^\infty \left(\frac{t}{\beta+x}\right)^{\alpha} e^{- t} dt<br />
    You're in the bell lap!


    = \frac{\beta^{\alpha}}{\Gamma(\alpha) \, (\beta + x)} ~ \frac{1}{(\beta+x)^{\alpha}} ~ \int_0^\infty t^{\alpha} e^{- t} \, dt


    = \frac{\beta^{\alpha}}{\Gamma(\alpha)} ~ \frac{1}{\beta^{\alpha+1} \left(1 + \frac{x}{\beta} \right)^{\alpha + 1}} ~ \int_0^\infty t^{\alpha} e^{- t} \, dt


    = \frac{1}{\Gamma(\alpha)} ~ \frac{1}{\beta \, \left(1 + \frac{x}{\beta} \right)^{\alpha + 1}} ~ \Gamma(\alpha + 1)


    = \frac{1}{\Gamma(\alpha)} ~ \frac{1}{\beta \, \left(1 + \frac{x}{\beta} \right)^{\alpha + 1}} ~ \alpha \, \Gamma(\alpha)


    and the result (given in my first reply) is easily seen .......
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