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Math Help - moment generating function of exponential dist.

  1. #1
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    moment generating function of exponential dist.

    I'm having trouble finding the moment generating function of a exponential distribution.

    \phi = \int e^{ty}f(x) dx = \int_{0}^{\infty} e^{ty}\frac{1}{\beta} e^{-y/\beta} dy

    \phi = \frac{1}{\beta} \int_{0}^{\infty}e^{y(t-1/\beta)} dy

    \phi = \frac{1}{\beta}\times \frac{1}{t-\frac{1}{\beta}} e^{y(t-1/\beta)}\bigg{|}^{\infty}_{0}

    \phi= \left(\frac{1}{\beta} \right) \left( \frac{1}{t-\frac{1}{\beta}} \right)

    \phi = \frac{1}{\beta t-1}

    but the table in the back of my book gives:

    \phi = \frac{1}{\color{red}{1-\beta t}}
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  2. #2
    Rhymes with Orange Chris L T521's Avatar
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    Quote Originally Posted by lllll View Post
    I'm having trouble finding the moment generating function of a exponential distribution.

    \phi = \int e^{ty}f(x) dx = \int_{0}^{\infty} e^{ty}\frac{1}{\beta} e^{-y/\beta} dy

    \phi = \frac{1}{\beta} \int_{0}^{\infty}e^{y(t-1/\beta)} dy

    \phi = \frac{1}{\beta}\times \frac{1}{t-\frac{1}{\beta}} e^{y(t-1/\beta)}\bigg{|}^{\infty}_{0}

    \phi= \left(\frac{1}{\beta} \right) \left( \frac{1}{t-\frac{1}{\beta}} \right)

    \phi = \frac{1}{\beta t-1}

    but the table in the back of my book gives:

    \phi = \frac{1}{\color{red}{1-\beta t}}
    The issue here in integrating this is the the value of e^{y(t-1/\beta)} as y\to\infty.

    We can fix this by writing e^{y(t-1/\beta)}=e^{-y(1/\beta-t)}

    In evaluating the integral for the moment generating function, we should have

    \phi = \frac{1}{\beta}\times -\frac{1}{\frac{1}{\beta}-t} e^{-y(1/\beta-t)}\bigg{|}^{\infty}_{0} =-\frac{1}{1-\beta t}\times\left[e^{-\infty}-e^0\right]=-\frac{1}{1-\beta t}\times\left(-1\right)=\color{red}\boxed{\frac{1}{1-\beta t}}

    Does this make sense?

    --Chris
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