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Math Help - Method of moment generating function

  1. #1
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    Method of moment generating function

    Suppose that Y follows an exponential distribution, with mean \theta. Use the method of moment generating functions to show that \frac{2Y}{\theta} is a pivotal quantity and has a distribution with 2 df.

    Attempt:


    <br />
f(y) = \left\{ \begin{array}{rcl}<br />
\frac{1}{\beta}e^{-y/\beta} & \mbox{for} & y>0 \\ <br />
0 & \mbox{otherwise} & <br />
\end{array}\right.<br />

    so substituting \theta=\frac{1}{\beta}, Y = \frac{U\theta}{2} and dy=\frac{\theta}{2} du I get:

    \int^\alpha_0 \theta \exp\left(-\theta \times \frac{U\theta}{2}\right) \frac{\theta}{2} du

    = \int^\alpha_0 \frac{\theta^2}{2} \exp\left(-\theta^2 \times \frac{U}{2}\right) du

    = \frac{\theta^2}{2} \int^\alpha_0  \exp\left(-\theta^2 \times \frac{U}{2}\right) du

    = \frac{\theta^2}{2} \times -\frac{2}{\theta^2} \times \exp\left(-\theta^2\times \frac{U}{2}\right) \bigg{|}^{\alpha}_0

    =-\exp\left(-\theta^2\times \frac{U}{2}\right) \bigg{|}^{\alpha}_0

    =1-\exp\left(-\theta^2 \cdot \frac{\alpha}{2} \right) \neq (1-2t)^{\nu/2}
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  2. #2
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    Quote Originally Posted by lllll View Post
    Suppose that Y follows an exponential distribution, with mean \theta. Use the method of moment generating functions to show that \frac{2Y}{\theta} is a pivotal quantity and has a distribution with 2 df.

    Attempt:


    <br />
f(y) = \left\{ \begin{array}{rcl}<br />
\frac{1}{\beta}e^{-y/\beta} & \mbox{for} & y>0 \\ <br />
0 & \mbox{otherwise} & <br />
\end{array}\right.<br />
Mr F says: No. Instead of {\color{red}\beta} you should have {\color{red}\theta}. Also, I'm sorry to say, all that follows is no good.

    so substituting \theta=\frac{1}{\beta}, Y = \frac{U\theta}{2} and dy=\frac{\theta}{2} du I get:

    \int^\alpha_0 \theta \exp\left(-\theta \times \frac{U\theta}{2}\right) \frac{\theta}{2} du

    = \int^\alpha_0 \frac{\theta^2}{2} \exp\left(-\theta^2 \times \frac{U}{2}\right) du

    = \frac{\theta^2}{2} \int^\alpha_0 \exp\left(-\theta^2 \times \frac{U}{2}\right) du

    = \frac{\theta^2}{2} \times -\frac{2}{\theta^2} \times \exp\left(-\theta^2\times \frac{U}{2}\right) \bigg{|}^{\alpha}_0

    =-\exp\left(-\theta^2\times \frac{U}{2}\right) \bigg{|}^{\alpha}_0

    =1-\exp\left(-\theta^2 \cdot \frac{\alpha}{2} \right) \neq (1-2t)^{\nu/2}
    Theorem (easy to prove): The moment generating function of X = aY + b is m_X(t) = e^{bt} \, m_Y (at) where m_Y(t) is the moment generating function of Y.

    Let X = \frac{2Y}{\theta}. Then using the above theorem, m_X(t) = M_Y\left( \frac{2}{\theta} t\right).

    It is well known (and easy to prove) that if Y follows an exponential distribution with mean \theta then m_Y(t) = \frac{1}{1 - \theta t}.

    Therefore m_X(t) = \frac{1}{1 - \theta \left( \frac{2}{\theta} \right) t} = \frac{1}{1 - 2t}.

    This is readily recognised as the moment generating function for an an exponential distribution with mean equal to 2.

    Therefore the pdf of X = \frac{2Y}{\theta} is f(x) = \frac{1}{2} e^{-x/2}.

    (I'm not sure where the two degrees of freedom comes from because that's not the case here).


    1. X is a function of Y and \theta.

    2. The pdf of X does not depend on \theta.

    Therefore X = \frac{2Y}{\theta} is a pivotal quantity.
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