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Math Help - Method of moment generating function normal distribution

  1. #1
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    Method of moment generating function normal distribution

    Suppose that Y is normally distributed with mean 0, and unknown variance \sigma^2. Using the method of moment generating functions, show that \frac{Y^2}{\sigma^2} has a \chi^2 distribution with 1 df.

    I know that the mgf of a normal distribution is \exp\left(\mu t +\frac{t^2 \cdot \sigma^2}{2} \right) , so if I was to replace the variable with what was given in the question then I would get:

    \exp\left(0 t +\frac{(Y^2/\sigma^2)^2 \cdot \sigma^2}{2} \right) = \exp\left(\frac{Y^4}{2\sigma^2} \right) which looks nothing like the mgf of a \chi^2 distribution.

    the other way I was thinking of doing solving this was using the definition, so I would get:

    m_{Y^2/\sigma^2} (t) = E[e^{t \cdot Y^2/\sigma^2}] = \int^{+\infty}_{-\infty} \frac{1}{\sigma\sqrt{2\pi}}\cdot \exp \left[ -\left(\frac{1}{2\sigma^2}\right)(y-\mu)^2\right] \cdot \exp\left(\frac{ty^2}{\sigma^2}\right) \ dy

    m_{Y^2/\sigma^2} (t) = E[e^{t \cdot Y^2/\sigma^2}] = \int^{+\infty}_{-\infty} \frac{1}{\sigma\sqrt{2\pi}}\cdot \exp \left[ -\left(\frac{1}{2\sigma^2}\right)(y)^2 \cdot \left(\frac{ty^2}{\sigma^2}\right)\right]  \ dy

    m_{Y^2/\sigma^2} (t) = E[e^{t \cdot Y^2/\sigma^2}] = \int^{+\infty}_{-\infty} \frac{1}{\sigma\sqrt{2\pi}}\cdot \exp \left[ -\left(\frac{ty^4}{2\sigma^4}\right)\right]  \ dy

    at which point I don't know how to proceed
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  2. #2
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    Quote Originally Posted by lllll View Post
    Suppose that Y is normally distributed with mean 0, and unknown variance \sigma^2. Using the method of moment generating functions, show that \frac{Y^2}{\sigma^2} has a \chi^2 distribution with 1 df.

    I know that the mgf of a normal distribution is \exp\left(\mu t +\frac{t^2 \cdot \sigma^2}{2} \right) , so if I was to replace the variable with what was given in the question then I would get:

    \exp\left(0 t +\frac{(Y^2/\sigma^2)^2 \cdot \sigma^2}{2} \right) = \exp\left(\frac{Y^4}{2\sigma^2} \right) which looks nothing like the mgf of a \chi^2 distribution.

    Mr F says: You need to stop thinking like this. You cannot just substitute like it's a change of variable. This is a grave mistake that is a common to a lot of your solutions.

    Mr F quarantines the above. ---------------------------------------------------------------------------------------------

    the other way I was thinking of doing solving this was using the definition, so I would get:

    m_{Y^2/\sigma^2} (t) = E[e^{t \cdot Y^2/\sigma^2}] = \int^{+\infty}_{-\infty} \frac{1}{\sigma\sqrt{2\pi}}\cdot \exp \left[ -\left(\frac{1}{2\sigma^2}\right)(y-\mu)^2\right] \cdot \exp\left(\frac{ty^2}{\sigma^2}\right) \ dy Mr F says: Good move!

    m_{Y^2/\sigma^2} (t) = E[e^{t \cdot Y^2/\sigma^2}] = \int^{+\infty}_{-\infty} \frac{1}{\sigma\sqrt{2\pi}}\cdot \exp \left[ -\left(\frac{1}{2\sigma^2}\right)(y)^2 \cdot \left(\frac{ty^2}{\sigma^2}\right)\right] \ dy Mr F says: NO! When you multiply exponentials with the same base, you ADD the powers (this is a basic index law). You do not multiply the powers.

    m_{Y^2/\sigma^2} (t) = E[e^{t \cdot Y^2/\sigma^2}] = \int^{+\infty}_{-\infty} \frac{1}{\sigma\sqrt{2\pi}}\cdot \exp \left[ -\left(\frac{ty^4}{2\sigma^4}\right)\right] \ dy

    at which point I don't know how to proceed
    Quote Originally Posted by lllll and edited in red by Mr F View Post
    [snip]
    m_{Y^2/\sigma^2} (t) = E[e^{t \cdot Y^2/\sigma^2}] = \int^{+\infty}_{-\infty} \frac{1}{\sigma\sqrt{2\pi}}\cdot \exp \left[ -\left(\frac{1}{2\sigma^2}\right)(y)^2 {\color{red}+} \left(\frac{ty^2}{\sigma^2}\right)\right] \ dy
    [snip]
    Then:

    m_{Y^2/\sigma^2} (t) = \int^{+\infty}_{-\infty} \frac{1}{\sigma\sqrt{2\pi}}\cdot \exp \left[ - y^2 \left(\frac{1}{2\sigma^2} - \frac{t}{\sigma^2}\right)\right] \, dy = \frac{1}{\sigma\sqrt{2\pi}} \int^{+\infty}_{-\infty} \exp \left[ - y^2 \left(\frac{1 - 2t}{2\sigma^2}\right)\right] \, dy


    Now make the substitution x = \frac{y}{\sigma} \sqrt{\frac{1 - 2t}{2}} \Rightarrow dy = \sigma \sqrt{\frac{2}{1-2t}} \, dx:


    = \frac{1}{\sqrt{\pi}} \, \frac{1}{\sqrt{1 - 2t}} \, \int^{+\infty}_{-\infty} \exp \left[ - x^2 \right] \, dx


    = \frac{1}{\sqrt{\pi}} \, \frac{1}{\sqrt{1 - 2t}} \, \sqrt{\pi}

    (using the well known result)


    = \frac{1}{\sqrt{1 - 2t}}.
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