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    Standard Normal Distribution




    Z ~ N(0,1)

    How to show that Z\stackrel{d}{=} -Z and \frac{Z^2}{2} ~ gamma(\frac{1}{2})? Also how to show that only the standard normal distribution satisfies these 2 properties? In other words, if you have Z\stackrel{d}{=} -Z and \frac{Z^2}{2} ~ gamma(\frac{1}{2}), then Z ~ N(0,1).

    To show that Z\stackrel{d}{=} -Z:
    I know that distribution of Z is F(z) = \frac{1}{2} + \phi(z), where \phi(z) = \frac{1}{\sqrt{2\pi}}\int_{0}^{z} e^{\frac{-t^2}{2}} dt. Also \phi(-z)=-\phi(z).
    P(-Z < z) = P(Z > -z) = 1 - P(Z < -z) = 1 - (\frac{1}{2} + \phi(-z)) = 1 - (\frac{1}{2} - \phi(z)) = \frac{1}{2} + \phi(z) = P(Z < z).
    Hence Z\stackrel{d}{=} -Z.

    To show that \frac{Z^2}{2} ~ gamma(\frac{1}{2}):
    P(\frac{Z^2}{2} < z) = P(-\sqrt{2z} < Z < \sqrt{2z}) = 2\phi(\sqrt{2z}).

    Now, this is where I am stuck. How do I continue to get gamma(\frac{1}{2})? Also how to show if you have Z\stackrel{d}{=} -Z and \frac{Z^2}{2} ~ gamma(\frac{1}{2}), then Z ~ N(0,1).
    Last edited by geniusacamel; 11-04-2016 at 02:24 AM.

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    Re: Standard Normal Distribution

    I'm not a great theoretician, but I think there may be something wrong with the question because the gamma distribution usually has two arguments, alpha and beta. However, the distribution of one z^2 is chi square with 1 df, and the chi square is a simplified gamma with alpha = 2 and n = 2.beta with n = 1 in this case. It all looks hopeful. Perhaps if you work through the derivation of the chi square backwards you may get somewhere. kat

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    Re: Standard Normal Distribution

    I did not learn chi square distribution so I think it can be found directly without resorting to going to chi square. I hope somebody who knows more about this will be able to help me.

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    Re: Standard Normal Distribution

    Ok, differentiating 2\phi(\sqrt{2z}) with respect to z gives me \frac{e^{-z}}{\sqrt{\pi}\sqrt{z}} = \frac{z^{\frac{1}{2}-1}e^{-z}}{\Gamma(\frac{1}{2})} which is the probability density function of gamma(\frac{1}{2}, 1). Hence, I believe I have solved it. Now, if someone could help me with the reverse direction, I would greatly appreciate it.

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    Re: Standard Normal Distribution


    Quote Originally Posted by geniusacamel View Post
    ... Now, if someone could help me with the reverse direction, I would greatly appreciate it.

    Well, in general, if a random variable X has a gamma distribution then it could actually depend on three (positive) parameters \left ( \alpha ,\beta ,\gamma  \right ) – it is of the Pearson Type III system.

    That said, most often, the two parameter system is used i.e. \left ( \alpha ,\beta  \right ).

    If you use the standardized form of the gamma distribution i.e. \left ( \alpha \right ), then it tends to the standard normal distribution as the parameter \left ( \alpha \right ) tends to infinity:

    \lim_{\alpha \rightarrow\infty}Pr\left [\left (X-\alpha  \right )\alpha ^{-\frac{1}{2}}\leq u  \right ]=\Phi \left ( u \right ),

    for all real values of u, where

    \Phi \left ( u \right ) is the standard normal cdf.

    That said, similar results hold for the general gamma distribution as well as the special case of the family of chi-square distributions.

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