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November 2nd, 2016, 05:11 PM   #1
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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 $Z\overset{d}{=} -Z$ 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 $\phi(z) = \int_{0}^{z} e^{\frac{-t^2}{2}} dt$. 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 $Z\overset{d}{=} -Z$ and $\frac{Z^2}{2}$ ~ gamma($\frac{1}{2}$), then $Z$ ~ $N$(0,1)?
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November 3rd, 2016, 05:02 AM   #2
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I am surprised that is no edit button. Anyway, I have corrected some typos.

$Z$ ~ $N$(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 $Z\overset{d}{=} -Z$ 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 $Z\overset{d}{=} -Z$ and $\frac{Z^2}{2}$ ~ gamma($\frac{1}{2}$), then $Z$ ~ $N$(0,1)?
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November 3rd, 2016, 07:06 PM   #3
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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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November 3rd, 2016, 11:27 PM   #4
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Corrected more typos:

$Z$ ~ $N$(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 $Z\overset{d}{=} -Z$ 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 $\phi(z) = \int_{0}^{z} e^{\frac{-t^2}{2}} dt$. 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($\frac{1}{2}$):
$P(\frac{Z^2}{2} < z) = P(-\sqrt{2z} < Z < \sqrt{2z}) = 2\phi(\sqrt{2z})$.
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$(\frac{1}{2}, 1)$.

I would greatly appreciate it if somebody could help me with the reverse direction.

Last edited by geniusacamel; November 3rd, 2016 at 11:32 PM.
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November 3rd, 2016, 11:37 PM   #5
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neat problem.

Ok you have a random variable $U$ which is distributed as $\Gamma\left(\dfrac 1 2 , 1\right) = \dfrac{e^{-u}}{\sqrt{\pi u}}$

now $Z = \sqrt{2U}$

$F_Z(z) = P[Z < z] = P[\sqrt{2U}<z] = P\left[U < \dfrac{z^2}{2}\right]$

let $F_U(u)$ be the CDF of the $\Gamma\left(\dfrac 1 2,1 \right)$ distribution.

$F_Z(z) = F_U\left(\dfrac{z^2}{2}\right)$

and thus

$f_Z(z) = \dfrac{d}{dz} F_Z(z) = \dfrac{d}{dz} F_U\left(\dfrac{z^2}{2}\right) = f_U\left(\dfrac{z^2}{2}\right) \cdot z$

$f_Z(z) = \dfrac{e^{-z^2/2}}{\pi\sqrt{z^2/2}}z=\sqrt{\dfrac{2}{\pi}}e^{-z^2/2},~z \geq 0$

Now were were told however that the distribution of $Z$ is symmetric about $0$, i.e. $f_Z(z) = f_Z(-z)$

Thus it must be that each half of the axis gets half the probability mass and thus we end up with

$f_Z(z) = \dfrac 1 2 \sqrt{\dfrac{2}{\pi}}e^{-z^2/2}=\dfrac{1}{\sqrt{2\pi}}e^{-z^2/2},~z \in \mathbb{R}$

and this is recognized as the standard normal distribution
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