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November 2nd, 2016, 05:11 PM  #1 
Newbie Joined: Oct 2016 From: Earth Posts: 16 Thanks: 0  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)? 
November 3rd, 2016, 05:02 AM  #2 
Newbie Joined: Oct 2016 From: Earth Posts: 16 Thanks: 0 
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)? 
November 3rd, 2016, 07:06 PM  #3 
Newbie Joined: Oct 2016 From: Earth Posts: 16 Thanks: 0 
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.

November 3rd, 2016, 11:27 PM  #4 
Newbie Joined: Oct 2016 From: Earth Posts: 16 Thanks: 0 
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. 
November 3rd, 2016, 11:37 PM  #5 
Senior Member Joined: Sep 2015 From: Southern California, USA Posts: 1,484 Thanks: 744 
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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