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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: USA Posts: 2,529 Thanks: 1389 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}

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