Distribution of ratio of 2 points drawn from normal distribution? Let's say we have a known normal distribution $N(\mu,\sigma^2)$.
I now draw 2 points $p1$ and $p2$ randomly from this Gaussian distribution for every observation, and repeat this process large number of times.


*

*What will the distribution of $\frac{p1}{p2}$ look like? Will it be normal? Can we say something about it's mean and standard deviation?

*What will the distribution of $\operatorname{max} (\frac{p1}{p2},\frac{p2}{p1})$ look like? Will it be normal? Can we say something about it's mean and standard deviation?

*What will the distribution of $\frac{e^p_1}{ e^p_2}$ and the distribution of $\operatorname{max} (\frac{e^p_1}{ e^p_2}, \frac{e^p_2}{e^p_1})$ look like? Will it be normal? Can we say something about it's mean and standard deviation?
 A: For Q 1 and Q 2: This is a gaussian ratio distribution, see https://en.wikipedia.org/wiki/Ratio_distribution#Gaussian_ratio_distribution and https://www.amazon.com/Probability-Distributions-Involving-Gaussian-Variables/dp/0387346570  and search this site.  Si I will concentrate an Q3. First part is straightforward, as noted in whuber's coooent, it is lognormal with parameters $\mu=0, 2\sigma^2$. The second part is the a maximum of two related lognormals:
$$
   \max\left( e^{p_1-p_2}, e^{p_2-p_1} \right)
$$
Note that this maximum is given by $e^{|p_1-p_2|}$ so we could call it a log-halfnormal distribution, in this case a half-normal (or absolute value of normal with zero-mean) based on normal with parameters $\mu=0, 2\sigma^2$ with density function
$$
   f(x) =\frac{2}{\sqrt{2\pi}\sqrt{2\sigma^2}}e^{-\frac12 \frac{x^2}{2\sigma^2}} = \frac1{\sqrt{\pi}\sigma} e^{-\frac14(\frac{x}{\sigma})^2}, \quad x\ge 0
$$
The density of such a half-normal variate exponentiated can be found from first principles.  The result becomes
$$
  f_Y(y) = \frac1{\sqrt{\pi}\sigma} e^{-\frac14 (\frac{\log y}{\sigma})^2}, \quad y> 1
$$
expectation and variance can be found to be 
$$
  2 e^{\sigma^2} \Phi(\sqrt{2}\sigma), \\
2e^{4\sigma^2} \Phi(2\sqrt{2}\sigma) - 4e^{2\sigma^2} \Phi(\sqrt{2}\sigma)^2
$$ (calculated with help of maple).
