What Ratio of Independent Distributions gives a Normal Distribution? The ratio of two independent normal distributions give a Cauchy distribution. The t-distribution is a normal distribution divided by an independent chi-squared distribution. The ratio of two independent chi-squared distribution gives an F-distribution.
I am looking for a ratio of independent continuous distributions that gives a normally distributed random variable with mean $\mu$ and variance $\sigma^2$?
There is probably an infinite set of possible answers. Can you give me some of these possible answers? I would particularly appreciate if the two independent distributions which ratio is computed are the same or at least have similar variance.
 A: Let $Y_1 = Z \sqrt{E}$ where $E$ has an exponential distribution with mean $2 \sigma^2$ and $Z = \pm 1$ with equal probability. Let $Y_2 = 1 / \sqrt{B}$ where $B \sim \mbox{Beta}(0.5, 0.5)$. Assuming $(Z, E, B)$ are mutually independent, then $Y_1$ is independent of $Y_2$ and $Y_1 / Y_2 \sim \text{Normal}(0, \sigma^2)$. Hence we have 


*

*$Y_1$ independent of $Y_2$; 

*Both continuous; such that

*$Y_1 / Y_2 \sim \text{Normal}(0, \sigma^2)$. 


I haven't figured out how to get a $\text{Normal}(\mu, \sigma^2)$. It is harder to see how to do this since the problem reduces to finding $A$ and $B$ which are independent such that 
$$
\frac{A - B \mu}{B} \sim \text{Normal}(0, 1)
$$
which is quite a bit harder than making $A/B \sim \text{Normal}(0,1)$ for independent $A$ and $B$. 
A: 
I would particularly appreciate if the two independent distributions which ratio is computed are the same 

There is no possibility that a normal variable can be written as a ratio of two independent variables with the same distribution or distribution family (such as the F-distribution which is the ratio of two scaled $\chi^2$ distributed variables or the Cauchy-distribution which is the ratio of two normal distributed variables with zero mean).


*

*Suppose that: for any $A, B \sim F$ where $F$ is the same distribution or distribution family we have $$X = \frac{A}{B} \sim N(\mu,\sigma^2)$$

*We must also be able to reverse $A$ and $B$ (if a normal variable can be written as a ratio of two independent variables with the same distribution or distribution family then the order can be reversed) $$\frac{1}{X} = \frac{B}{A} \sim N(\mu,\sigma^2)$$ 

*But if $X \sim N(\mu,\sigma^2)$ then $X^{-1} \sim N(\mu,\sigma^2)$ can not be true (the inverse of a normal distributed variable is not another normal distributed variable).



Broader conclusion: If the variables in any distribution family $\mathcal{F}_X$ can be written as a ratio of variables in another distribution family $\mathcal{F}_Y$ then it must be that family $\mathcal{F}_X$ is closed under taking the reciprocal (ie. for any variable whose distribution is in $\mathcal{F}_X$ the distribution of it's reciprocal will also be in $\mathcal{F}_X$). 
E.g. the inverse of a Cauchy distributed variable is also Cauchy distributed. The inverse of an F-distributed variable is also F-distributed.


*

*This 'if' is not an 'iff', the converse is not true. When $X$ and $1/X$ are in the same distribution family then it may not always be possible to be written as a ratio distribution with nominator and denominator from the same distribution family. 
Counterexample: We can imagine distribution families for which for any $X$ in the family we have $1/X$ in the same family but we do not have $P(X=1)=0$. This is contradicting with the fact that for a ratio distribution where the denominator and nominator have the same distribution we must have $P(X=1) \neq 0$ (and something similar can be expressed for continuous distributions like the integral along the line X/Y=1 in a scatterplot of X,Y has some non zero density when X and Y have the same distribution and are independent).
