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.
Best Answer
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
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 $B$.