What Ratio of Independent Distributions gives a Normal Distribution?

Question

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.

Answer 1

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

1. $Y_1$ independent of $Y_2$;
2. Both continuous; such that
3. $Y_1 / Y_2 \sim \text{Normal}(0, \sigma^2)$.

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 $B$.

Answer 2

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).

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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).

Answer 3

Well, here is one but I will not prove it, only show it in simulation.

Make two beta distributions with equal large shape parameters $\text{Beta}(200,200)$ (here, $n=40,000$), subtract 1/2 from the $x$-values of one of them and call it "numerator." That gives us a PDF that has a maximum range of $\left(-\frac{1}{2},\,\frac{1}{2}\right)$, but because the shape parameters are so large, we never get to the maximum values of the range. Here is a histogram of an $n=40,000$ "numerator"

Next, we call the second beta distribution "denominator" without subtracting anything, so it has the usual beta distribution range of $(0,1)$ and one of those looks like this

Again, because the shapes are so large, we do not approach the maximum range with the values. Next we plot the quotient $\frac{\text{numerator}}{\text{denominator}}$ as a PDF with the superimposed normal distribution.

Now in this case the normal distribution result has $\mu \to -0.0000204825,\sigma \to 0.0501789$ and tests for normality that look like this

$$\left( \begin{array}{ccc} \text{} & \text{Statistic} & \text{P-Value} \\ \text{Anderson-Darling} & 0.799786 & 0.481181 \\ \text{Baringhaus-Henze} & 1.40585 & 0.0852017 \\ \text{Cramér-von Mises} & 0.123145 & 0.482844 \\ \text{Jarque-Bera ALM} & 4.48103 & 0.106404 \\ \text{Kolmogorov-Smirnov} & 0.00452328 & 0.386335 \\ \text{Kuiper} & 0.00798063 & 0.109127 \\ \text{Mardia Combined} & 4.48103 & 0.106404 \\ \text{Mardia Kurtosis} & 1.53849 & 0.123929 \\ \text{Mardia Skewness} & 2.09399 & 0.147879 \\ \text{Pearson }\chi ^2 & 134.353 & 0.571925 \\ \text{Watson }U^2 & 0.113831 & 0.211187 \\ \end{array} \right)$$

In other words, we cannot prove the ratio is not normal even trying very hard to do so.

Now why? Intuition on my part, which I have in overabundance. Proof left to reader, if any exists (maybe via limit of method of moments, but again that is just intuition).

Hint: If I use only $\text{Beta}(20,20)$ in denominator and $\text{Beta}(20,20)-\frac{1}{2}$ in the numerator and I get Student's $t$ with $\mu \to -0.000251208,\sigma \to 0.157665,\text{df}\to 33.0402$

$$\begin{array}{l|ll} \text{} & \text{Statistic} & \text{P-Value} \\ \hline \text{Anderson-Darling} & 0.275262 & 0.955502 \\ \text{Cramér-von Mises} & 0.0351108 & 0.956524 \\ \text{Kolmogorov-Smirnov} & 0.00320936 & 0.804486 \\ \text{Kuiper} & 0.00556501 & 0.657146 \\ \text{Pearson }\chi ^2 & 145.077 & 0.323168 \\ \text{Watson }U^2 & 0.0351042 & 0.878202 \\ \end{array}$$

Another hint $\dfrac{\mathcal{N}(0,1)}{\mathcal{N}(10,1/1000)}\to$ Student's $t$ $\mu \to -0.0000535722,\sigma \to 0.0992765,\text{df}\to 244.154$

$$\left( \begin{array}{ccc} \text{} & \text{Statistic} & \text{P-Value} \\ \text{Anderson-Darling} & 0.501677 & 0.745102 \\ \text{Cramér-von Mises} & 0.0696824 & 0.753515 \\ \text{Kolmogorov-Smirnov} & 0.00355688 & 0.692225 \\ \text{Kuiper} & 0.00608382 & 0.501133 \\ \text{Pearson }\chi ^2 & 142.88 & 0.370552 \\ \text{Watson }U^2 & 0.0603207 & 0.590369 \\ \end{array} \right)$$

Edit "Under certain conditions a normal approximation is possible" See link. Moreover, the main text of that Wikipedia article shows interesting material on the ratio distribution topic. Another reference.