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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.

Alexis
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Remi.b
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    While the [Wikipedia article on ratio disributions](https://en.wikipedia.org/wiki/Ratio_distribution) does not provide examples of the case for which you seek, it is an interesting read. – Avraham Oct 28 '14 at 17:27
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    A rather special case is $X$ a standard normal, and $Y$ independently $\pm1$ each with probability $\frac12$, then $X$, $Y$ and $\frac{X}{Y}$ have the same mean and variance and $\frac{X}{Y}$ is normally distributed. – Henry Oct 29 '14 at 21:40
  • To make sure I understand the special case your describe. $Y$ is a discrete random variable that takes values $-1$ or $1$ with equal probabilities, is that right? Simple and interesting. I should have said that I am looking for countinuous r.v. I edited my post to add this information. Thanks for your comment – Remi.b Oct 29 '14 at 21:50
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    "*The ratio of two independent chi-squared distribution gives a F-distribution*" --- well, not quite. It gives a beta-prime distribution. To get an F you need to scale each chi-square by its df. – Glen_b Oct 30 '14 at 01:12
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    A number of things make me not at all convinced that it is necessarily possible to fulfill all your conditions. – Glen_b Oct 30 '14 at 01:19
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    taking the generation of normal variables method (e.g Box-Muller) as example (which uses the circle method) i would say there are no ratios of **uniform distributions** that give a normal distribution (assuming uniform distributions are asked for) – Nikos M. Oct 31 '14 at 07:04
  • @Glen_b Not sure about that, see my intuition below. – Carl Jun 08 '18 at 23:36
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    @Remi.b All your examples had numerator and denominator i.i.d. and I have assumed that you intended the identically distributed case. Is that so, or was that just happenstance and you're after any numerator and denominator pair? If the latter, there are trivial solutions that yield a normal result (e.g. half-normal numerator and a random sign for the denominator) – Glen_b Jun 09 '18 at 02:34
  • related: https://stats.stackexchange.com/questions/417121/are-there-two-distributions-whose-product-equals-a-gaussian – Sextus Empiricus Jul 27 '19 at 23:51

3 Answers3

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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 $B$.

guy
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    If this is true, this is awesome. – Neil G Jun 09 '18 at 04:02
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    @NeilG it is true; the product of my beta and exponential is a gamma with shape 1/2 (because of how you can build the beta and an independent gamma using gammas). Then the square root of that is half-normal using the fact that the square of a normal is chi-square. – guy Jun 09 '18 at 15:22
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    We had recently a question asking for a product of two variables that is normal distributed (I can not find it back). That question had a comment or answer relating to the [Box-Muller transform](https://en.wikipedia.org/wiki/Box%E2%80%93Muller_transform) which computes a normal distribution (or more precisely a bivariate normal distribution) from the *product* of two transformed uniform distributed variables. This answer relates a lot to that but takes the inverse of one of those variables in the Box-Muller transform. cc: @kjetilbhalvorsen – Sextus Empiricus Jul 27 '19 at 23:30
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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).

Sextus Empiricus
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  • Don't see it. Seems to me that just because $A/D$ and $B/C$ are normal that does not make $\frac{A/D}{B/C}$ normal. – Carl Jun 09 '18 at 00:57
  • Better. Now it makes sense. – Carl Jun 09 '18 at 01:07
  • Carl, I changed my question now into something simpler but in my previous answer I did not write that $\frac{A/D}{B/C}$ is normal distributed. It is distributed as the *ratio* of two Gaussian distributed variables (and also as the *product* on the other hand, which creates a contradiction). – Sextus Empiricus Jun 09 '18 at 01:10
  • Agreed, but your answer did not make sense to me, and now it does. – Carl Jun 09 '18 at 01:12
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    I don't understand how the second statement follows from the first. If there exists some $A, B$ such that their quotient is normal, why does it follow that their quotient in the other order should also be normal? The question didn't ask for a distribution family such that the quotient of all pairs of elements is normal. – Neil G Jun 09 '18 at 04:00
  • @NeilG I think you would have to see before answer edit to follow the thread. – Carl Jun 09 '18 at 04:11
  • I don't see it, but anyway, it's a worthwhile point that “there is no distribution family such that the quotient of any pair of its elements is normally distributed”. I think this is what you are showing. – Neil G Jun 09 '18 at 04:13
  • @NeilG Well, maybe not entirely. See second hint below. – Carl Jun 09 '18 at 04:43
  • @NeilG it is in a previous edit where I compared (A/B)*(C/D)=(A/D)/(B/C). If those A,B,C,D have the same distribution then the resulting ratio distribution is such that if you multiply two independent variables with this distribution you get something which has the same distribution in comparison to when you would take the ratio. I did not change it because it is incorrect, but instead the current explanation is easier, more intuitive. – Sextus Empiricus Jun 09 '18 at 08:07
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    I don't understand what you're saying. Ideally, your answer would be a coherent argument without requiring someone to read the edits. Right now, it seems like your second statement ("we must also have") doesn't follow from the first. – Neil G Jun 09 '18 at 08:11
  • ah, I see, my mistake. I thought that you where referring Carl's statement which was about the pre-editted version, that is why I mentioned the older version. Note that I am not answering the question by stating that there can not be a/b=c such that c is normal distributed. I am just saying that those a and b can not be distributed the same, and in this way I provide an answer in a negative sense, what it can not be instead of what it can be. – Sextus Empiricus Jun 09 '18 at 09:09
  • and aside from a and b not being able to be distributed the same you could also say they can not be the same family. For any ratio distribution c=a/b with a and b from the same family you will have that 1/c is from the same family as c (since it can be expressed as a similar ratio). For instance the F distribution. If $X \sim F(d_i,d_j)$ then $1/X \sim F(d_j,d_i)$. But the inverse of a normal distributed variable is not itself normal distributed. – Sextus Empiricus Jun 09 '18 at 09:12
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    +1: Straightforward, clear, elegant, and insightful. – whuber Jul 26 '19 at 13:00
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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" enter image description here

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

enter image description here

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.

enter image description here

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 $\mu \to -0.000251208,\sigma \to 0.157665,\text{df}\to 33.0402$

enter image description here

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

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

enter image description here

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

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.

Carl
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  • @whuber How is this for a show-me? Can you prove or disprove it? – Carl Jun 08 '18 at 23:39
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    You are clearly very close to a normal distribution. However, that isn't at all the same thing as having a normal distribution, and I don't believe the ratio of a centered symmetric beta to an ordinary symmetric beta with the same parameters is ever to be actually normal. I'd be very interested in being wrong about this though. – Glen_b Jun 09 '18 at 02:24
  • @Glen_b If I had to guess I would say it is Student's-t with 200 df and the difference between ND and Student's-t may be unimportant. – Carl Jun 09 '18 at 03:12
  • @Glen_b Put it in ans. as somewhat less of a guess. – Carl Jun 09 '18 at 04:00
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    Your solution definitely is not Normal. You could generalize this approach: take any distribution that is approximately Normal and divide it by a distribution with its probability concentrated near a nonzero number. The result (obviously) will be close to Normal--but it still will not be Normal. Applying a bunch of tests is unconvincing because all it shows it that you didn't generate sufficiently large samples to demonstrate the non-Normality. – whuber Jun 09 '18 at 18:34
  • @whuber You only looked at the last hint which, BTW, I took from you. It is not the same as what I did above that. This is $n=40,000$, you are suggesting that is not "enough" for what? The denominator for that last hint is just almost a constant, and it is you who pointed out that a normal divided by a shifted Dirac $\delta$ is normal. What comes before that is not so "trivial." – Carl Jun 10 '18 at 01:53
  • Repeat your test with $n=10^8.$ Not only will this definitively demonstrate your data are not Normal, it will help reveal their slightly heavy tails. – whuber Jun 10 '18 at 20:25
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    @whuber At $10^8$ one could also show in normal machine precision that noise will cause anything to be not normal. I do not have a super computer to do that in extended precision. What you could show, and why I wanted you to look at this, was to prove or disprove these things mathematically, not just criticize around the edges with unachievable goals. – Carl Jun 11 '18 at 14:25
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    Let me get to the heart of the matter, then: (1) disproving normality is a simple exercise in integral approximation--no need to give the details here. You can, *e.g.*, readily prove the 200th moment is infinite. (2) Your answer confuses *distributions* with *samples.* It is this fundamental confusion that I object to; it's the reason why I think this answer is more misleading than helpful. BTW, I did not write my last comment lightly: I performed that test. I did it not with a supercomputer, but with a decade-old PC workstation, and the whole process took just seconds. – whuber Jun 11 '18 at 14:27
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    @whuber Which approximation are you testing? The first, the second or the third? BTW, if they are only approximations, so be it. I suggest only that in the limiting case that they might be exact. All of statistics is an approximation so I do not share your apprehension. – Carl Jun 11 '18 at 16:15
  • @whuber For example, $\lim_{\delta\to 0}\dfrac{\mathcal{N}(0,1)}{\mathcal{N}(10,\delta)}\to \dfrac{\mathcal{N}(0,1)}{10}$, which limit is exactly normal. – Carl Jun 11 '18 at 16:20
  • @whuber Try $10^8$ iterations on $\dfrac{\mathcal{N}(0,1)}{\mathcal{N}(10,10^{-32})}$. In limited machine precision one cannot distinguish between 10 and $\mathcal{N}(10,10^{-32})$, which was rather my point. – Carl Jun 11 '18 at 16:35
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    @whuber Let me make one final point. I answered here to stimulate interest, which I have done. Without my interest, this conversation, which lay dormant since 2014, would still be unanswered. I see my role as stimulating thought, not as being pluperfect or walking on water. – Carl Jun 11 '18 at 16:41