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For a pair of random variables $Y$ and $Z$, is it possible that their ratio $X:=\frac{Y}{Z}$ is (exactly, not asymptotically) normally distributed?
If so, could you offer an example of the distributions of $Y$ and $Z$ and the relationship between them (besides the obvious case where $Y$ is normal and $Z$ is a constant, as suggested by @gunes)?

P.S. A special case of my question has been answered here: What Ratio of Independent Distributions gives a Normal Distribution?. My question is more general than that.

Richard Hardy
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  • Here is a related question: https://stats.stackexchange.com/questions/162483/is-the-ratio-distribution-of-two-normally-distributed-variables-ever-normal with a useful comment just under it. It seems the trivial case is the only case for independent RVs. – gunes Apr 03 '20 at 16:58
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    @Gunes there are slightly less trivial cases. One is the ratio of a half-normal and an independent Rademacher variable. Apart from requiring the variables be continuous (and obviously one of them *must* be), https://stats.stackexchange.com/questions/121752/what-ratio-of-independent-distributions-gives-a-normal-distribution is exactly the same question. – whuber Apr 03 '20 at 17:31
  • It's a very good trick, i.e. Y half-normal and Z being 1,-1 with 0.5 prob. – gunes Apr 03 '20 at 20:36
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    You probably want to say independent RVs to exclude things like $Z^2/Z$. – Hasse1987 Apr 04 '20 at 01:39
  • @Hasse1987, thanks. I wanted the general case, including dependent $Y$ and $Z$. Once I got some good examples on that, the remaning interest is of course on independent cases. But that was not the sole focus of the question *before* I got the examples of the dependent cases. – Richard Hardy Apr 04 '20 at 10:58
  • @kjetilbhalvorsen, no, it does not. My question is more general as it allows for dependence between $Y$ and $Z$. – Richard Hardy Apr 06 '20 at 16:37
  • But that's no generalization, because the answer is trivial: multiply any normal variable $X$ by any variable $Z$ whatsoever (provided only it has zero chance of equalling zero) to produce $Y.$ – whuber Apr 06 '20 at 17:18
  • @whuber, the applicability of the qualifier *trivial* depends on the user and more frequently holds ex post than ex ante... And to be precise, a trivial generalization is still a generalization. – Richard Hardy Apr 06 '20 at 19:03

1 Answers1

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A trivial case: Let $Y$ be a normal RV, and $Z$ be a constant RV, then $X$ is going to be normally distributed.

Another one: let $A,B$ normal RVs, and $C=A/B,D=1/B$ are two other RVs that belonging to Cauchy and Reciprocal Normal Distributions. Their ratio will be $C/D=A$ normally distributed.

gunes
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    It would be more interesting with some less obvious examples ... – kjetil b halvorsen Apr 03 '20 at 16:36
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    Thank you! Techincally you are right. I did not exclude the constant RV case explicitly as I thought it was too obvious, but obviously it was not. I will edit my post. (P.S. I did not downvote your answer.) – Richard Hardy Apr 03 '20 at 16:37
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    couldn't understand why I'm downvoted. (thank you, I guessed that you didn't, it was a general question). It'd be interesting if we have two Rvs that are independent and their ratio is normally distributed. Otherwise, we can define any dependence relation we want just as in the example case. – gunes Apr 03 '20 at 16:38
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    The latter idea is a nice one. The case of independent $Y$ and $Z$ is even more interesting, but there is less hope there, I guess. – Richard Hardy Apr 03 '20 at 17:36
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    The answer would be improved if the contrived trivial case were removed, and the substance left. – wolfies Apr 06 '20 at 15:01