# Can a ratio of random variables be normal? [duplicate]

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.

• Here is a related question: stats.stackexchange.com/questions/162483/… with a useful comment just under it. It seems the trivial case is the only case for independent RVs. – gunes Apr 3 '20 at 16:58
• @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), stats.stackexchange.com/questions/121752/… is exactly the same question. – whuber Apr 3 '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 3 '20 at 20:36
• You probably want to say independent RVs to exclude things like $Z^2/Z$. – Hasse1987 Apr 4 '20 at 1:39
• @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 4 '20 at 10:58

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.
• 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 3 '20 at 17:36