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Let $Z = X / Y$ where $X$ and $Y$ are normally distributed random variables. Is $Z$ normally distributed for any $X$ and $Y$?

how
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    Assuming independence, yes: if and only if $Y$ is a nonzero constant. Closely related threads shed more light on this: http://stats.stackexchange.com/questions/121752, http://stats.stackexchange.com/questions/157557, http://stats.stackexchange.com/questions/148983, http://stats.stackexchange.com/questions/58800. (There's a duplicate here somewhere--it appeared recently--but I can't find it with a search.) – whuber Jul 21 '15 at 15:25
  • @whuber Do you mean when $Y$ is a nonzero constant? – Danica Jul 21 '15 at 15:28
  • @Dougal Yes, thank you: I was still in the process of writing that comment when you caught the typo! – whuber Jul 21 '15 at 15:30
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    I would comment but do not have the reputation. I do not understand whuber's comment, surely if **Y** is a nonzero constant, it is not a normally distributed random variable? Assuming **X** and **Y** are iid *standard* normal (that is normal(0,1)), then their ratio is a cauchy random variable. To the best of my knowledge, in general, the ratio of two normal variables is not anything special, and is NOT normal. See here: https://en.wikipedia.org/wiki/Cauchy_distribution#Properties – John Madden Jul 21 '15 at 16:37
  • Whether or not a constant $\mu$ can be regarded as a normal random variable with mean $\mu$ and _zero_ variance has been discussed extensively in the comments on a question that I cannot find, and the consensus was that it should be so regarded so as to avoid making special cases when writing definitions such as: "$X_1, X_2, \ldots, X_n$ are said to have a multivariate normal density if $\sum_i a_iX_i$ is a normal random variable for _all_ choices of $a_i \in \mathbb R$." – Dilip Sarwate Jul 21 '15 at 17:21
  • Very interesting, I had not considered that case. Still, the question was asked in regards to normally distributed RV's in general, and so would still not be satisfied by whuber's comment. – John Madden Jul 21 '15 at 17:26
  • John, declaring constants not to be normal would be similar to a convention adopted by some schoolteachers of insisting that squares are not rectangles: as a convention it will work fine provided everyone understands it, but--as @Dilip says--it leads to awfully contorted and redundant statements of geometric theorems. It is also analogous to refusing to work with real numbers that are not rational. Just as reals are important as limits of rationals, a constant RV is an important (and natural) limit of normal distributions, filling out the "ends" of this family. – whuber Jul 21 '15 at 18:21
  • Whuber, I see your point, and concede to it. – John Madden Jul 21 '15 at 18:31
  • I read @whuber's original comment as answering the question asked by the OP: "If $X$ and $Y$ are normal random variables, can $X/Y$ also be normal?" with a resounding "NO, except in the trivial case when $Y$ is a constant, which can be termed a _trivial_ normal random variable." Indeed, expanding on this notion, consider the following: If $X$ and $Y$ are _trivial_ normal random variables, then $X/Y$ is also a _trivial_ normal random variable. – Dilip Sarwate Jul 21 '15 at 21:09
  • @DilipSarwate Actually, there is nothing trivial about a Dirac $\delta(t\neq0)$ as a limiting value of a normal distribution. In fact, it is likely that $X/Y$ can be made as close to normal as desired for a standard deviation of $Y$ sufficiently small; $\sigma_Y\to0, \text{ and } ND\to\delta$ – Carl Jun 09 '18 at 01:52

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It is not normal, but it can be approximated with a normal distribution if the coefficient of variation of $Y$ is sufficiently small (<0.1). See

On the existence of a normal approximation to the distribution of the ratio of two independent normal random variables.

Trinity
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  • Also, see [Ratio distribution for noncentral normal distributions](https://en.wikipedia.org/wiki/Ratio_distribution#Uncorrelated_noncentral_normal_ratio) – Carl Sep 26 '21 at 19:52