# What are the mean and variance of the ratio of two normal variables, with non-zero means?

If X,Y are normal independent N(a,s), N(b,s') what are means and variances of the ratio X/Y ?

• Possible starting point: Wikipedia article. – COOLSerdash Mar 26 '15 at 18:00
• It seems neither mean, nor variance exist, at least if X is not a normal, but constant, then non of them exist: math.stackexchange.com/questions/646428/… Hope some one confirms. – Alexander Chervov Mar 26 '15 at 19:21
• @AlexanderChervov: this is the only necessary argument, see my answer below. – Xi'an Mar 26 '15 at 19:37
• A ratio of zero-mean Normals is a scaled Cauchy (Student $t(1)$) distribution, known to have no mean or variance.. – whuber Mar 26 '15 at 19:53
• @whuber, if the means aren't zero, does the ratio have a name? Is it simply a Cauchy with scale and location? Or something more complex? (Update: seems to be answered on Wikipedia, as expected! I should have checked first) – Aaron McDaid May 19 '16 at 13:33

## 1 Answer

Since, as pointed by Alexander Chervov, the mean of $1/X$ does not exist when $X\sim\mathcal{N}(\mu,\sigma^2)$, the mean of $Y/X$, which, were it to exist, would be equal to the mean of $Y$ times the mean of $1/X$ does not exist either. Since the mean does not exist, the variance does not exist either.

To make the above more precise (in connection with whuber's criticism), the integral

$$\int_{\mathbb{R}^2} \frac{y}{x}\,\varphi(x-a;\sigma)\,\varphi(y-b;\tau)\,\text{d}x\text{d}y$$

is defined iff the integral

$$\int_{\mathbb{R}^2} \frac{|y|}{|x|}\,\varphi(x-a;\sigma)\,\varphi(y-b;\tau)\,\text{d}x\text{d}y$$

is finite, which is not the case since

$$\int_{\mathbb{R}} |y|\,\varphi(y-b;\tau)\,\text{d}y\,\int_{\mathbb{R}} \frac{1}{|x|}\,\varphi(x-a;\sigma)\,\text{d}x=+\infty\,.$$

• This argument seems incomplete, because it does not address what happens when the mean of $Y$ is zero. In the most extreme case $Y$ is an atom at $0$, making $Y/X$ defined and equal to $0$ a.e.--and obviously its mean does exist. – whuber Mar 26 '15 at 19:57
• What's the matter with the counterexample I proposed? I would agree that an atom at $0$ is not normal--but the point of the counterexample is that you must appeal to more than the expectations; you need to use additional properties of $Y$ to draw your conclusion validly. – whuber Mar 26 '15 at 19:58
• If the mean and variance are undefined, are there other quantities we can compute that might be useful? The median perhaps? And perhaps the median absolute deviation? – Aaron McDaid May 19 '16 at 13:35
• @AaronMcDaid: the median and the median absolute deviation are always defined, indeed. – Xi'an Jun 19 '18 at 10:56