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

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    $\begingroup$ Possible starting point: Wikipedia article. $\endgroup$ – COOLSerdash Mar 26 '15 at 18:00
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    $\begingroup$ 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. $\endgroup$ – Alexander Chervov Mar 26 '15 at 19:21
  • $\begingroup$ @AlexanderChervov: this is the only necessary argument, see my answer below. $\endgroup$ – Xi'an Mar 26 '15 at 19:37
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    $\begingroup$ A ratio of zero-mean Normals is a scaled Cauchy (Student $t(1)$) distribution, known to have no mean or variance.. $\endgroup$ – whuber Mar 26 '15 at 19:53
  • $\begingroup$ @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) $\endgroup$ – Aaron McDaid May 19 '16 at 13:33

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

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

  • $\begingroup$ 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. $\endgroup$ – whuber Mar 26 '15 at 19:57
  • $\begingroup$ 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. $\endgroup$ – whuber Mar 26 '15 at 19:58
  • $\begingroup$ 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? $\endgroup$ – Aaron McDaid May 19 '16 at 13:35
  • $\begingroup$ @AaronMcDaid: the median and the median absolute deviation are always defined, indeed. $\endgroup$ – Xi'an Jun 19 '18 at 10:56

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