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I want to calculate $\newcommand{\var}{\mathrm{var}}\var(X/Y)$. I know that the solution is $$\var(X) + \var(Y) - 2 \var(X) \var(Y) \mathrm{corr}(X,Y) \>,$$ but, how do I derive it from "common" rules of variance calculations?

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What makes you say that is the solution? – cardinal May 22 '11 at 19:18
$\var(X-Y) = \var(X) + \var(Y) - 2 \var(X) \var(Y) \mathrm{corr}(X,Y)$ and not $\var(X/Y)$ – user17762 May 22 '11 at 19:21
Sanity check your formula for $Y=X$, then your formula should vanish which is not the case here. – Listing May 22 '11 at 19:22
Don't believe everything which you see in a video feed! – Fabian May 22 '11 at 19:34
Strictly $\var(X−Y)=\var(X) + \var(Y) - 2 \sqrt{\var(X) \var(Y)} \mathrm{corr}(X,Y)$. Otherwise there is a dimension problem. – Henry May 22 '11 at 20:42

As others have noted, the formula you provide is incorrect. For general distributions, there is no closed formula for the variance or a ratio. However, you can approximate it by using some Taylor series expansions. The results are presented (in strange notation) in the this pdf file.

You can work it out exactly yourself by constructing the Taylor series expansion of $f(X,Y)=X/Y$ about the expected $X$ and $Y$ (E$[X]$ and E$[Y]$). Then look at E$[f(X,Y)]$ and E$[\left(f(X,Y)-\right.$E$\left.[f(X,Y)]\right)^2]$ using those approximations.

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This is particularly useful if you are working on asymptotics. – Vokram Nov 23 '12 at 12:27
A useful form of the result you linked to, expressed in terms of expected values, is $$ \operatorname{Var}\frac CD\approx\frac{\langle CC\rangle\langle D\rangle\langle D\rangle-2\langle CD\rangle\langle C\rangle\langle D\rangle+\langle DD\rangle\langle C\rangle\langle C\rangle}{\langle D\rangle^4}\;. $$ – joriki Jan 19 '13 at 10:13

The ratio of two random variables does not in general have a well-defined variance, even when the numerator and denominator do. A simple example is the Cauchy distribution which is the ratio of two independent normal random variables. As Sivaram has pointed out in the comments, the formula you have given (with the correction noted by Henry) is for the variance of the difference of two random variables.

Henry's wiki link includes a formula for approximating the ratio distribution with a normal distribution. It seems like this can be used to create the sort of formula you are looking for.

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The Cauchy distribution the ratio of two independent standard normal random variables: the important part is that they have zero means. Otherwise you get the extremely complicated Gaussian ratio distribution – Henry May 22 '11 at 20:45
Technically only the standard Cauchy distribution is the ratio of two standard normal random variables. Not sure what is the best way to phrase this distinction in my answer. – Dan Brumleve May 22 '11 at 21:10
And to nitpick even more, the standard Cauchy distribution is the ratio of two independent zero-mean normal random variables with identical variance. They don't have to be standard normal random variables as long as they are zero-mean (cf. @Henry's comment) and have equal variance. Also, the standard Cauchy random variable is the ratio of any two random variables whose joint density has circular symmetry about the origin. Such random variables are dependent except when they are marginally normal. – Dilip Sarwate Apr 18 '12 at 19:47

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