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Suppose we have two random variables $X$ and $Y$ with means $\mu_x, \mu_y$ and variances $\sigma_{X}^{2}$ and $\sigma_{Y}^{2}$. How would we derive $\text{Var} \left(\frac{X}{Y} \right)$?

Edit. $X$ and $Y$ are normally distributed.

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do you have the distribution of $X$ and $Y$? – Seyhmus Güngören Oct 17 '12 at 17:50

As soon as the distribution of a random variable $Z$ has positive continuous density at zero, $\frac1Z$ is not integrable.

In the case at hand, $\frac1Y$ is not integrable. Since $X$ is independent of $Y$, $\frac{X}Y$ is not integrable either, a fortiori the variance of $\frac{X}Y$ does not exist, except in the degenerate case when $\sigma_Y^2=0\ne\mu_Y$.

To show the first assertion, consider $Z$ with density at least $\varepsilon\gt0$ on the interval $(-z,z)$. Then $$ \mathbb E\left(\frac1{|Z|}\right)\geqslant\int_{-z}^z\frac\varepsilon{|t|}\,\mathrm dt=+\infty. $$

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can one define a density for $X/Y$ on the extended real axis? Does it make sense? – Seyhmus Güngören Oct 17 '12 at 18:10
@SeyhmusGüngören There is nothing to further define, the density of X/Y does exist, on the non extended real axis (since X/Y is almost surely finite). – Did Oct 17 '12 at 20:50

To illustrate the problem let's look at a little R code. I'll define a routine that samples $n$ times from a normal distribution (calling the results $X$) and $n$ times from another normal distribution (calling the result $Y$) and then returns the variance of $Z=X/Y$.

f <- function(n) {
  X <- normr(n);     # the operator '<-' is assignment to a variable
  Y <- normr(n);
  Z <- X / Y;

Let's look at the output for a few different random samples:

> f(1e6)
[1] 14135397
> f(1e6)
[1] 706438.6
> f(1e6)
[1] 5685218
> f(1e6)
[1] 11334216
> f(1e6)
[1] 2090359

You can see that we're getting results from as low as 700,000 up to more than 14,000,000. The variance is completely dominated by large values of $Z$, corresponding to values of $Y$ near zero. This is what non-integrability looks like "in practice".

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This is what non-integrability looks like. $X/Y$ has a density. – Nate Eldredge Oct 28 '12 at 3:39
Thanks Nate; corrected. – Chris Taylor Oct 28 '12 at 8:16
This is the special case of the ratio of two independent standard normals, which has a Cauchy distribution – Henry yesterday

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