Variance of Ratio of Two Random Variables

Question

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.

Answer 1

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 $$ \mathbb E\left(\frac1{|Z|}\right)\geqslant\int_{-z}^z\frac\varepsilon{|t|}\,\mathrm dt=+\infty. $$

Answer 2

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;
  var(Z)
}

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".