0

Say I have two independent random variables X ~ N($μ_{x}$,$σ_{x}$) and Y ~ N($μ_{y}$,$σ_{y}$). I know that I can calculate the variance of their difference with $Var(X-Y) = σ_{x} - σ_{y}$. But I want to find $Var(\frac{X-Y}{Y})$ for which I went only this far:

$Var(\frac{X-Y}{Y}) = Var(\frac{X}{Y})-Var(1) = Var(\frac{X}{Y})-0 =Var(\frac{X}{Y})$

So how can I calculate $Var(\frac{X-Y}{Y})$ or $Var(\frac{X}{Y})$, is there a straightforward formula to calculate one of them?

Thank you!

gülsemin
  • 65
  • 4
  • The variance is infinite, because it's the same as the variance of $(X-Y)/Y+1 = X/Y$ and that ratio always has an infinite variance. (One general demonstration is at https://stats.stackexchange.com/questions/299722, which shows the expectation is undefined, whence the variance must be infinite.) – whuber May 09 '21 at 18:02

1 Answers1

1

The variance $Var(\frac{X}{Y})$ I think it can be expanded as

$$Var(\frac{X}{Y}) = \mathbb{E}[\frac{X^{2}}{Y^{2}}]-[\mathbb{E}(\frac{X}{Y})]^{2} \\ = \frac{\mathbb{E}[X^{2}]}{\mathbb{E}[Y^{2}]}-\frac{\mathbb{E}[X]^{2}}{\mathbb{E}[Y]^{2}}$$

Also, in order to derive that result we have to consider the independence of $X$ with $Y$, otherwise we wouldn't be able to manipulate the expected value of the product $X\frac{1}{Y}$

If $X$ and $Y$ are independent then also $X$ is independent with the random variable $\frac{1}{Y}$. So, you can write that $\mathbb{E}[X\frac{1}{Y}]= \frac{\mathbb{E}[X]}{\mathbb{E}[Y]}$. In similar manner, if $X$ and $Y$ are independent random variables then also their squares are independed and then you can use the same argument that $X^{2}$ is independent of $\frac{1}{Y^{2}}$ and write their expectation $\mathbb{E}[X^{2}\frac{1}{Y^{2}}]=\frac{\mathbb{E}[X^{2}]}{\mathbb{E}[Y^{2}]}$.

Fiodor1234
  • 1,679
  • 6
  • 15