# Distribution of percent difference of two normal variables

I have two performance measures in a given experiment that I know to be approximately normally distributed:

$X_1\sim \mathcal{N}\left(\mu_1,\sigma_1^2\right)$

$X_2\sim \mathcal{N}\left(\mu_2,\sigma_2^2\right)$

I am interested in the distribution of the percent difference between these two variables. If I define:

$Y = X_1 - X_2$

$W = \sqrt{X_1^2}$

Then I can define the percent difference I am interested in by either of two forms:

$D_1 = \frac{Y}{X_1}$

or, if $X_1$ is likely to have negative values:

$D_2 = \frac{Y}{W}$

I know that the ratio of two independent normal variables with zero mean is distributed as a Cauchy variable, but in my case the two variables are neither independent nor have zero mean. I also know that the distribution of $W$ will be a folded normal distribution.

Does anyone out there knows the distribution of either form? I would be very thankful for a few pointers or ideas.

Cheers,

• Yes, $X_1$ can have negative values, so you're not interested in $D_1$. Considering that $$D_2 = \frac{Y}{W} = \frac{X_1-X_2}{|X_1|} = \operatorname{sign}(X_1)\left(1 - \frac{X_2}{X_1}\right),$$ and this has the same distribution as $U\left(1 - \frac{X_2}{X_1}\right)$ for an independent Rademacher variable $U$, your question is effectively reduced to finding the ratio of two Normal variables. – whuber Sep 21 '16 at 15:44
• Thanks, @whuber. However, $U$ being a Rademacher variable seems to be contingent on $X_1$ having zero mean (the only case in which the probability of negative and positive values would be equal to 0.5). For an $X_1$ with arbitrary mean $\mu_1$ I believe $sign(X_1)$ would be better modeled as $1 - 2Bern(p)$, with $Bern(p)$ a Bernoulli variable with unknown coefficient $p$ given as $P(X_1 < 0)$. I'm starting to think about bootstrapping my way out of this particular problem. ;-) Any other ideas? Cheers, – Felipe Campelo Sep 26 '16 at 13:14