What is the probability that the magnitude of one normal random variable is greater than the magnitude of another normal random variable?

Question

Using notation, I am trying to determine:

Given $X \sim N(\mu_x, \sigma_x^2)$ and $Y \sim N(\mu_y, \sigma_y^2)$, what is $P(|X| > |Y|)$?

Without the absolute values, I know that I could use the variable

$Z = X - Y \sim N(\mu_z = \mu_x - \mu_y, \sigma_z^2 = \sigma_x^2 + \sigma_y^2 - 2\rho\sigma_x\sigma_y)$

and then $P(X > Y) = P(Z > 0) = \Phi(\frac{\mu_z}{\sigma_z})$

However, I don't expect $Z' = |X| - |Y|$ to have a nice distribution, in which case this trick wouldn't work anymore. Does anyone know some other way to approach this problem?

Thanks in advance. :)

Answer 1

Assuming that $X$ and $Y$ are independent (that is not granted by knowing the distributions only), the wanted probability is given by the integral over the region $|x|>|y|$ of the joint distribution function $$ f(x,y) = \frac{1}{2\pi \sigma_x \sigma_y}\exp\left(-\frac{(x-\mu_x)^2}{2\sigma_x^2}-\frac{(y-\mu_y)^2}{2\sigma_y^2}\right) $$ By switching to polar coordinates, that becomes

$$ \frac{1}{2\pi\sigma_x \sigma_y}\left(\int_{-\pi/4}^{\pi/4}+\int_{3\pi/4}^{5\pi/4}\right)\int_{0}^{+\infty}\rho \exp\left(-\frac{(\rho\cos\theta-\mu_x)^2}{2\sigma_x^2}-\frac{(\rho\sin\theta-\mu_y)^2}{2\sigma_y^2}\right)\,d\rho\,d\theta $$ that can be computed in terms of the error function.
The normal ratio distribution can be quite complicated if $X$ and $Y$ do not have mean zero.

Answer 2

Let $A_1 =\{X<0;Y<0\}$, $A_2 =\{X<0;Y\geq 0\}$, $A_3 =\{X\geq0;Y<0\}$, $A_4 =\{X\geq0;Y\geq 0\}$ and $C=\{|X|>|Y|\}$. Then

$$ P(C) = P(C|A_1) P(A_1) + P(C|A_2) P(A_2) + P(C|A_3) P(A_3) + P(C|A_4) P(A_4).$$

Note that $P(C|A_2)=0$ and that $P(C|A_3) = 1$.

If $X$ and $Y$ are independent all the other probabilities can be computed using the cumulative distribution function of the normal distribution law.

If $X$ and $Y$ are correlated, it does not seem possible to get an explicit solution to your problem.