Probability that one folded normal is bigger than another? What is the probability that one folded normal distribution is bigger than another?
In other words, if $Z_1=\mathcal{N}(\mu_1,\sigma_1)$ and $Z_2=\mathcal{N}(\mu_2,\sigma_2)$, what is $\mathcal{P}(|Z_1|>|Z_2|)$?
 A: Looking at the solution provided in Probability of a point taken from a certain normal distribution will be greater than a point taken from another?
Consider $P(|Z_1| > |Z_2|)$ in 4 parts
:
$P(Z_1 > Z_2) = (1 - \Phi(\frac{-\mu_1+\mu_2}{\sigma_1+\sigma_2})) * P(Z_1 > 0) * P(Z_2 > 0)$
$P(-Z_1 > Z_2) = (1 - \Phi(\frac{\mu_1+\mu_2}{\sigma_1+\sigma_2})) * P(Z_1 < 0) * P(Z_2 > 0)$
$P(Z_1 > -Z_2) = (1 - \Phi(\frac{-\mu_1-\mu_2}{\sigma_1+\sigma_2}))  * P(Z_1 > 0) * P(Z_2 < 0)$
$P(-Z_1 > -Z_2) = (1 - \Phi(\frac{\mu_1-\mu_2}{\sigma_1+\sigma_2}))  * P(Z_1 < 0) * P(Z_2 < 0)$
My guess is that the absolute sign also changes the comparison and the compound distribution. If so, the sum of these probabilities gives the desired answer.
Comments are welcome.
R code to check the solution:
mu_1 = -1; sigma_1 = 2; mu_2 = 2; sigma_2 = 3
par(mfrow=c(4,1))
curve(dnorm(x, mu_1, sigma_1), 0, 11, col='red', ylab='density')
curve(dnorm(x, mu_2, sigma_2), 0, 11, add=T)
(p1 = (1 - pnorm((-mu_1+mu_2)/(sigma_1+sigma_2))) * (1-pnorm(0, mu_1, sigma_1)) * (1-pnorm(0, mu_2, sigma_2)))
curve(dnorm(-x, mu_1, sigma_1), 0, 11, col='red', ylab='density')
curve(dnorm(x, mu_2, sigma_2), 0, 11, add=T)
(p2 = (1 - pnorm((mu_1+mu_2)/(sigma_1+sigma_2))) * (pnorm(0, mu_1, sigma_1)) * (1-pnorm(0, mu_2, sigma_2)))
curve(dnorm(x, mu_1, sigma_1), 0, 11, col='red', ylab='density')
curve(dnorm(-x, mu_2, sigma_2), 0, 11, add=T)
(p3 = (1 - pnorm((-mu_1-mu_2)/(sigma_1+sigma_2))) * (1-pnorm(0, mu_1, sigma_1)) * (pnorm(0, mu_2, sigma_2)))
curve(dnorm(-x, mu_1, sigma_1), 0, 11, col='red', ylab='density')
curve(dnorm(-x, mu_2, sigma_2), 0, 11, add=T)
(p4 = (1 - pnorm((mu_1-mu_2)/(sigma_1+sigma_2))) * (pnorm(0, mu_1, sigma_1)) * (1-pnorm(0, mu_2, sigma_2)))
p1+p2+p3+p4

