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I know the probability of event A is given by:

$$\Phi(f(x)+g(y)) - \Phi(f(x)-g(y)), $$

and the probability of event B is

$$\Phi(m(y)+n(x)) - \Phi(m(y)-n(x)).$$ where $\Phi$ is the cumulative distribution function of the standard normal.

I want to find the more likely of those two events as a function of $x$ and $y$. I tried a few different approaches, but I couldn't make it much simpler than this:

$$\Pr(A|x,y) > \Pr(B|x,y) \\ \Leftrightarrow \frac{1}{2 \pi} \int\limits_{ f(x) - g(y) }^{f(x) + g(y) } e^{-t^2/2} dt > \frac{1}{2 \pi} \int\limits_{ m(y) - n(x) }^{ m(y) + n(x) } e^{-t^2/2} dt $$

An obvious sufficient condition for $\Pr(A|x,y) > \Pr(B|x,y)$ is $f(x) - g(y) < m(y) - n(x)$ and $f(x) + g(y) > m(y) + n(x)$. But I can't figure out a tractable necessary condition. I would appreciate any hints!

Edit: I can assume that $f$ and $m$ are linear, and $g$ and $n$ quadratic, if that is of any use.

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