# Choice probabilities with Frechet distributions that have different minimums

For two independent Frechet distributed variables that have the same shape parameter but different scale parameters: $$F_i(x) = e^{-\psi_i x^{-\epsilon}}, i=1,2$$ The probability of one variable being larger than the other has a simple closed form solution: $$Pr(x_1 > x_2 ) = \frac{\psi_1}{\psi_1 + \psi_2}$$

I am trying to obtain a similar result for Frechet variables that have different minimum location parameters, with distribution functions

$$F_i(x) = e^{-\psi_i (x-m_i)^{-\epsilon}}, i=1,2$$

The probability, in this case, can be obtained by calculating

\begin{align} Pr(x_1 > x_2 ) &= F_2(m_1) + \int_{max(m_1,m_2)}^{\infty} F_2(x) f_1(x) dx \\ &= F_2(m_1) + \int_{max(m_1,m_2)}^{\infty} e^{-\psi_2 (x-m_2)^{-\epsilon}} e^{-\psi_1 (x-m_1)^{-\epsilon}} \epsilon \psi_1 (x-m_1)^{-\epsilon-1} dx \end{align}

But I haven't found a closed form solution for this integral. This paper suggests that to have closed form choice probabilities the distribution functions need to satisfy $F_1^a(x) = F_2^b(x)$ for some $a,b$ which the latter distributions don't satisfy, but I'm not sure if this is a necessary condition.