Difference of Frechet variables Let
$$ X \sim Frechet(\alpha, s_1, m)\\
Y \sim Frechet(\alpha, s_2, m)
$$
I'm trying to compute $Prob(X > Y$). This is equivalent of computing $Prob(X - Y > 0)$. Unfortunately, this is where my insights end. Is there any cute trick using Frechet, or how else would I approach this?
 A: The statistical understanding of the parameters--$m$ is a location, $s$ is a scale, and $\alpha$ is a power transformation--tells us how to proceed.

Consider this generalization of the problem.  Let $F$ be any distribution function.  Let $\{t_\alpha\,|\, \alpha\in A\subset\mathbb{R}^p\}$ be a parameterized family of strictly monotonic transformation functions that "play nicely" with rescaling in the following sense: there is a function $g$ such that for any positive number $s$
$$ t_\alpha(s\,t_\alpha^{-1}(y)) = g(s, \alpha) y.$$
This looks pretty abstract, so to fix the idea let's consider a common example where $p=1$ and $t_{(\alpha)}$ is the negative power transformation $x \to x^{-\alpha}$, $A = \{(\alpha)\,|\,\alpha \gt 0\}$.  Then (dropping the distinction between the $1$-vector $(\alpha)$ and its component $\alpha$),
$$t_\alpha(s\,t_\alpha^{-1}(y)) = (s\,y^{-1/\alpha})^{-\alpha} = s^{-\alpha} y.\tag{1}$$
In this case we see
$$g(s,\alpha) = s^{-\alpha}.$$
Define a location-scale-shape family by means of parameters $\mu$, $\sigma$, and $\alpha$ via
$$F_{\mu, \sigma, \alpha}(x) = F\left(t_\alpha\left(\frac{x-\mu}{\sigma}\right)\right)$$
for $\mu\in\mathbb{R}$, $\sigma\gt 0$, and $\alpha\in A$.  This means that any variable $X$ with such a distribution is obtained from a variable with an $F$ distribution by means of a $t_\alpha$ transformation, a rescaling by $\sigma$, and a shift by $\mu$.
Suppose $X$ has the distribution $F_{\mu, \sigma_1,\alpha}$ and the independent variable $Y$ has the distribution $F_{\mu, \sigma_2,\alpha}$.  That is, they have the same shape and location but their scales might differ.  Specifically,
$$X = \sigma_1 t_\alpha^{-1}(U) + \mu, \quad Y =  \sigma_2 t_\alpha^{-1}(V) + \mu$$
for two independent variables $U, V$ distributed according to $F$.
Using this, the event $X - Y \gt 0$ may be rewritten as 
$$t_\alpha^{-1}(U) \gt \sigma\, t_\alpha^{-1}(V)$$
for $\sigma = \sigma_2/\sigma_1$.  The relationship $(1)$ simplifies this inequality to
$$ U \gt g(\sigma,\alpha) V.$$
(When $t_\alpha$ is decreasing, the $\gt$ changes to a $\lt$.  In that case we should swap $U$ and $V$--which does nothing, since $U$ and $V$ are identically distributed--and we must change $g(\sigma,\alpha)$ to $1/g(\sigma,\alpha)$ in what follows.) 
Because $U$ has an $F$ distribution, the chance of this relationship is
$$\Pr( U \gt g(\sigma,\alpha) V) = 1 - F(g(\sigma, \alpha)V).$$
Its expectation gives the answer:
$$\Pr(X - Y \gt 0) = \int_{\mathbb{R}} \left(1 - F(g(\sigma, \alpha)v)\right) dF(v).\tag{2}$$

The beauty of this solution is that it reduces the calculation to one involving only $F$.  For instance, the Frechet distribution family is obtained from a negative power transformation of an exponential variable.  Thus $$F(x) = 1 - \exp(-x);\quad dF(x) = \exp(-x)dx$$ (for $x\gt 0$ only) and (according to $(1)$)
$$t_\alpha(y) = y^{-\alpha}, \quad g(s,\alpha) = s^{-\alpha}.$$
Because this $t_\alpha$ is decreasing in $y$ for any $\alpha \gt 0$, we must invariably use $1/g(\sigma,\alpha) = \sigma^\alpha$ in the calculations.  The value of $(2)$ therefore is
$$\int_0^\infty \exp(-\sigma^\alpha v)\exp(-v)dv = \int_0^\infty \exp(-(\sigma^\alpha + 1) v)dv = \frac{1}{1 + \sigma^\alpha} = \frac{\sigma_1^\alpha}{\sigma_1^\alpha + \sigma_2^\alpha}.$$
The actual amount of calculation needed to obtain this result is remarkably little.
A: I rename the variables $X_1=X$ and $X_2 = Y$ so the question is: What is the probability that $X_1>X_2$ given that $X_j$ is Frechet$(\alpha,s_j,m)$? 
First note that this problem can be considered as the problem that
$$X_1 = \max\{X_1,X_2\},$$
this problem is well known in the theory of extreme values. First I note that
$$Pr(X_1>X_2) = Pr(X_1 - m > X_2 - m) = Pr(Z_1>Z_2),$$
defining $Z_j = X_j - m$ such that $Z_j$ is Frechet$(\alpha,s_j,0)$.
I then note that
$$ Pr(Z_1>Z_2) = Pr(Z_1 = \max\{Z_1,Z_2\}) = \int_{z_1} \int_{z_2} I[z_1 = \max\{z_1,z_2\}] f_{Z_2}(dz_2)f_{Z_1}(dz_1).$$
Because the Frechet with location $0$ has positive support on $(0,\infty)$ the first integral goes from $0$ to $\infty$. The same thing goes for the second integral however the indicator is 0 unless $z_2\leq z_1$ so the integral runs from $0$ to $z_1$. Making the limits explicit I therefore have
$$= \int_{0}^{\infty} \int_{0}^{z_1}  f_{Z_2}(dz_2)f_{Z_1}(dz_1)$$
and it follows that 
$$= \int_{0}^{\infty}  F_{Z_2}(z_1)f_{Z_1}(z_1)dz_1, \  \ \ (eq. 1)$$
into which the Frechet c.d.f. and p.d.f. can be inserted to get the solution. 
The Frechet c.d.f. is given as
$$F_{Z_j}(z) = Pr(Z_j\leq z) = \exp\left(  - \left(\frac{z}{s_j}\right)^{-\alpha}  \right),$$
which I choose to reparameterize in order to get
$$F_{Z_j}(z) = \exp\left(  -  \Phi_j z^{-\alpha}  \right),$$
having defined $\Phi_j :=s_j^\alpha$. By differentiation it follows that
$$f_{Z_j}(z) = \exp\left(  -  \Phi_j z^{-\alpha}  \right) \Phi_j \alpha z^{-\alpha -1} \ \ \ (eq. 2).$$
Insert these into (eq.1) above to get 
$$  Pr(Z_1 = \max\{Z_1,Z_2\}) = \int_{0}^{\infty}  \exp\left(  -  \Phi_2 z^{-\alpha}  \right) \times   \exp\left(  -  \Phi_1 z^{-\alpha}  \right) \Phi_1 \alpha z^{-\alpha -1} dz$$
paying attention to the fact that the $\Phi_j$ parameters come from respective distributions of $Z_1$ and $Z_2$. This integral is easily solved by defining $\Phi= \Phi_1 + \Phi_2$ and rewriting
$$  Pr(Z_1 = \max\{Z_1,Z_2\}) =\frac{\Phi_1}{\Phi} \int_{0}^{\infty}  \exp\left(  -  \Phi z^{-\alpha}  \right) \Phi \alpha z^{-\alpha -1} dz$$
where I move $\Phi_1$ outside the integral and then divide by $\Phi$ outside the integral and multiply with $\Phi$ inside the integral. It is now easy to see that the function under the integral is the p.d.f. as given in (eq. 2) and hence it integrates 1 implying that
$$ Pr(Z_1 = \max\{Z_1,Z_2\}) =\frac{\Phi_1}{\Phi},$$ 
which offcourse can be written using the parameterization $$\Phi_j :=s_j^\alpha$$ and $\Phi := \sum_j s_j^\alpha$. 
This result can be generalized to get
$$Pr(i \in \arg \max_j \{X_j\}) = Pr(X_i \geq \max_j \{X_j\}) = \frac{\Phi_i}{\sum_j \Phi_j}$$
assuming independence and $X_j \sim Frechet(\alpha,\Phi_j,0)$ for all $j$.
