Expectation of the min of two independent random variables?

$F_{X,Y}(x,y)$ be the joint cumulative distribution function. Then, for $Z= \min(X,Y)$ $$ \begin{eqnarray} 1-F_Z(z) &=& \mathbb{P}\left(\min(X,Y) > z\right) = \mathbb{P}\left(X > z, Y>z\right) \\ &=& 1 - \mathbb{P}\left(X\leqslant z\right) - \mathbb{P}\left(Y\leqslant z\right) + \mathbb{P}\left(X\leqslant z, Y\leqslant z\right) \end{eqnarray} $$ where the inclusion exclusion principle was applied to get the last equality. Thus $$ F_Z(z) = \mathbb{P}\left(X\leqslant z\right) + \mathbb{P}\left(Y\leqslant z\right) - \mathbb{P}\left(X\leqslant z, Y\leqslant z\right) = F_X(z) + F_Y(z) - F_{X,Y}(z,z) $$ Notice that we have not used the information about the correlation of $X$ and $Y$.

Let's consider an example. Let $F_{X,Y}(x,y) = F_X(x) F_Y(y) \left(1+ \alpha (1-F_X(x)) (1-F_Y(y))\right)$, known as Farlie-Gumbel-Morgenstern copula, and let $F_X(x)$ and $F_Y(y)$ be cdfs of uniform random variables on the unit interval. Then, for $0<z<1$ $$ F_Z(z) = 2 z - z^2 \left(1 + \alpha (1-z)^2 \right) $$ leading to $$ \mathbb{E}\left(Z\right) = \int_0^1 z F_Z^\prime(z) \mathrm{d}z = \frac{1}{3} \left(1 + \frac{\alpha}{10} \right) $$

Let: $U = \min(X,Y)$, where $\min(X,Y)\leq z$.

$Pr(\min(X,Y) > z) = Pr((X>z) \cap (Y>z))$.

$Pr(U>z) = Pr(X>z)*Pr(Y>z)$.

$Pr(U\geq z) = (1 - Fx(z))*(1 - Fy(z))$.

$Fu(z) = 1 - (1 - Fx(z))*(1 - Fy(z))$.

Thus,

$F_\min(x,y) = Fx(z) + Fy(z) - Fx(z)*Fy(z)$.