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A partial answer. First, $$ F_{U_i}(u_i) = P(U_i\leq u_i) = P(U_i-w_i\leq u_i-w_i) = P(\epsilon_i\leq u_i-w_i) = e^{-e^{-(u_i - w_i)}} \, . $$ Also, $$ F_Z(z) = P(Z\leq z) = P(\max \{U_0,U_1\} \leq z) = P(U_0\leq z, U_1\leq z) $$ $$ = P(U_0\leq z)P(U_1\leq z) = F_{U_0}(z)F_{U_1}(z) = e^{-\left(e^{-(z - w_0)}+e^{-(z - w_1)}\right)} \, . $$ Therefore, $$ \mathrm{E}[Z] = \int_0^\infty (1 - F_Z(z))dz - \int_{-\infty}^0 F_Z(z)\,dz $$ $$ = \int_0^\infty \left( 1 - e^{-\left(e^{-(z - w_0)}+e^{-(z - w_1)}\right)} \right) dz - \int_{-\infty}^0 e^{-\left(e^{-(z - w_0)}+e^{-(z - w_1)}\right)} dz \, . $$ May someone with access to Mathematica (Huber?) do these integrals?

A partial answer. First, $$ F_{U_i}(u_i) = P(U_i\leq u_i) = P(U_i-w_i\leq u_i-w_i) = P(\epsilon_i\leq u_i-w_i) = e^{-e^{-(u_i - w_i)}} \, . $$ Also, $$ F_Z(z) = P(Z\leq z) = P(\max \{U_0,U_1\} \leq z) = P(U_0\leq z, U_1\leq z) $$ $$ = P(U_0\leq z)P(U_1\leq z) = F_{U_0}(z)F_{U_1}(z) = e^{-\left(e^{-(z - w_0)}+e^{-(z - w_1)}\right)} \, . $$ Therefore, $$ \mathrm{E}[Z] = \int_0^\infty (1 - F_Z(z))dz - \int_{-\infty}^0 F_Z(z)\,dz $$ $$ = \int_0^\infty \left( 1 - e^{-\left(e^{-(z - w_0)}+e^{-(z - w_1)}\right)} \right) dz - \int_{-\infty}^0 e^{-\left(e^{-(z - w_0)}+e^{-(z - w_1)}\right)} dz \, . $$ These integrals can probably be expressed in terms of the Exponential Integral.

A partial answer. First, $$ F_{U_i}(u_i) = P(U_i\leq u_i) = P(U_i-w_i\leq u_i-w_i) = P(\epsilon_i\leq u_i-w_i) = e^{-e^{-(u_i - w_i)}} \, . $$ Also, $$ F_Z(z) = P(Z\leq z) = P(\max \{U_0,U_1\} \leq z) = P(U_0\leq z, U_1\leq z) $$ $$ = P(U_0\leq z)P(U_1\leq z) = F_{U_0}(z)F_{U_1}(z) = e^{-\left(e^{-(z - w_0)}+e^{-(z - w_1)}\right)} \, . $$ Therefore, $$ \mathrm{E}[Z] = \int_0^\infty (1 - F_Z(z))dz - \int_{-\infty}^0 F_Z(z)\,dz $$ $$ = \int_0^\infty \left( 1 - e^{-\left(e^{-(z - w_0)}+e^{-(z - w_1)}\right)} \right) dz - \int_{-\infty}^0 e^{-\left(e^{-(z - w_0)}+e^{-(z - w_1)}\right)} dz \, . $$

A partial answer. First, $$ F_{U_i}(u_i) = P(U_i\leq u_i) = P(U_i-w_i\leq u_i-w_i) = P(\epsilon_i\leq u_i-w_i) = e^{-e^{-(u_i - w_i)}} \, . $$ Also, $$ F_Z(z) = P(Z\leq z) = P(\max \{U_0,U_1\} \leq z) = P(U_0\leq z, U_1\leq z) $$ $$ = P(U_0\leq z)P(U_1\leq z) = F_{U_0}(z)F_{U_1}(z) = e^{-\left(e^{-(z - w_0)}+e^{-(z - w_1)}\right)} \, . $$ Therefore, $$ \mathrm{E}[Z] = \int_0^\infty (1 - F_Z(z))dz - \int_{-\infty}^0 F_Z(z)\,dz $$ $$ = \int_0^\infty \left( 1 - e^{-\left(e^{-(z - w_0)}+e^{-(z - w_1)}\right)} \right) dz - \int_{-\infty}^0 e^{-\left(e^{-(z - w_0)}+e^{-(z - w_1)}\right)} dz \, . $$ May someone with access to Mathematica (Huber?) do these integrals?

A possible path: $F_{U_i}(u_i)=e^{-e^{-(u_i + w_i)}} I_{(0,\infty)}(u_i)$. But $$ F_Z(z) = P(Z\leq z) = P(\max \{U_0,U_1\} \leq z) = P(U_0\leq z, U_1\leq z) $$ $$ = P(U_0\leq z)P(U_1\leq z) = F_{U_0}(z)F_{U_1}(z) $$ $$ = e^{-\left(e^{-(z + w_0)}+e^{-(z + w_1)}\right)} I_{(0,\infty)}(z) \, . $$ Since $Z>0$, we have $$ \mathrm{E}[Z] = \int_0^\infty (1 - F_Z(z))dz = \int_0^\infty \left( 1 - e^{-\left(e^{-(z + w_0)}+e^{-(z + w_1)}\right)} \right) dz \, . $$

A partial answer. First, $$ F_{U_i}(u_i) = P(U_i\leq u_i) = P(U_i-w_i\leq u_i-w_i) = P(\epsilon_i\leq u_i-w_i) = e^{-e^{-(u_i - w_i)}} \, . $$ Also, $$ F_Z(z) = P(Z\leq z) = P(\max \{U_0,U_1\} \leq z) = P(U_0\leq z, U_1\leq z) $$ $$ = P(U_0\leq z)P(U_1\leq z) = F_{U_0}(z)F_{U_1}(z) = e^{-\left(e^{-(z - w_0)}+e^{-(z - w_1)}\right)} \, . $$ Therefore, $$ \mathrm{E}[Z] = \int_0^\infty (1 - F_Z(z))dz - \int_{-\infty}^0 F_Z(z)\,dz $$ $$ = \int_0^\infty \left( 1 - e^{-\left(e^{-(z - w_0)}+e^{-(z - w_1)}\right)} \right) dz - \int_{-\infty}^0 e^{-\left(e^{-(z - w_0)}+e^{-(z - w_1)}\right)} dz \, . $$