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For your first question, your derivative is incorrect. It instead is $\frac{\partial C^2}{\partial x \partial y} = 1+\theta(1-2x-2y+4xy)$. Note also that $x+y-2xy \geq x^2 + y^2 -2xy = (x-y)^2 \geq 0$. That is, $1-2x-2y+4xy \leq 1$. On the other hand, $1-2x-2y+4xy = 2(1-x)(1-y)+2xy - 1 \geq -1$. Then, $\frac{\partial C^2}{\partial x \partial y} \geq 0$, for ...


Note that $$P(X_i >s)= \exp\Big(-\int_0^s \lambda_i(u) du \Big),$$ for $i=1, 2$. Then, $$P(\min(X_1, X_2) >s) = P((X_1>s)\cap (X_2>s)) = P(X_1>s)P(X_2>s) = \exp\Big(-\int_0^s (\lambda_1(u)+\lambda_2(u)) du \Big).$$ That is, the hazard function for $\min(X_1, X_2)$ is $\lambda_1(s)+\lambda_2(s)$. Alternatively, note that $$\lambda_i(s) = ...

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