How to prove the differentiability of $(a,b) \mapsto \Bbb E [ \max(a+X_1, b+X_2) ]$? Let $X_1$ and $X_2$ be two IID random variables whose CDF $F$ is twice differentiable. Let $a$ and $b$ be two parameters. Is the function $$H(a,b) := \Bbb E [ \max(a+X_1, b+X_2) ]$$ differentiable in $(a,b)$? How do I prove it?

My incomplete answer:
Denote $Y:=\max(a+X_1, b+X_2)$. Then the c.d.f of random variable Y is $$G(y)=Pr\{Y\le y\}=Pr\{\max(a+X_1, b+X_2)\le y\}=Pr\{a+X_1\le y, b+X_2\le y\}=Pr\{X_1\le y-a\}Pr\{X_2\le y-b\}=F(y-a)F(y-b)$$
Therefore, $$H(a,b)= \Bbb E [ \max(a+X_1, b+X_2) ]=\int_{-\infty}^{+\infty} y dF(y-a)F(y-b)$$
This is what I have got so far. Starting from here, I don't know how to rigorously prove this function is differentiable. Or say, to prove $H(a,b)$ is differentiable when $X_i$ follows type-I extreme value distribution or Normal distribution if this question only makes sense when we impose specific distribution functions.
 A: You are given that the cdf $F$ is twice differentiable. Let $g$ denote the pdf of $Y$.
So $g(y)=\frac{dG(y,a,b)}{dy}$
Then $$f(a,b)=\mathbb{E}(Y)=\int_{-\infty}^{\infty}y\frac{dG(y,a,b)}{dy}\,dy=\int_{-\infty}^{\infty}y\left(F(y-a)\frac{dF(y-b)}{dy}+\frac{dF(y-a)}{dy}F(y-b)\right)\,dy$$.
Now the lets denote $G'(y,a,b)=\left(F(y-a)\frac{dF(y-b)}{dy}+\frac{dF(y-a)}{dy}F(y-b)\right)$
Then $\frac{\partial G'(y,a,b)}{\partial a}$ is continuous and uniformly bounded as $F$ is twice differentiable.
Similarly    $\frac{\partial G'(y,a,b)}{\partial b}$ is continuos and uniformly bounded.
So By dominated Convergence Theorem we can interchange the order of Expectation and Derivative .
And so $\frac{\partial f}{\partial a}$ and $\frac{\partial f}{\partial a}$ both exist and equals $\displaystyle \int_{-\infty}^{\infty}y\frac{\partial G'(y,a,b)}{\partial a}\,dy$ and $\displaystyle \int_{-\infty}^{\infty}y\frac{\partial G'(y,a,b)}{\partial b}\,dy$ . respectively. And they are both continuous.
So we have the differentiability of $f(a,b)$ as the partial derivatives are continuous.
