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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.

PiHal
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1 Answers1

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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.

Mr. Gandalf Sauron
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  • Thank you very much! I have a question about whether we need to first prove that $\int_{-\infty}^{+\infty} y dF(y-a)F(y-b)$ is Riemann-integrable as it is an improper integral. And I am not sure how to prove it. – PiHal Jan 29 '22 at 15:05
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    Well it might not even be an improper integral. The function might just have a compact support( I mean it might be zero outside some interval of finite length). In that case you do not even need to prove anything. Everything is justified by continuity. Also I think you can safely assume $F$ to be $C^{2}$ , that is twice continuously differentiable. After all , loosely speaking , this is not real analysis. You have some leeway. – Mr. Gandalf Sauron Jan 29 '22 at 15:09
  • If $X_i$ follows normal distribution or Type I extreme value distribution, the function f(a, b) would be improper integral, isn't it? In these cases, do I need to worry about the integrability? – PiHal Jan 29 '22 at 15:19
  • The Gaussian function is $C^{\infty}$ which is even better and you can actually do integrate differentiate as many times as you want with the Gaussian. I don't think you need to justify the integrability. But that is upto you and how much rigour you want to have in your answer. – Mr. Gandalf Sauron Jan 29 '22 at 15:23