Assume there are $N$ iid random variables $x_1,x_2,\ldots,x_N$. I'm wondering that assumptions do we need to guarantee that $$ \mathbb{E}[\max(x_n)](N) $$ is concave in $N$?
The expression I'm looking at is $$ \mathbb{E}[\max(x_n)](N)=\int_x NF(x)^{N-1}f(x)\cdot x \,dx $$ Seems that to make the function concave in $N$, we need the condition imposed on distribution $x\sim F(x)$.
Thanks in advance! Of course, it would be even better to have necessary and sufficient condition.
Many thanks to @Did in his answer. Here is how the result is proved.
Start from $$ a(N)\equiv\mathbb{E}[\max \{x_n\}](N)=\int_x x dF(x)^N = \int_x x d(-(1-F(x)^N)) $$ Integration by part gives $$ \mathbb{E}[\max \{x_n\}](N)=-(1-F(x)^N))|_0^\infty+\int_x (1-F(x)^N)dx=\int_x (1-F(x)^N)dx $$
The concavity can be observed directly. And seems that the expression gives us an more general result which is $$ \frac{\partial^m a(N)}{\partial N^m} = -\int_x F(x)^N (\ln F(x))^m dx $$ So we has determined convexity or concavity of all orders of derivative.
Am I right? Let me know if I'm wrong. Thanks!
A useful formula for $$u_n=E(\max\{X_1,X_2,\ldots,X_n\})$$ for nonnegative i.i.d. random variables $(X_k)$ with CDF $F$, is $$u_n=\int_0^\infty P(\max\{X_1,X_2,\ldots,X_n\}>x)\,dx=\int_0^\infty(1-F(x)^n)\,dx$$ For every $x$, the sequence $(F(x)^n)_n$ is convex, hence the sequence $(u_n)_n$ is indeed concave.