TITLE: Density function of $\max(X_1,\dots,X_n)$.
QUESTION [7 upvotes]: I'm making this statistics exercise and I'm not sure about my solution. 
Find the density function of $Y=\max(X_1,\dots,X_n)$ if they are all i.i.d.
This was my take on this question: $F_Y(a)=P(X_1 \leq a, \dots, X_n \leq a)$. Using that they are independent this gives $F_Y(a)=P(X_1 \leq a) \cdot P(X_2 \leq a) \dots P(X_n \leq a)= (P(X_1 \leq a))^n= (F_{X_1}(a))^n$. So the density function is $f_Y(x)= \frac{\partial F_Y(x)}{\partial x}= n\cdot f_{X_1}(x) \cdot F_{X_1}^{n-1}(x)$.
What do you think of this argument?
How would you calculate $E[Y]$?
Thanks!

REPLY [2 votes]: Your answer looks like a text-book answer.  To calculate $E[Y]=\int yf_Y(y) \mathrm{d}y$ you just plug in your expression for $f(y)$, i.e.
$$
E[Y]=n\int y f_{X_1}(y)F_{X_1}^{n-1}(y)\mathrm{d}y.
$$
If for example $X_i\sim U(0,1)$ then $f_{X_1}(x)=1$ if $0\leq x\leq 1$ and $F_{X_1}(x)=x$, so
$$
E[Y] = n \int_0^1 y^n \mathrm{d}y = \frac{n}{n+1}.
$$