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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 $E[Y]$?


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Looks solid to me. As far as I can remember, we did exactly this in lectures on extreme value theory. You can calculate $E(Y)$ using this density if its not to complicated to integrate. – Dima McGreen Aug 15 '13 at 20:59
up vote 2 down vote accepted

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 $$ E[Y] = n \int_0^1 y^n \mathrm{d}y = \frac{n}{n+1}. $$

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