If X1...Xn are observed completion times of an experiment with value is [0,1]. Each of these random variables is uniformly distributed on [0,1]. If Y is the maximum observed completion time, then the mean of Y is?
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2$\begingroup$ en.wikipedia.org/wiki/… $\endgroup$– Steve HuntsmanJun 14, 2011 at 12:23
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This amounts to finding the integral over the unit cube of $\max(x_1,x_2,...,x_n)$. Partition the unit cube into $n!$ equal parts according to how the magnitudes of the variables are ordered. Let us consider the part where $x_1>x_2>...>x_n$. For this part, we end up with the integral $$\int_0^1 x_1\int_0^{x_1}\int_0^{x_2}...\int_0^{x_{n-1}}\,dx_n\,dx_{n-1}...dx_2\,dx_1.$$ This integral is equal to $$\int_0^1 \frac{x_1^n}{(n-1)!}\,dx_1=\frac{1}{(n+1)(n-1)!}.$$ Multiply by $n!$ and you get $n/(n+1)$.
answered Jun 14, 2011 at 11:04
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$\begingroup$ Simpler: $P(Y\le y)=y^n$ by independence of the $X_i$s and $E(Y)$ is the (one-dimensional) integral of $P(Y\ge y)=1-y^n$ from $y=0$ to $y=1$. This is $1-1/(n+1)$. $\endgroup$– DidJun 14, 2011 at 17:48