Let $X$ be a uniformly distributed random variable on $[0,\lambda]$. Let $\{x_i\}$ be a random sample of size $n$ from the population.
Is there any way to compute $E[max\{x_1,...,x_n\}]$ with this information?
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1$\begingroup$ Hint: Compute the cdf for $max\{X_1,...,X_n\}$ using the fact that $max\{X_1,...,X_n\} \leq \alpha \iff X_1 \leq \alpha$ and $...$ and $X_n \leq \alpha$. $\endgroup$Loafy Loafer– Loafy Loafer2019-11-01 09:28:46 +00:00Commented Nov 1, 2019 at 9:28
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$\begingroup$ Thanks. Computed the cdf and pdf. Then, how do I compute the integral for the expected value? $\endgroup$econ86– econ862019-11-01 09:39:25 +00:00Commented Nov 1, 2019 at 9:39
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$\begingroup$ Have you tried to compute the integral? $\endgroup$Loafy Loafer– Loafy Loafer2019-11-01 09:42:55 +00:00Commented Nov 1, 2019 at 9:42
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$\begingroup$ Got it. Thank you! $\endgroup$econ86– econ862019-11-01 10:10:52 +00:00Commented Nov 1, 2019 at 10:10
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1 Answer
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Let $M:=\max\{X_1,\ldots,X_n\}$. Then for $x\in(0,\lambda)$, $$ \mathsf{P}(M\le x)=(\mathsf{P}(X\le x))^n=(x/\lambda)^n, $$ and $$ \mathsf{E}M=\int_0^{\infty}\mathsf{P}(M> x)\,dx=\int_0^{\lambda}1-\left(\frac{x}{\lambda}\right)^n\,dx=\lambda\times \frac{ n}{n+1}. $$
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$\begingroup$ Thank you. Where does the expression for $ E(M)$ come from (the first equality)? $\endgroup$econ86– econ862019-11-01 10:10:35 +00:00Commented Nov 1, 2019 at 10:10
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1$\begingroup$ math.stackexchange.com/questions/63756/tail-sum-for-expectation $\endgroup$user140541– user1405412019-11-01 10:11:07 +00:00Commented Nov 1, 2019 at 10:11