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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1Hint: 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 $X_n \leq \alpha$. – Loafy Loafer Nov 01 '19 at 09:28
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Thanks. Computed the cdf and pdf. Then, how do I compute the integral for the expected value? – econ86 Nov 01 '19 at 09:39
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Have you tried to compute the integral? – Loafy Loafer Nov 01 '19 at 09:42
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Got it. Thank you! – econ86 Nov 01 '19 at 10:10
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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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Thank you. Where does the expression for $ E(M)$ come from (the first equality)? – econ86 Nov 01 '19 at 10:10
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