I'm struggling to find when $X_1,\ldots,X_n \sim Uniform(\theta,2\theta)$, how the expected value of the largest order statistic is $E[X_{(n)}]=\dfrac{2n+1}{n+1}\theta$. I can find that the density of the largest order statistic is $f(x_{(n)})=\dfrac{n}{\theta^n}(x_{(n)}-\theta)^{n-1}$ for $\theta \le x_{(n)} \le 2\theta$. But the expected value of $X_{(n)}$ doesn't seem to take such a nice form when I'm trying to do it.
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4$\begingroup$ Hint: If you define $Y = X_{(n)}/(2\theta)$, you should be able to recognize the density of $Y$ as a well-known distribution. $\endgroup$– knrumseyCommented Mar 2, 2022 at 19:38
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1$\begingroup$ You can find this problem asked and answered, in various guises, in quite a few threads here on CV. For instance, it's worked out at stats.stackexchange.com/questions/130075. stats.stackexchange.com/questions/465477 also answers your question. Here's the site search that turned these up. Please notice that upon recentering by $-\theta$ and scaling by $1/\theta$ your problem is reduced to the case of a uniform distribution on $(0,1),$ $\endgroup$– whuber ♦Commented Mar 2, 2022 at 19:42
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$\begingroup$ Thank you so much. I set $Y=X_{(n)}/\theta$ and it all simplified! $\endgroup$– Blain WaanCommented Mar 2, 2022 at 20:18
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