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I'm given the following exercise; Let $X_1$,$X_2$,...,$X_n$ a sample from Uniform distribution $U(\theta,2\theta)$, $\theta>0$, with PDF $$f(x)=\frac{1}{\theta}, \theta<x<2\theta$$ and $$T=\frac{n+1}{2n+1}X_{(n)}$$where $X_{(n)}=max{X_i}, i=1,2,....,n$. I 'm instructed to show that $T$ is an unbiased estimator for $\theta$, and determine its variance ($Var(T)$).

The only step I 'm sure of, is that I need to show $E[T]=\theta$, the definiton of an unbiased estimator. From there, I am not sure how to continue. Normally I would have to simplify $E[T]$, which might go like this; $$E[T]=E[\frac{n+1}{2n+1}X_{(n)}]=\frac{n+1}{2n+1}E[X_{(n)}]$$... and that's it. Even if I take $E[X_{(n)}]=E[X]=\frac{3\theta}{2}$ (where the first equality is probably not correct / the second equality derives from the fact that $E[X]=\frac{a+b}{2}$ for the uniform distribution in a domain $(a,b)$), I don't see the path that would lead me to $E[T]=\theta$.

Any help would be really appreciated, I 'm just getting started with statistics on Uni and only at the beginning of getting the hang of it.

John0207
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    To find $E[X_{(n)}]$ you can start with finding the distribution (particularly the CDF) of the $X_{(n)}=\max_{1 \leq i \leq n}X_i$. It's a useful exercise. – perepelart Jul 25 '23 at 19:53
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    Your third displayed equation is correct. Next, try to estimate $E(X_{(n)})$ by considering for $t\in [\theta, 2\theta]$ the cumulative distribution function $F(t)=P(X_{(n)}\leq t)=(\frac{t-\theta}{\theta})^n$. Then, the density function of $X_{(n)}$ is $f=F^\prime(t)=\frac{n}{\theta}(\frac{t-\theta}{\theta})^{n-1}$. Finally, $E(X_{(n)})=\int_{\theta}^{2\theta} t f(t) ; dt= \frac{n}{\theta}\int_{\theta}^{2\theta} t (\frac{t-\theta}{\theta})^{n-1}; dt$. Evaluate the integral by doing $y=t-\theta$ substitution. – C614 Jul 25 '23 at 20:01
  • This question has been answered in the comments (closing vote) – Snoop Jul 25 '23 at 20:02
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    @Snoop Is there any movement on meta for closing due to being answered in the comments? – Brian Moehring Jul 25 '23 at 20:10
  • @Snoop - you are always free to copy a comment into an answer yourself (giving credit to the comment if appropriate) – Henry Jul 26 '23 at 00:47

1 Answers1

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Please see Distribution of maximum of n uniformly distributed random variables

with $a=\theta$ and $b=2\theta$, $P_{X_{(n)}}(y) = \frac{n(y-\theta)^{n-1}}{\theta^n}$.

$E[X_{(n)}] = \int_\theta^{2\theta} {yP_{X_{(n)}}(y)dy}=\int_\theta^{2\theta} {y\frac{n(y-\theta)^{n-1}}{\theta^n}dy}=\int_0^{\theta} {(y+\theta)\frac{ny^{n-1}}{\theta^n}dy}=\frac{n}{\theta^n}\int_0^{\theta} {y^n dy}+\frac{n}{\theta^{n-1}} \int_0^{\theta} {y^{n-1}dy}=\frac{n\theta }{n+1}+\theta=\frac{2n+1}{n+1}\theta$.

$E(T)=\frac{n+1}{2n+1}E[X_{(n)}]=\frac{n+1}{2n+1}\times \frac{2n+1}{n+1}\theta = \theta$

Mehdi
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  • thanks for the analytical response, this' been really helpful and easy to grasp, put this way! Cheers! – John0207 Jul 26 '23 at 05:38