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Hello all,

Here is a question I am struggling to understand,

Let Y1, Y2, ....... Yn denote a random sample from the uniform 
distribution on the interval (0, Θ). 
Prove the unbiased  estimators for Θ are 

 Θ1 = 2Ybar (Sorry dont know how to make the symbol)

 and  Θ2 = (n+1)/n Y(n)     (Where Y(n) = max (Y1, Y2, .....Yn)

I understand how to prove Θ1 is unbiased

E(2*(Ybar)) = 2(E(Ybar)) = 2(Θ/2) = Θ

However I am not too sure what to do for Θ. How does Y(n) affect things ? I guessing it some how produces n/n+1.

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up vote 1 down vote accepted

When you "chop up" an interval into pieces, the lengths of the pieces are distributed identically. You can imagine each number $Y_i$ as being a cut of the number interval. Therefore the length of the last piece will on average be $$\frac{\theta}{n+1}$$ because $n$ cuts produce $n+1$ pieces and the interval has length $\theta$. The max number will be $$\theta-\text{length of last piece}=\theta-\frac{\theta}{n+1}=\theta\left(1-\frac{1}{n+1}\right)=\theta\left(\frac{n}{n+1}\right)$$ on average. Therefore the expected value of the statistic $$\frac{n+1}{n}Y(n)$$ will be $$E(\Theta_2)=E\left(\frac{n+1}{n}Y(n)\right)=\frac{n+1}{n}E(Y(n))=\frac{n+1}{n}\cdot \theta\left(\frac{n}{n+1}\right)=\theta$$ as required.

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Thanks for the explanation. Can you explain why we are interested in the last piece ? Also I know (1- 1/n+1)) = (n/n+1) but I do not intuitively see why, can you clear this up ? – user126497 Apr 8 '14 at 2:34
Because the number at the left end of the last piece will be the biggest number of all the Y's. It is the largest number (max) that we are interested in taking, so we want to know how large the biggest number will be on average. – user140943 Apr 8 '14 at 2:36

Those two unbiased estimators of $\theta$ are not the only ones.

The way to deal with the second estimator begins by finding the distribution of $\max\{Y_1,\ldots,Y_n\}$.

We have $$ \begin{align} \Pr(\max\{Y_1,\ldots,Y_n\} \le y) & = \Pr(Y_1\le y\ \&\ \cdots\ \&\ Y_n\le y) \\[8pt] & = \Pr(Y_1\le y)\cdots\Pr(Y_n\le y) \\[8pt] & = \Big(\Pr(Y_1\le y)\Big)^n = \left(\frac y \theta\right)^n\text{ if }0<y<\theta. \end{align} $$

The probability density function of $\max\{Y_1,\ldots,Y_n\}$ is therefore $$ f(y)=\frac{d}{dy}\left(\frac y\theta\right)^n = \frac{ny^{n-1}}{\theta^n} \text{ for }0<y<\theta. $$

Hence the expected value of $\max\{Y_1,\ldots,Y_n\}$ is $$ \int_0^\theta yf(y)\,dy = \int_0^\theta y\frac{ny^{n-1}}{\theta^n}\,dy = \frac{n}{n+1}\theta. $$

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