Proving Unbiased estimators 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.
 A: 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
$$
\int_0^\theta yf(y)\,dy = \int_0^\theta y\frac{ny^{n-1}}{\theta^n}\,dy = \frac{n}{n+1}\theta.
$$
A: 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.
