How do I find the mean and variance of $Y_{(1)}$ in a random sample? I have that $Y_1, \dots, Y_n$ is a random sample from a uniform population on $(\theta, \theta+1)$. I need to show that $Y_{(1)}-\frac{1}{n+1}$ is an unbiased estimator of $\theta$. I know that would mean that its mean is should equal $\theta$ but I'm confused on finding the mean of $Y_{(1)}$ as opposed to $\bar{Y}$?

\begin{align}
F_{Y_{(1)}} (x)=\Pr(Y_{(1)} \le x) = 1-\Pr(Y_{(1)} > x) & = 1 - \Pr(Y_1>x\ \&\ \cdots\ \&\ Y_n>x) \\[10pt]
& = 1-\Pr(Y_1>x)\cdots\Pr(Y_n>x) \\[10pt]
& = 1-\Big(\Pr(Y_1>x)\Big)^n.
\end{align}
And then $f_{Y_{(1)}}(x) = \dfrac d {dx} F_{Y_{(1)}}(x).$ Once you have the density, you can evaluate an integral from $\theta$ to $\theta+1$ to find the expected value.
This is a case of a simple unbiased estimator that is clearly not the best unbiased estimator, since the best would be $(Y_{(1)} + Y_{(n)} -1)/2.$
