How do I find the mean and variance of $Y_{(1)}$ in a random sample?

Question

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 $\bar{Y}$?

Answer 1

\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 $(Y_{(1)} + Y_{(n)} -1)/2.$

Answer 2

In order to compute the mean of $Y_{(1)} = \min_{i=1,\cdots,n}{Y_i}$, you first need to find the CDF/PDF of $Y_{(1)}$. In particular, $$\Pr(Y_{(1)} \ge y) = \Pr(Y_1 \ge y, Y_2 \ge y, \cdots, Y_n \ge y) = \left\{ \begin{matrix}1, & y\le \theta;\\(\theta+1-y)^n, & \theta < y \le \theta+1;\\0, & y > \theta +1.\end{matrix}\right.$$ Consequently, $$f_{Y_{(1)}}(y) = \left\{\begin{matrix}n(\theta+1-y)^{n-1}, & \theta < y \le \theta +1; \\0, & \text{otherwise.}\end{matrix}\right.$$ Your assertion immediately follows by noting that $$\int_{\theta}^{\theta+1} y f_{Y_{(1)}}(y) dy = \theta + \frac{1}{n+1}.$$