# Uniform Distribution Estimator

Let $$X_1,..,X_n$$ be For a random sample $$X_1,…,X_n$$ from a Uniform$$[\theta,θ+1]$$distribution, with probability density function $$f(x;\theta) = \begin{cases} 1, & \theta \le x \le \theta+1 \\ 0, & \text{otherwise} \end{cases}$$ and $$X_{(n)}=max(X_1,X_2,…,X_n)$$ and $$X_{(1)}=min(X_1,...,X_n)$$. I want to show:

1)$$X_{(1)}$$ has the same distribution as $$\theta - X_{(n)}$$

2)$$\hat{\theta}=X_{(1)}+X_{(n)}$$ is an unbiased estimator of θ.

My work for 1) I know that $$X_{(1)}$$ has cdf $$F_{X_{(1)}}(x)=1-(1-F(x))^n=\begin{cases} 0, & x\le\theta \\ 1-(1-(x-\theta))^n, & \theta< x<\theta+1 \\ 1, &\theta+1 \le x \end{cases}$$.

• $X_{(1)}$ does not have the same distribution as $\theta-X_{(n)}$, though I think $X_{(1)}- \theta$ has the same distribution as $\theta+1-X_{(n)}$. – Henry May 4 at 23:35
• Similarly $X_{(1)}+X_{(n)}$ is not an unbiased estimator of $\theta$, though I think $\frac12(X_{(1)}+X_{(n)}-1)$ is an unbiased estimator of $\theta$. – Henry May 4 at 23:36
• Just consider for example $\theta=7$. Then $7 \le X_{(1)} \le X_{(n)}\le 8$ while $-1 \le \theta-X_{(n)} \le 0$ and $14 \le X_{(1)}+X_{(n)} \le 16$ – Henry May 4 at 23:38
• What you say is true for $U(0,\theta)$ but not $U(\theta,\theta+1)$. – StubbornAtom May 7 at 21:38

The numbers seem to be a little off. You can say, however, that $$X_{(1)}$$ is distributed like $$2\theta + 1 - X_{(n)}$$, while $$\frac{1}{2}(X_{(1)} + X_{(n)} - 1)$$ is an unbiased estimator of $$\theta$$ (as @Henry pointed out in the comments).

The first problem can be solved using a direct computation, but there's no need to expand any expressions or do any complicated algebra.

$$F_{X_{(1)}}(x) = 1 - (1 - F(x))^n.$$

as you noted. We can also compute,

$$F_{X_{(n)}}(x) = P(X_{(n)} < x) = P(X_k < x\text{ for } k=1,\dots,n) = (F(x))^n$$

Then,

$$F_{2\theta+1-X_{(n)}}(x) = P(2\theta + 1 - X_{(n)} < x) = P(X_{(n)} > 2\theta + 1 -x) = 1 - (F(2\theta+1-x))^n$$

However, we know that $$F(x) = x - \theta$$ for $$\theta \leq x \leq \theta+1$$, so

$$F(2\theta + 1 - x) = \theta + 1- x = 1 - (x - \theta) = 1 - F(x)$$

Conclude,

$$F_{2\theta+1-X_{(n)}}(x) = 1 - (F(2\theta+1-x))^n = 1 - (1 - F(x))^n = F_{X_{(1)}}(x).$$

For the second problem, just apply the first:

$$E[X_{(1)} + X_{(n)}] = E[2\theta + 1 - X_{(n)}] + E[X_{(n)}] = 2\theta + 1.$$