Q:

Show that $\left(X_{(1)} + X_{(n)}\right)/2$ is a consistent estimator for $\theta$

Let $X_1, \ldots , X_n$ be a random sample from the uniform distribution on the interval
$(\theta − 1/2, \theta + 1/2)$, where $\theta$ is unknown. Let $X_{(1)} = \min(X_1, \ldots , X_n)$ $X_{(n)} = \max(X_1, \ldots , X_n).$
Show that $\left(X_{(1)} + X_{(n)}\right)/2$ is a consistent estimator for $\theta$.
Not really sure where to start with this. I tried finding the MLE and saying that is was a consistent estimator but found that the fisher information is 0. I also tried using the MME but that got me no where.

A:

First, find the variance $\sigma_n^2$ of this estimator $\widehat\theta_n$. (The variances of the maximum and minimum separately are not enough, since they are not independent.) If you can show that $\sigma_n^2\to0$ as $n\to\infty$, then by Chebyshev's inequality you can say for every $\varepsilon>0$,
$$
\Pr( |\widehat\theta_n - \theta| > \varepsilon) = \Pr\left( \left| \frac{\widehat\theta_n - \theta}{\sigma_n} \right| > \frac\varepsilon{\sigma_n} \right) \le \frac{\sigma_n^2}{\varepsilon^2} \to 0 \text{ as } n\to\infty.
$$
And there you have convergence in probability of $\widehat\theta_n$ to $\theta$.
Postscript:
\begin{align}
F_{X_{(1)},X{(n)}}(u,v) & = \Pr( X_{(1)} \le u\ \&\ X_{(n)} \le v) \\[10pt]
& = \Pr(X_{(n)} \le v)  - \Pr(u>X_{(1)}\ \&\ X_{(n)} \le v) \\[10pt]
& = \Pr(\text{All of } X_1,\ldots,X_n \text{ are} \le v) - \Pr(\text{All are between } u \text{ and } v.) \\[10pt]
& = (v-(\theta - \tfrac 1 2))^n - (v-u)^n).
\end{align}
From this you can find the density
$$
f_{X_{(1)},X{(n)}}(u,v) = \frac{\partial^2}{\partial u\,\partial v} F_{X_{(1)},X{(n)}}(u,v)
$$
and then you can use that to find $\sigma_n^2$.
Note that if $\theta - \dfrac 1 2 =0$ then the variance is the same as if $\theta-\dfrac 1 2$ is anything else, so for simplicity you may as well put $0$ there.
PPS:
$$
\operatorname{var}(X_{(n)} + X_{(1)}) = \operatorname{E}\Big((X_{(n)} + X_{(1)})^2 \Big) - \Big( \operatorname{E}(X_{(n)} + X_{(1)}) \Big)^2 = \operatorname{E}\Big((X_{(n)} + X_{(1)})^2 \Big) - \theta^2.
$$
$$
\operatorname{E}\Big((X_{(n)} + X_{(1)})^2 \Big) = \iint\limits_{\left[ \theta - \frac 1 2,\, \theta + \frac 1 2 \right]^2} (u+v)^2 f_{X_{(1)},\,X_{(n)}} (u,v)\, d(u,v).
$$

