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

Question

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.

Answer 1

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: $$ \operatorname{var}(X_{(n)} + X_{(1)}) = \operatorname{E}\Big((X_{(n)} + X_{(1)})^2 \Big) - \Big( \operatorname{E}(X_{(n)} + X_{(1)}) \Big)^2 = \operatorname{E}\Big((X_{(n)} + X_{(1)})^2 \Big) - \theta^2. $$ $$ \operatorname{E}\Big((X_{(n)} + X_{(1)})^2 \Big) = \iint\limits_{\left[ \theta - \frac 1 2,\, \theta + \frac 1 2 \right]^2} (u+v)^2 f_{X_{(1)},\,X_{(n)}} (u,v)\, d(u,v). $$

Answer 2

As an alternative of computing the variance (more direct, but still not very easy), we can do this by definition. (Not the common way, but it can be instructive).

We need to show that the estimator converges in probability to the true value. For the sake of notation simplicity, let's take $\theta =0$ (you should replace for $\theta$ for the general case).

Then we need to show that for any fixed $\epsilon>0$, $$\lim _{n \to \infty}P(|\frac{X_{(1)}+X_{(n)}}{2}|>\epsilon)\to 0 \tag{1}$$

Intutively, we guess that, as $n$ grows, $ X_{(1)}$ will concentrate near $-\frac12$ and $ X{(n)}$ near $\frac12$

Formally: let $A_{\delta,n}$ ($B_{\delta,n}$) be the event that the distance from $ X_{(1)}$ to $-\frac12$ (resp from $ X_{(n)}$ to $\frac12$) is greater than $ \delta$ . Then, for $1 > \delta>0$ , $$P(A_{\delta,n})=P(B_{\delta,n})=(1-\delta)^n \to 0 \tag{2}$$

To show that this implies $(1)$:

$$\left|\frac{X_{(1)}+X_{(n)}}{2}\right|=\left|\frac{X_{(1)}+\frac12+X_{(n)}-\frac12}{2}\right|\le \frac12\left|X_{(1)}+\frac12\right| + \frac12 \left|X_{(n)}-\frac12\right| \tag{3} $$

Then $$P(|\frac{X_{(1)}+X_{(n)}}{2}|>\epsilon)\le P(A_{\epsilon,n} \cap B_{\epsilon,n})\le (1-\epsilon)^n \to 0 \tag{4}$$