如何证明δ=(minX_i+maxX_i)/2是θ的极小极大估计？

Step 1: Show $\delta$ has constant risk #

Let $X_i' = X_i - \theta$, so $X_i' \sim U(-1/2, 1/2)$ independently. Let $Y' = \min X_i'$, $Z' = \max X_i'$. Then:
$$\delta - \theta = \frac{(\theta + Y') + (\theta + Z')}{2} - \theta = \frac{Y' + Z'}{2}$$
The risk of $\delta$ is:
$$R(\theta, \delta) = E\left[ |\theta - \delta| \right] = E\left[ \left| \frac{Y' + Z'}{2} \right| \right]$$
Since the distribution of $Y'$ and $Z'$ does not depend on $\theta$, this risk is constant for all $\theta$. Denote this constant risk as $R_0$.

Step 2: Bayes Estimator under Prior $\Pi_\alpha = U(-\alpha, \alpha)$ #

The likelihood function is non-zero iff $\theta \in [\max X_i - 1/2, \min X_i + 1/2]$. Combining with the prior $\Pi_\alpha$, the posterior distribution of $\theta$ is uniform over the interval:
$$[\max(\max X_i - 1/2, -\alpha), \min(\min X_i + 1/2, \alpha)]$$
For absolute loss, the Bayes estimator is the median of the posterior distribution, which for a uniform interval $[c, d]$ is $\frac{c + d}{2}$. Thus the Bayes estimator $\delta_\alpha$ is:
$$\delta_\alpha = \frac{1}{2}\left( \max(\max X_i - 1/2, -\alpha) + \min(\min X_i + 1/2, \alpha) \right)$$

Step 3: Bayes Risk Converges to $R_0$ as $\alpha \to \infty$ #

As $\alpha \to \infty$, for any fixed $\theta$, the terms $\max(\max X_i -1/2, -\alpha)$ and $\min(\min X_i +1/2, \alpha)$ converge to $\max X_i -1/2$ and $\min X_i +1/2$ respectively. Thus:
$$\delta_\alpha \to \frac{1}{2}\left( (\max X_i -1/2) + (\min X_i +1/2) \right) = \delta$$
pointwise for all $X_1,...,X_n$.

Now, consider the Bayes risk $r(\Pi_\alpha, \delta_\alpha) = \frac{1}{2\alpha} \int_{-\alpha}^\alpha R(\theta, \delta_\alpha) d\theta$. We split the integral into three parts:

1. $\theta \in [-\alpha, -\alpha + 1]$: $R(\theta, \delta_\alpha) \leq 1$ (since $\delta_\alpha$ is bounded within $[-\alpha, \alpha]$ and $\theta$ is near $-\alpha$, the absolute difference is at most 1). The integral over this interval is $\leq 1$.
2. $\theta \in [\alpha -1, \alpha]$: Similarly, the integral is $\leq1$.
3. $\theta \in [-\alpha +1, \alpha -1]$: For large enough $\alpha$, $\delta_\alpha = \delta$, so $R(\theta, \delta_\alpha) = R_0$. The integral over this interval is $(2\alpha - 2)R_0$.

Combining these:
$$r(\Pi_\alpha, \delta_\alpha) = \frac{1}{2\alpha}\left( \text{integral from } -\alpha \text{ to } -\alpha+1 + \text{integral from } -\alpha+1 \text{ to } \alpha-1 + \text{integral from } \alpha-1 \text{ to } \alpha \right)$$
$$\leq \frac{1}{2\alpha}(1 + (2\alpha-2)R_0 +1) = \frac{2 + (2\alpha-2)R_0}{2\alpha} = R_0 + \frac{1 - R_0}{\alpha}$$
As $\alpha \to \infty$, this converges to $R_0$. Additionally, since $\delta_\alpha$ is the Bayes estimator, $r(\Pi_\alpha, \delta_\alpha) \leq r(\Pi_\alpha, \delta) = R_0$. Thus $\lim_{\alpha \to \infty} r(\Pi_\alpha, \delta_\alpha) = R_0$.

Step 4: Conclude Minimaxity #

Suppose there exists an estimator $\delta'$ such that $\sup_\theta R(\theta, \delta') < R_0$. Then for every $\alpha$, the Bayes risk $r(\Pi_\alpha, \delta') \leq \sup_\theta R(\theta, \delta') < R_0$. But since $\delta_\alpha$ is the Bayes estimator, $r(\Pi_\alpha, \delta_\alpha) \leq r(\Pi_\alpha, \delta') < R_0$. Taking the limit as $\alpha \to \infty$ gives $R_0 \leq \sup_\theta R(\theta, \delta') < R_0$, which is a contradiction.

Therefore, no such estimator $\delta'$ exists, so $\delta$ is the minimax estimator for $\theta$ under absolute loss.

内容的提问来源于stack exchange，提问作者Heydude