1. **Problem Statement:** Given $X_1, X_2, \ldots, X_n$ are i.i.d. Uniform$(0, \theta)$ random variables, and an estimator $\hat{\theta} = \frac{k}{n} \sum_{i=1}^n X_i$, determine which options about $\hat{\theta}$ are correct. 2. **Recall properties of Uniform$(0, \theta)$:** - $E[X_i] = \frac{\theta}{2}$ - $\mathrm{Var}(X_i) = \frac{\theta^2}{12}$ 3. **Calculate $E[\hat{\theta}]$:** $$ E[\hat{\theta}] = E\left[ \frac{k}{n} \sum_{i=1}^n X_i \right] = \frac{k}{n} \sum_{i=1}^n E[X_i] = \frac{k}{n} \cdot n \cdot \frac{\theta}{2} = \frac{k \theta}{2} $$ 4. **Check unbiasedness:** $\hat{\theta}$ is unbiased if $E[\hat{\theta}] = \theta$, so $$ \frac{k \theta}{2} = \theta \implies k = 2 $$ Thus, $\hat{\theta}$ is unbiased if and only if $k=2$. 5. **Calculate $\mathrm{Var}(\hat{\theta})$:** Since $X_i$ are independent, $$ \mathrm{Var}(\hat{\theta}) = \mathrm{Var}\left( \frac{k}{n} \sum_{i=1}^n X_i \right) = \left( \frac{k}{n} \right)^2 \sum_{i=1}^n \mathrm{Var}(X_i) = \frac{k^2}{n^2} \cdot n \cdot \frac{\theta^2}{12} = \frac{k^2 \theta^2}{12 n} $$ 6. **Evaluate risk for $k=2$:** Risk (mean squared error) for unbiased estimator equals variance, $$ \mathrm{Risk} = \mathrm{Var}(\hat{\theta}) = \frac{4 \theta^2}{12 n} = \frac{\theta^2}{3 n} $$ This is not equal to $\frac{\theta^2}{9}$ unless $n=3$. 7. **Check the options:** - $E[\hat{\theta}] = k \cdot \frac{n(n+1)}{2} E[X_1]$ is incorrect because $E[\hat{\theta}] = \frac{k \theta}{2}$, no $n(n+1)/2$ factor. - $\hat{\theta}$ is unbiased if and only if $k=2$ is correct. - $\mathrm{Var}(\hat{\theta}) = \frac{k^2}{n} \cdot \frac{\theta^2}{12}$ is correct. - $E[\hat{\theta}] = \theta$ for all $k$ is incorrect. - Risk $= \frac{\theta^2}{9}$ for $k=2$ is incorrect unless $n=3$, so generally false. **Final answers:** Only the second and third options are correct.