## 1.

Since $$\mathbb{E}_\lambda[\hat{\lambda}] = \mathbb{E}_\lambda[X_1]$$, the estimator is unbiased. Moreover, $$\operatorname{se}(\hat{\lambda})^2 = \mathbb{V}_\lambda(X_1) / n = \lambda / n$$. By the bias-variance decomposition, the MSE is equal to $$\operatorname{se}(\hat{\lambda})^2$$.

## 2.

If $$y$$ is between $$0$$ and $$\theta$$,

$\begin{equation} \mathbb{P}_\theta(\hat{\theta} \leq y) = \mathbb{P}_\theta(X_1 \leq y)^n = (y/\theta)^n. \end{equation}$

Differentiating yields the PDF of $$\hat{\theta}$$ between $$0$$ and $$\theta$$ as $$y \mapsto n(y/\theta)^n / y$$. Therefore,

$\begin{equation} \mathbb{E}_\theta[\hat{\theta}] = \int_0^\theta n(y/\theta)^n dy = \theta n / (n + 1). \end{equation}$

It follows that the bias of this estimator is $$-\theta/(n+1)$$ Moreover,

$\begin{equation} \operatorname{se}(\hat{\theta})^2 = \int_0^\theta ny(y/\theta)^n dy - \mathbb{E}_\theta[\hat{\theta}]^2 = \theta^2 n / (n+2) - \mathbb{E}_\theta[\hat{\theta}]^2. \end{equation}$

By the bias-variance decomposition, the MSE is $$\theta^2 n / (n+2) - \theta^2 (n^2 - 1) / (n+1)^2$$.

Remark. $$\hat{\theta} (n+1)/n$$ is an unbiased estimator.

## 3.

Since $$\mathbb{E}_\theta[\hat{\theta}] = 2 \mathbb{E}_\theta[X_1] = \theta$$, the estimator is unbiased. Moreover,

$\begin{equation} \operatorname{se}(\hat{\theta})^2 = 4 \mathbb{V}_\theta(X_1) / n = \theta^2 / (3n). \end{equation}$

By the bias-variance decomposition, the MSE is equal to $$\operatorname{se}(\hat{\theta})^2$$.