1.

TODO (Computer Experiment)

2.

TODO (Computer Experiment)

3.

TODO (Computer Experiment)

4.

This is a stars and bars problem (or, equivalently, an “indistinguishable balls in distinct buckets” problem). For example, the configuration ★|★★★||★ corresponds to sampling \(X_1\) once, sampling \(X_2\) three times, sampling \(X_3\) zero times, and sampling \(X_4\) once. In general, there are \(n\) stars and \(n-1\) bars, and hence the total number of configurations is \((2n - 1)!/(n!(n-1)!)\).

5.

First, note that

\[\begin{equation} \mathbb{E}\left[\overline{X}_{n}^{*}\mid X_{1},\ldots,X_{n}\right] =\mathbb{E}\left[X_{1}^{*}\mid X_{1},\ldots,X_{n}\right]=\overline{X}_{n}. \end{equation}\]

Therefore, by the tower property, \(\mathbb{E}[\overline{X}_{n}^{*}]=\mathbb{E}[X_{1}]\). Next, note that

\[\begin{equation} \mathbb{V}(\overline{X}_{n}^{*}\mid X_{1},\ldots,X_{n})=\frac{1}{n}\mathbb{V}(X_{1}^{*}\mid X_{1},\ldots,X_{n})=\frac{1}{n^{2}}\sum_{i}\left(X_{i}-\overline{X}_{n}\right)^{2}. \end{equation}\]

The above can also be expressed as \(S_{n}(n-1)/n^{2}\) where \(S_{n}\) is the unbiased sample variance of \((X_{1},\ldots,X_{n})\). Next, note that

\[\begin{equation} \mathbb{E}\left[\left(\overline{X}_{n}\right)^{2}\right]=\frac{1}{n^{2}}\mathbb{E}\left[\sum_{i}X_{i}^{2}+\sum_{i\neq j}X_{i}X_{j}\right]=\frac{1}{n}\left(\sigma^{2}+\mu^{2}\right)+\frac{n-1}{n}\mu^{2}=\frac{\sigma^{2}}{n}+\mu^{2} \end{equation}\]

where \(\mu = \mathbb{E}[X_{1}]\) and \(\sigma^{2} = \mathbb{V}(X_{1})\). Now, recall that for any random variable \(Y\),

\[\begin{equation} \mathbb{V}(Y\mid\mathcal{H})=\mathbb{E}\left[Y^{2}\mid\mathcal{H}\right]-\mathbb{E}\left[Y\mid\mathcal{H}\right]^{2}. \end{equation}\]

Therefore, by the tower property,

\[\begin{equation} \mathbb{E}\left[Y^{2}\right]=\mathbb{E}\left[\mathbb{V}(Y\mid\mathcal{H})+\mathbb{E}\left[Y\mid\mathcal{H}\right]^{2}\right]. \end{equation}\]

Applying this to our setting,

\[\begin{equation} \mathbb{E}\left[\left(\overline{X}_{n}^{*}\right)^{2}\right]=\mathbb{E}\left[\frac{n-1}{n^{2}}S_{n}+\left(\overline{X}_{n}\right)^{2}\right]=\frac{2n-1}{n^{2}}\sigma^{2}+\mu^2. \end{equation}\]

As such, we can conclude that

\[\begin{equation} \mathbb{V}(\overline{X}_{n}^{*}) = \frac{2n-1}{n^{2}} \sigma^{2} = \frac{2n-1}{n} \mathbb{V}(\overline{X}_n) \sim 2\mathbb{V}(\overline{X}_{n}) \end{equation}\]

where the asymptotic is in the limit of large \(n\).

6.

TODO (Computer Experiment)

7.

a)

The distribution of \(\hat{\theta}\) is given in the solution of Question 2 of Chapter 6.

TODO (Computer Experiment)

b)

Let \(\hat{\theta}^*\) be a bootstrap resample. Then,

\[\begin{equation} \mathbb{P}(\hat{\theta}^* = \hat{\theta} \mid \hat{\theta}) = 1 - \mathbb{P}(\hat{\theta}^* \neq \hat{\theta} \mid \hat{\theta}) = 1 - \left( 1 - 1/n \right)^n \rightarrow 1 - \exp(-1) \approx 0.632. \end{equation}\]

8.

TODO