Issues with Exercise 9.6.5

[Pipe-Vash]

In the sections a) and b) (and by extension in the rest of the exercise) there are some issues with the meaning of $\mathbb{E}\left(\overline{X_n^\star}\vert X_1, ... ,X_n\right )$ and $\mathbb{V}\left(\overline{X_n^\star} \vert X_1, ... ,X_n\right)$. As these expressions are conditioned over random variables, the outcomes should also be random variables, not values (as it is expressed in this solution). My solution is the following:

$$\mathbb{E}\left(\overline{X_n^\star}\vert X_1, ... ,X_n\right ) = \frac{1}{n} \sum_i \mathbb{E}\left(X_i^\star \vert X_1, ... ,X_n\right ) = \frac{1}{n} \sum_i \frac{1}{n}\sum_j X_j = \frac{1}{n} \sum_i \overline{X_n} = \overline{X_n} $$

$$\begin{eqnarray} \mathbb{V}\left(\overline{X_n^\star} \vert X_1, ... ,X_n\right) &=& \frac{1}{n^2} \sum_i \mathbb{V}\left(X_i^\star \vert X_1, ... ,X_n\right ) = \frac{1}{n} \mathbb{V}\left(X_1^\star \vert X_1, ... ,X_n\right ) = \frac{1}{n} \left[ \mathbb{E}\left({X_1^\star}^2 \vert X_1, ... ,X_n\right ) - \mathbb{E}^2\left({X_1^\star} \vert X_1, ... ,X_n\right) \right] \\ &=& \frac{1}{n} \left[ \frac{1}{n} \sum_i X_i^2 - \overline{X_n}^2 \right]= \frac{1}{n^2} \sum_i \left(X_i - \overline{X_n} \right)^2 = \frac{1}{n} \hat{\sigma} = \frac{n-1}{n^2} Sn^2 \end{eqnarray}$$

As I previously said, these changes produce some effects over the rest of the solution. However, the modifications I propose lead to the same final results of c) and d).