We can then wrap up
\begin{align} n_{j+1} \overline{X}_{j+1} &= \sum_{i=1}^{n_{j+1}} (\hat{X}_{j+1})_i = \frac{1}{2}\sum_{i=1}^{n_{j}/2} (\hat{X}_{j})_{2i-1} + (\hat{X}_{j})_{2i} \nonumber \\ &= \frac{1}{2}\left[ (\hat{X}_j)_1 + (\hat{X}_j)_2 + \cdots + (\hat{X}_j)_{n_j} \right] = \underbrace{\frac{n_j}{2}}_{=n_{j+1}} \overline{X}_j = n_{j+1}\overline{X}_j. \tag{27} \end{align}By repeated use of this equation we get V(\overline{X}_i) = V(\overline{X}_0) = V(\overline{X}) for all 0 \leq i \leq d-1 . This has the consequence that
\begin{align} V(\overline{X}) = \frac{\sigma_k^2}{n_k} + e_k \qquad \text{for all} \qquad 0 \leq k \leq d-1. \tag{28} \end{align}Flyvbjerg and Petersen demonstrated that the sequence \{e_k\}_{k=0}^{d-1} is decreasing, and conjecture that the term e_k can be made as small as we would like by making k (and hence d ) sufficiently large. The sequence is decreasing (Master of Science thesis by Marius Jonsson, UiO 2018). It means we can apply blocking transformations until e_k is sufficiently small, and then estimate V(\overline{X}) by \widehat{\sigma}^2_k/n_k .
For an elegant solution and proof of the blocking method, see the recent article of Marius Jonsson (former MSc student of the Computational Physics group).