We can then wrap up
$$ \begin{align} n_{j+1} \overline{X}_{j+1} &= \sum_{i=1}^{n_{j+1}} (\hat{X}_{j+1})_i = \frac{1}{2}\sum_{i=1}^{n_{j}/2} (\hat{X}_{j})_{2i-1} + (\hat{X}_{j})_{2i} \nonumber \\ &= \frac{1}{2}\left[ (\hat{X}_j)_1 + (\hat{X}_j)_2 + \cdots + (\hat{X}_j)_{n_j} \right] = \underbrace{\frac{n_j}{2}}_{=n_{j+1}} \overline{X}_j = n_{j+1}\overline{X}_j. \tag{6} \end{align} $$By repeated use of this equation we get \( \mathrm{var}(\overline{X}_i) = \mathrm{var}(\overline{X}_0) = \mathrm{var}(\overline{X}) \) for all \( 0 \leq i \leq d-1 \). This has the consequence that
$$ \begin{align} \mathrm{var}(\overline{X}) = \frac{\sigma_k^2}{n_k} + e_k \qquad \text{for all} \qquad 0 \leq k \leq d-1. \tag{7} \end{align} $$