Suppose $g(\alpha)=-\alpha\log\left(E_{\nu^0_{s,a}}[e^{-r/a}]\right)-\alpha\delta$ and denote $\alpha^*=\arg\max_{a\geq 0}g(\alpha)$.
Given $2^N$ samples of $r_i$ based on the distribution of $\nu^0_{s,a}$, let $\hat{\nu}_{2^N}(r)=\frac{1}{2^N}\sum_{i=1}^{2^N}\mathbb{1}_{\{r=r_i\}}$.
$\hat{\nu}_{2^{N}_{O}}$ (resp. $\hat{\nu}_{2^{N}_{E}}$) is defined by using the subset of $2^N$ samples in which the index of every element is odd (resp. even). We use $g_{2^N}$ to represent the function defined by replacing $\nu^0_{s,a}$ by $\hat{\nu}_{2^{N}}$ in $g$ and $\alpha_{2^N}^{*}=\arg\max_{\alpha\geq 0}g_{2^N}(\alpha)$. $g_{2^N_O}, g_{2^N_E}, \alpha_{2^N_O}^{*}, \alpha_{2^N_E}^{*}$ are defined similarly.
Hence, we have $$ g_{2^{n+1}_{O}}(\alpha_{2^{n+1}_O}^{*})=\sup_{\alpha\geq 0}\left\{-\alpha\log\left(\frac{1}{2^n}\sum_{i=1}^{2^n}e^{-\frac{r_{2i-1}}{\alpha}}\right)-\alpha\delta\right\} $$ and $$ g_{2^{n+1}_{E}}(\alpha_{2^{n+1}_E}^{*})=\sup_{\alpha\geq 0}\left\{-\alpha\log\left(\frac{1}{2^n}\sum_{i=1}^{2^n}e^{-\frac{r_{2i}}{\alpha}}\right)-\alpha\delta\right\} $$
How can I show that $\frac{1}{2}(g_{2^{n+1}_{O}}[\alpha_{2^{n+1}_O}^{*})+g_{2^{n+1}_{E}}(\alpha_{2^{n+1}_E}^{*})] = g_{2^{n}}(\alpha_{2^{n}}^{*})$?
This is a part of the proof written on page 13 of https://proceedings.mlr.press/v162/liu22a/liu22a.pdf . I tried with some small examples of $n$ but it does not hold. I wonder what I overlooked and what the proof is trying to do.