$w$ for generalist predictions. The associated utility function is $\hat U_{\text{ws}}(w;(1/2,1/2))$ and $$\hat U_{\text{ws}}(w;(1/2,1/2)) = w^\intercal \hat \Sigma_{\text{ws}} w - 2 \hat b_{\text{ws}}^\intercal w + \frac{\overline{y_1^2} + \overline{y_2^2}}{2},$$ where $$\small
\hat \Sigma_{\text{ws}} = M^{-1}\left[
\begin{array}{cc}
\sum\limits_{m=1}^2 \bar y_{1,-m}^2& \sum\limits_{m=1}^2 \bar y_{1,-m}\bar y_{2,-m}\\
\sum\limits_{m=1}^2 \bar y_{1,-m}\bar y_{2,-m}& \sum\limits_{m=1}^2 \bar y_{2,-m}^2\\
\end{array}\right],~~\hat b_{\text{ws}} = (2M)^{-1}\left[
\begin{array}{c}
\sum\limits_{m=1}^2(\bar y_{1,m} + \bar y_{2,m})\bar y_{1,-m}\\
\sum\limits_{m=1}^2(\bar y_{1,m} + \bar y_{2,m})\bar y_{2,-m}
\end{array}\right].
\normalsize$$ Note by construction, $\bar y_k = (M-1)/M\bar y_{k,-m} + 1/M\bar y_{k,m}$. Therefore, $$\sum_{m} \bar y_{k,-m}y_{k',-m} = \frac{M^3 - 2M^2 + M}{(M-1)^2} \bar y_k\bar y_{k'} + \sum_m\left(\bar y_{k,m}\bar y_{k',m} - \bar y_k\bar y_{k'}\right),$$ for any $k,k' \in \{1,2\}$. It is straightforward to show that $${\text{var}}\left(\sum_m\left(\bar y_{k,m}\bar y_{k',m} - \bar y_k\bar y_{k'}\right)\right) = \frac{1}{4}{\text{var}}\left((\bar y_{k,1} - \bar y_{k,2})(\bar y_{k',1} - \bar y_{k',2})\right) = \frac{1 + \mathbb I(k=k')}{n^2}.$$ Therefore $
\sum_{m} \bar y_{k,-m}y_{k',-m} = 2\bar y_k\bar y_{k'} + O_p(1/n).
$ Similarly, we can prove that $\sum_m (\bar y_{k,m} + \bar y_{k',m})\bar y_{k,-m} = 2M\bar y \bar y_k + O_p(1/n)$ for $k,k'\in\{1,2\}$ and $k\neq k'$. Based on these results, if $w$ is bounded by a finite constant, we have $
|\hat U_{\text{ws}}(w;(1/2,1/2)) - \hat U(w;(1/2,1/2))| \leq O_p(1/n).
$

On the other hand, the limit of $\hat 