consider applying ${\text{CV}_{\text{cs}}}$ for generalist predictions. Note that under exchangeable $P_k$ distributions, $k=1,\ldots$, $\hat U_{\text{cs}}^{(t)}(w)$ is an unbiased estimator of $U_g(w^{(t)})$, where $w^{(t)}$ is equal to $w$ except for components $w_{\ell,t}$ $\ell = 1,\ldots,L$, which are set to zero: $$\mathbb E\left[\hat U^{(t)}_{\text{cs}}(w)\right] = \mathbb E\left[U_g(w^{(t)})\right],$$ where the expectation is taken over $\mathcal S$. For $\{\nu_t;t\leq K\}\in \Delta_{K-1}$, it follows that $$\mathbb E\left[\sum_{t=1}^K \nu_t \hat U^{(t)}_{\text{cs}}(w)\right] = \sum_{t=1}^K \nu_t\mathbb E\left[U_g(w^{(t)})\right].$$

Consider the Taylor expansion of $U_g(w^{(t)}),~t=1,\ldots,K$, around ${\mathbf{0}}$. Since $\sum_t \nu_t = 1$, $$\sum_t \nu_t U_g(w^{(t)}) = U_g({\mathbf{0}}) + \frac{\partial U_g}{\partial w^\intercal}({\mathbf{0}}) \sum_t \nu_k w^{(t)} + o(\|w\|).$$ By construction $\sum_t \nu_t w^{(t)} = ((1-\nu_t)w_{\ell,t};\ell=1,\ldots,L,t = 1,\ldots,K)$. Let $S$ be a $KL\times KL$ diagonal matrix with the diagonal term corresponding to $w_{\ell,t}$ as $1-\nu_t$, we have $\sum_t \nu_t w^{(t)} = Sw$.

Based on the results above, we know that $$\mathbb E\left[\nu_t\sum_{t=1}^K \hat U^{(t)}_{\text{cs}}(w)\right] = \mathbb E U_g({\mathbf{0}}) + \mathbb E\left[\frac{\partial U_g}{\partial w^\intercal}({\mathbf{0}})\right]Sw + o(\|w\|).$$ If $w$ is defined close to $\mathbf{0}$, e.g. $W = \{w:\|w\|_1\leq 1\}$, the linear term in the above expansion dominates higher order terms of $w$ and we have $$\mathbb E\left(\sum_k\nu_k\hat U^{(k)}_{\text{cs}}(w)\right) = \mathbb EU_g(Sw) + o(\|w\|).$$ Therefore, a nearly unbiased estimator of $U_g(w)$ is $$\hat U_{\text{cs}}(w;\nu) = \sum_k\nu_k\hat 