(\delta_k - 1)\frac{\sigma_k^2}{\sigma_k^2 + \nu_k^2(1 + (\delta_k-1)\theta_k/n)}, \nonumber \\
\newomega_i &= \theta_i (\delta_i - 1)\frac{\frac{1}{n} \sum_{k = 1}^n \delta_k \rho_k}{1 + \frac{1}{n} \sum_{k = 1}^n \theta_k(\delta_k -1)  }- \delta_i \rho_i, \nonumber \\
\newC_i &= \epsilon_i^{-\delta_i} \left(\left(\prod_{k = 1}^n \epsilon_k^{\delta_k} \right)^{1/n}\right)^{\theta_i(\delta_i - 1) / (1 + \frac{1}{n} \sum_{k = 1}^n \theta_k(\delta_k-1))}, \label{def:C_i}\end{aligned}$$ where we define also $(\rho_i)_{i=1}^n$ by $$\begin{aligned}
\rho_i = (1 - 1/\delta_i) \Bigg\{ & \frac{(1 - \theta_i/n)(\mu_i - \sigma_i\theta_i(1 - 1/\delta_i)\frac{1}{n}\sum_{k \ne i} \sigma_k \pi^{k,*})^2}{2 (\sigma_i^2 + \nu_i^2)(1-(1 -\theta_i/n)(1 - 1/\delta_i) )}  \\
  &+ \frac{1}{2}\Big(\Big(\frac{1}{n} \sum_{k\ne i} \sigma_k \pi^{k,*} \Big)^2 + \frac{1}{n^2} \sum_{k \ne i} (\nu_k \pi^{k,*})^2\Big) \theta_i^2(1 - 1/\delta_i) \\
  &-  \theta_i \frac{1}{n} \sum_{k \ne i} \mu_k \pi^{k,*} + \frac{\theta_i}{2n} \sum_{k \ne i} ( \sigma^2_k + \nu_k^2)(\pi^{k,*})^2 \Bigg\},\end{aligned}$$ Moreover, we have the identity $$\begin{aligned}
\frac{1}{n} \sum_{k=1}^{n} \sigma_k \pi^{k,*} = \frac{\phi}{1 + \psi}. \label{def:vol-identity}\end{aligned}$$

Note that $\delta_i=1$ implies $\beta_i=0$, which means that log-investors always use the second form of $c^{i,*}_t$ given in . The form of the equilibrium does not seem to simplify much further, except in the single stock case:

(Single stock) Assume that for all $i = 1, ..., n$ we have $\mu_i = \mu > 0$, $\sigma_i = \sigma > 0$, and $\nu_i =0$. Then there 