competitors’ strategies. Then, we will resolved the resulting fixed point problem.

Define $Y_t := (\prod_{k \ne i} X_t^k)^{1/n}$, where $X_t^k$ solves (\[processX\]) subject to the strategies $(\pi_k,c_k)$, with $X_0^k = x^k_0$. We use the following abbreviations: $$\begin{array}{ccc}
\Sigma_k = \sigma_k^2 + \nu_k^2 & \quad \widehat{\mu \pi}_{-i} = \frac{1}{n} \sum_{k \ne i} \mu_k \pi_k, & \quad  \widehat{\sigma \pi}_{-i} = \frac{1}{n} \sum_{k \ne i} \sigma_k \pi_k, \\
\widehat{\Sigma \pi^2}_{-i} = \frac{1}{n} \sum_{k \ne i} \Sigma_k \pi_k^2, & \quad \widehat{(\nu \pi)^2}_{-i} = \frac{1}{n} \sum_{k \ne i} \nu_k^2 \pi_k^2, & \quad \widehat{c}_{-i}(t) = \frac{1}{n} \sum_{k \ne i} c_k(t). \\
\end{array}$$ A straightforward calculation with Itô’s formula (cf. the proof of Theorem 14 in [@lacker2017mean]) shows that the process $Y_t$ satisfies $$\label{processY}
\frac{dY_t}{Y_t} = (\eta_i - \widehat{c}_{-i}(t)) dt + \frac{1}{n} \sum_{k \ne i} \nu_k \pi_k dW^k_t + \widehat{\sigma \pi}_{-i} dB_t,$$ where we define also $$\eta_i = \widehat{\mu \pi}_{-i} - \frac{1}{2}\left( \widehat{\Sigma \pi^2}_{-i} - \widehat{\sigma \pi}_{-i}^2 - \frac{1}{n} \widehat{(\nu \pi)^2}_{-i}\right).$$ The $i^{\mathrm{th}}$ agent then solves the optimization problem $$\label{opt}
\sup_{(\pi^i, c^i) \in \A} \E\left[\int_0^T  U\left((c^i_t X^i_t)^{1 - \theta_i /n} (\bar{c}_{-i}(t)Y_t)^{- \theta_i}; \delta_i \right)dt + \epsilon_i U\left(\left(X^i_T\right)^{1 - \theta_i / n} Y_T^{- \theta_i}; \delta_i \right) \right],$$ where $\bar{c}_{-i}(t) = \left( \prod_{k \ne i} c_k(t) \right)^{1/n}$ and[^2] $$dX^i_t 