whether or not $\delta_i = 1$, we treat these two cases separately in the next part of the proof.\
[ ]{}

**The case** $\bm{\delta_i \neq 1}$: The utility function takes the form $$U\left((cx)^{\left(1 - \theta_i/n\right)}(\bar{c}_{-i}(t)y)^{-\theta_i}; \delta_i \right) = \left(1 - \frac{1}{\delta_i} \right)^{-1} (cx)^{(1 - \theta_i/n)(1 - 1 / \delta_i)}(\bar{c}_{-i}(t)y)^{- \theta_i(1- 1 / \delta_i)}.$$ Applying the first order conditions, the suprema in are attained by $$\label{optcontrolNot}
    \pi^{i,*}(x,y,t) = - \frac{\mu_i x v_x(x,y,t) + \sigma_i \widehat{\sigma \pi}_{-i} xy v_{xy}(x,y,t)}{\Sigma_i x^2 v_{xx}(x,y,t)},$$ and $$c^{i,*}(x,y,t) = \frac{1}{x}\left(\frac{(1 - \frac{\theta_i}{n})(\bar{c}_{-i}(t)y)^{- \theta_i(1- 1 / \delta_i)}}{ v_x(x,y,t)} \right)^{\frac{1}{1-(1 - \theta_i/n)(1- 1/\delta_i)}}.
    \tag{\theequation'}$$ Let us introduce the following constants: $$\begin{aligned}
\label{gammai}
\gamma_i &= \frac{1}{1-(1 - \theta_i/n)(1- 1/\delta_i)}, \\
\Gamma_i &= \left(1 - \frac{\theta_i}{n}\right)^{ \gamma_i}\left(\frac{1}{ \gamma_i - 1} \right). \nonumber \end{aligned}$$ Using these constants and the expressions and (\[optcontrolNot\]’), the HJB equation becomes $$\label{HJBnot}
\begin{split}
0 = v_t &- \frac{1}{2}\frac{(\mu_i x v_x + \sigma_i \widehat{\sigma \pi}_{-i} xy v_{xy})^2}{\Sigma_i x^2 v_{xx}} + \frac{1}{2}\big(\widehat{\sigma \pi}_{-i}^2 + \frac{1}{n}\widehat{(\nu \pi)^2}_{-i}\big)y^2 v_{yy} + (\eta_i - \widehat{c}_{-i}(t)) y v_y \\
    &+ (v_x)^{1 - \gamma_i}(\bar{c}_{-i}(t)y)^{- \gamma_i \theta_i(1- 1 / \delta_i)}\Gamma_i.
    \end{split}$$ We now make the ansatz $$\begin{aligned}
v(x,y,t) = \epsilon_i \left( 1 - \frac{1}{\delta_i} \right)^{-1}  x^{(1 - \theta_i/n)(1 - 1 / \delta_i)} y^{-\theta_i(1 - 1 / 