ds \right). \label{pf:f-expression}\end{aligned}$$ Now notice that is equivalent to $$c_i(t)^{1- \gamma_i(\theta_i/n)(1 - 1/\delta_i)} = \epsilon_i^{-\gamma_i} \bar{c}(t)^{- \gamma_i \theta_i(1 - 1/\delta_i)} f_i(t)^{-\gamma_i},$$ where $\bar{c}(t)$ denotes the full geometric average, $\bar{c}(t) := \left(\prod_{k = 1}^n c_k(t) \right)^{1/n}$. Hence, recalling the definition of $\gamma_i$ in , $$c_i(t) = \left(\epsilon_i f_i(t)\right)^{- \frac{\gamma_i}{1 - \gamma_i(\theta_i/n)(1 - 1/\delta_i)}}\bar{c}(t)^{- \frac{ \gamma_i \theta_i(1 - 1/\delta_i)}{1 - \gamma_i(\theta_i/n)(1 - 1/\delta_i)}}  = \left(\epsilon_i f_i(t)\right)^{-\delta_i}\bar{c}(t)^{- \theta_i (\delta_i - 1)} .$$ Next, plug in the expression for $f_i$ from to get $$c_i(t) = \epsilon_i^{-\delta_i} \bar{c}(t)^{- \theta_i (\delta_i - 1)} \exp \left( - \delta_i \int_t^T \left(\rho_i +  \theta_i \Big(1 - \frac{1}{\delta_i}\Big)\widehat{c}(s) + \frac{1}{\delta_i} c_i(s)\right) ds \right),$$ which is equivalent to $$\label{ci0}
c_i(t) \exp \left( \int_t^T  c_i(s) ds \right) = \epsilon_i^{-\delta_i} \bar{c}(t)^{- \theta_i (\delta_i - 1)} e^{- \delta_i \rho_i (T-t)} \exp \left(- \theta_i (\delta_i - 1) \int_t^T \widehat{c}(s) ds \right).$$ Taking the geometric mean over $i=1,\ldots,n$, we get $$\bar{c}(t) \exp \left( \int_t^T  \widehat{c}(s) ds \right) = \left(\overline{\epsilon^{\delta}}\right)^{-1} \bar{c}(t)^{- \widehat{\theta(\delta -1)}}  e^{- \widehat{\delta \rho} (T-t)}\exp \left(- \widehat{\theta (\delta - 1)}\int_t^T \widehat{c}(s) ds \right),$$ where we defined $$\begin{array}{lll}
\widehat{\theta (\delta - 1)} := \frac{1}{n} \sum_{k = 1}^n \theta_k (\delta_k -1),  &  \widehat{\delta \rho} := \frac{1}{n} \sum_{k = 1}^n \delta_k \rho_k, \text{ and} & \overline{\epsilon^{\delta}} := \left(\prod_{k 