I am struggling with a calculation for my thesis and was wondering, whether you could assist me.
I have the following payoff function: $$X_i=N_i R-c, N_i \sim \textrm{Poisson} (\lambda) $$ And the following exponential utility: $$u(x)\equiv\frac{1}{\rho}(1-e^{-\rho x})$$ My goal is to calculate the certainty equivalent: $CE_i\equiv u^{-1}(\mathbb E[u(X_i)])$, however my result $CE_i=\lambda R$ contradicts my source which comes up with $CE_i=\lambda u(R)$. Looking forward to any help :)
First note that $$ u^{-1}(y)= -\frac{\log(1-\rho y)}{\rho}, $$ and that $$ E(U(X_i)) = \frac{1}{\rho}\left(1-e^{\rho c} E\left(e^{-\rho R N_i}\right) \right) =\frac{1}{\rho}\left(1- e^{\rho c+\lambda(\exp({-\rho R})-1)} \right), $$ where the expectation is recognized as the moment generating function of $N_i$ at $-\rho R$. Inserting the latter into the former we get that \begin{align} u^{-1} \left( E(U(X_i))\right) &= -\frac{\log(1-\rho \frac{1}{\rho}\left(1- e^{\rho c+\lambda(\exp({-\rho R})-1)} \right))}{\rho} \\ &= -\frac{\rho c+\lambda(\exp({-\rho R})-1) )}{\rho} \\ &= -c +\lambda \left( \frac{1-e^{-\rho R}}{\rho} \right) \\ &= -c +\lambda u(R). \end{align} This would align with your source if they don't have any cost $c\in \mathbb{R}$ in their payoff function.