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Suppose the consumer Solves $\max -e^{-\gamma W}$ where $W=X^T D -X^Tp R_f$ where $X$ is the vector invested in a risky asset and $D\sim N(E[D],\Sigma^2_D)$ and $R=\sim N(E[R],\Sigma^2_R)$. Then ${ X=(\gamma \Sigma_R)^{-1}(E[R-R_f])}$. Is this formula correct?

My reasoning is as follows: $e^{-\gamma W}=e^{-\gamma X(E[D]-p R_f)+\frac{1}{2}\gamma^2 X \Sigma X^T}$ Hence $ X=(\gamma \Sigma_D)^{-1}(E[D-p R_f])$ Hence $ X=(\gamma \Sigma_R)^{-1}(E[R-R_f])$

Here $\Sigma_D$ and $\Sigma_R$ refer to variance vector for dividend and returns.

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    $\begingroup$ Hey, can you please explain what $p$ is, and what's the difference betwen $R$ and $R_f$? Where does $R$ enter the formula in the first place? $\endgroup$ Oct 13, 2022 at 6:16
  • $\begingroup$ The optimal solution, though, looks good at first sight :) $\endgroup$ Oct 13, 2022 at 7:50
  • $\begingroup$ @Kermittfrog $R_f$ is risk free rate and $R=D/p$ is return on risky assets. $\endgroup$ Oct 18, 2022 at 16:42

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