The certainty equivalent is a guaranteed return that someone would accept now, rather than taking a chance on a higher, but uncertain, return in the future. Otherwise, some definitions say that the certainty equivalentis the mean income of a gamble. If we see the vast literature of microeconomic and game theory papers, then we shall see that they assume CARA Gaussian normal preferences (or negative expnential utility, i.e. $\mathbb{U}(\widetilde{W})=-e^{\rho\widetilde{W}}$). Specifically the certainty equivalent in such case is $$CE(\widetilde{W})=\mathbb{E}(\widetilde{W})-\frac{\rho}{2}\mathbb{V}ar(\widetilde{W})$$ How can someone claim that this return is guaranteed, since we need to use $\mathbb{V}ar(\widetilde{W})$ to measure the utility gain of an individual?
1 Answer
How can someone claim that this return is guaranteed, since we need to use $\mathbb{V}ar(\widetilde{W})$ to measure the utility gain of an individual?
Maybe I misunderstand your point, but though $\widetilde{W}$ is a random variable, $\mathbb{E}(\widetilde{W})-\frac{\rho}{2}\mathbb{V}ar(\widetilde{W})$ and hence $CE(\widetilde{W})$ is a real number. So $CE(\widetilde{W})$ is a measure of a certain income which to the consumer has the same utility as $\widetilde{W}$.
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$\begingroup$ Why $CE(\widetilde{W})$ is a real number? What is the intuition behind this? $\endgroup$ Commented Jun 6, 2020 at 22:52
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1$\begingroup$ The "intuition" is that $\mathbb{E}(\widetilde{W})$ is a number, $\frac{\rho}{2}$ is a number and $\mathbb{V}ar(\widetilde{W})$ is a number, so performing the mathematical operations $$\mathbb{E}(\widetilde{W})-\frac{\rho}{2}\mathbb{V}ar(\widetilde{W})$$ gives you a number, which you defined as $CE(\widetilde{W})$. $\endgroup$– GiskardCommented Jun 7, 2020 at 9:25