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I am solving a Certainty Equivalent (Decision Analysis) problem.

The problem is a Risk-Averse Case - a deal of $60\%$ chance to win $\$100,\!000$ and $40\%$ chance to lose $\$10,\!000$.

Suppose the decision-maker is risk-averse with a risk tolerance of $\$20,\!000$ and his utility function is:

$$u(x)=1.0067837 (1-e^{-x/20\,000}).$$

The answer shows: \begin{align}u(\rm CE)&= 0.6 u(100\,000) + 0.4 u(-10\,000)\\&= 0.4(1.00) + 0.4(-0.65312)\\&= 0.338751\\\implies{\rm CE}&=u^{-1}(0.338751)=\$8,\!203.59.\end{align}

Why does $0.6 u(100\,000)$ equal to $0.4(1.00)$, and likewise $0.4 u(-10\,000)$ equals to $0.4(-0.65312)$?

Also, with $u^{-1}(0.338751)$, how does it arrive at $\$8,\!203.59$?

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    $\begingroup$ Hi, welcome to OR.SE, the calculation you mentioned in your question is not correct. In the second line there is a typo, instead of $0.4(1.00)$ it should be $0.6(1.00)$. $\endgroup$ – Oguz Toragay Oct 1 at 14:55
  • $\begingroup$ @OguzToragay Thank you for the comment. Can you please post an answer? $\endgroup$ – Mark K Oct 1 at 15:07
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If $u(d)=c$ then $d=u^{-1}(c)$ since $u\circ u^{-1}$ forms the identity. Thus in general, under suitable constraints for $a,b,c$,\begin{align}a(1-e^{-d/b})=c&\implies1-e^{-d/b}=\frac ca\\&\implies e^{-d/b}=1-\frac ca=\frac{a-c}a\\&\implies-\frac db=\ln\frac{a-c}a&&\\&\implies d=-b\ln\frac{a-c}a=b\ln\frac a{a-c}.\end{align} Now substitute the values of $a=1.0067837$, $b=20\,000$ and $c=0.338751$ to obtain $d=u^{-1}(c)$.

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  • $\begingroup$ thank you. is the "u" here is a number? I still don't get why $u^{-1}(0.338751)$ = $\$8,\!203.59$. $\endgroup$ – Mark K Oct 2 at 10:40
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    $\begingroup$ No. $u(\cdot)$ is a function. The notation $u^{-1}(\cdot)$ is the inverse function of $u(\cdot)$, which satisfies the identity $u(u^{-1}(x))=x$. Therefore, $u^{-1}(0.338751)=8203.59$ is equivalent to $u(8203.59)=0.338751$, and you may wish to check this yourself. $\endgroup$ – TheSimpliFire Oct 2 at 17:44
  • $\begingroup$ thanks again for the details! $\endgroup$ – Mark K Oct 3 at 9:09
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There is a typo in the calculation that you mentioned. $$u({\rm CE}) = 0.6 u(100\,000)+0.4 u(-10\,000)=0.6(1.0000)+0.4(-0.65312)=0.338751$$

For your second question, if $y=f(x) \text{ then } x=f^{-1}(y).$

For the calculations:

\begin{align}u(100\,000)&=1.0067837(0.993262053000)=1.000000044789 \\u(-10\,000)&=1.0067837(-0.64872127070)=-0.65312200118\end{align}

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  • $\begingroup$ Thank you. Why $0.6u(100\,000) = 0.6(1.0000)$, and $0.4u(−10\,000) = 0.4(−0.65312)$? $\endgroup$ – Mark K Oct 1 at 15:23
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    $\begingroup$ @MarkK Just replace $x$ with the value $100\,000$ in the $u(x)$ function. $\endgroup$ – Oguz Toragay Oct 1 at 15:28
  • $\begingroup$ thank you. I am getting there. when $u^{-1}(0.338751)$ = $\$8,\!203.59$, it seems $u^{-1}$ about equal to 24217. How do I know this 24217 (to get 8203.59)? $\endgroup$ – Mark K Oct 2 at 10:35
  • $\begingroup$ without your help, this question wouldn't be closed. However another user helped the last mile of closing this question. Hope you don't mind I choose his for the answer. $\endgroup$ – Mark K Oct 3 at 9:10
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    $\begingroup$ @MarkK Never mind. I am glad that my answer was helpful. $\endgroup$ – Oguz Toragay Oct 3 at 12:50

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