6

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 $\$8,\!203.59$?

ə̷̶̸͇̘̜́̍͗̂̄︣͟
  • 5,412
  • 5
  • 22
  • 55
Mark K
  • 429
  • 2
  • 10
  • 1
    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 $0.6(1.00)$. – Oguz Toragay Oct 01 '19 at 14:55
  • @OguzToragay Thank you for the comment. Can you please post an answer? – Mark K Oct 01 '19 at 15:07

2 Answers2

3

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}

ə̷̶̸͇̘̜́̍͗̂̄︣͟
  • 5,412
  • 5
  • 22
  • 55
Oguz Toragay
  • 8,652
  • 2
  • 13
  • 41
  • Thank you. Why $0.6u(100,000) = 0.6(1.0000)$, and $0.4u(−10,000) = 0.4(−0.65312)$? – Mark K Oct 01 '19 at 15:23
  • 1
    @MarkK Just replace $x$ with the value $100,000$ in the $u(x)$ function. – Oguz Toragay Oct 01 '19 at 15:28
  • 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)? – Mark K Oct 02 '19 at 10:35
  • 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. – Mark K Oct 03 '19 at 09:10
  • 1
    @MarkK Never mind. I am glad that my answer was helpful. – Oguz Toragay Oct 03 '19 at 12:50
3

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 $d=u^{-1}(c)$.

ə̷̶̸͇̘̜́̍͗̂̄︣͟
  • 5,412
  • 5
  • 22
  • 55