# Solving a Certainty Equivalent (Decision Analysis) problem

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 $$\8,\!203.59$$?

• 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 1 '19 at 14:55
• @OguzToragay Thank you for the comment. Can you please post an answer? – Mark K Oct 1 '19 at 15:07

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 $$d=u^{-1}(c)$$.
• thank you. is the "u" here is a number? I still don't get why $u^{-1}(0.338751)$ = $\$8,\!203.59$. – Mark K Oct 2 '19 at 10:40 • 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. – TheSimpliFire Oct 2 '19 at 17:44 • thanks again for the details! – Mark K Oct 3 '19 at 9:09 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} • Thank you. Why0.6u(100\,000) = 0.6(1.0000)$, and$0.4u(−10\,000) = 0.4(−0.65312)$? – Mark K Oct 1 '19 at 15:23 • @MarkK Just replace$x$with the value$100\,000$in the$u(x)$function. – Oguz Toragay Oct 1 '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 2 '19 at 10:35