I have a question on an exercise from chapter 9 of D. Luenberger, Investment Science, International Edition, where I suspect there may be a typo.

Exercise 8 (Certainty approximation)

There is a useful approximation to the certainty equivalent that is easy to derive. A second-order expansion near $\bar x=E(x)$ gives $$U(x)\approx U(\bar x)+U^{'}(\bar x)(x-\bar x)+\frac12U^{''}(\bar x)(x-\bar x)^2$$

Hence, $$E[U(x)]\approx U(\bar x)+\frac 12 U^{''}(\bar x)var(x)$$ On the other hand, if we let c denote the certainty equivalent and assume it is close to $\bar x$, we can use the first-order expansion $$U(c)\approx U(\bar x)+U^{'}(\bar x)(c-\bar x)$$ Using these approximations, show that $$c \approx \bar x+{U^{''}(\bar x) \over U^{'}(\bar x)}var(x)$$

Now, I used general methods of algebra along with the fact that $E[U(x)] = U(c)$ to show directly that $$c \approx \bar x+\frac 12 ({U^{''}(\bar x) \over U^{'}(\bar x)})var(x)$$ as follows:

Take the third equation and transform it into $$ c\approx \bar x + {U(c)-U(\bar x) \over U^{'}(\bar x)}$$ Now all I have to do is show that the numerator in the fraction part is $\frac 12 U^{''}(\bar x)var(x)$ which is done by putting $E[U(x)] = U(c)$ into the second formula and you can see the result is immediately there.

On top of this work, I wrote out an example of an investment with the log utility function and showed that my approximation for c worked whereas the book's formula without the "2" didn't.

However, I would like to post all this here just to verify that this is a typo from the book and not some misunderstanding on my part.

Thanks in advance for any feedback.


1 Answer 1


Your calculation seems to be correct. I found this document here:http://home.uchicago.edu/rmyerson/teaching/util206.pdf. You can see that in P10, the certainty equivalence formula has that 1/2 factor there.

  • $\begingroup$ Thanks for confirming. These typos from this book are awfully annoying and that's not the only one I've come across. You end up spending about an hour trying to figure something out and then it ends up being a typo. Good work thanks again. $\endgroup$
    – James
    Sep 4, 2015 at 22:47

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