# How did we get $W_g=W_b$ from $\dfrac{U'(W_g)}{U'(W_b)}=1$?

My question is from Nicholson-Snyder's text , E-book here.

My question is here, from page 217 of the book. (I can't post image as my reputation is not enough.)

How did we get $W_g=W_b$ from $\dfrac{U'(W_g)}{U'(W_b)}=1$ ?

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@ honso : maybe from the injectivity of $U'$ ? –  TheBridge Dec 22 '13 at 18:31

## 1 Answer

$U$ is usually strictly concave. So $U'$ is strictly decreasing and is therefore injective.

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OK, but it is not assumed ! and we are perhaps going to prove risk aversion, another name of concave utility, isn't it? –  honso Dec 23 '13 at 2:32
@honso: It is assumed. See the footnote on page 216 before the proposition in question. It is bad that the book put such important assumption in an obscure footnote. Also the concavity is needed for proving the sufficiency and uniqueness of the concluding price being truly utility maximizing. The book is somewhat mathematically deficient. –  Hansen Dec 23 '13 at 4:23
If U'>0 then U is increasing not decreasing. –  user6820 Dec 23 '13 at 9:28
@Nic: Yes, you are right but it does not contradict what I said. I said $U'$ (not $U$) is strictly decreasing. –  Hansen Dec 23 '13 at 14:55
@Hansen, I am the same user and can't log in and can't accept yours! Please pardon me! I AM TRYING MY BEST AND HAVE ALSO ASKED FOR HELP AT META.STACKOVERFLOW –  honso Dec 26 '13 at 4:31