Let $\tilde{E}$ be the risk neutral expectation, and $X_t$ the wealth that time t and $R$ the return of a risk-free investment. Consider maximizing the function $EU(X_N)$ subject to $\tilde{E}\frac{X_n}{R^N}=X_0$.

The solution is discussed in Chapter 3 of Shreve volume 1, and question 3.8.i asks to show:

Fix $y$, and show that the function of $x$ given by $f(x)=U(x)-yx$ is maximized by $y=I(x)$. ($I(x)=\left[U'(x)\right]^{-1}$)

I might be misunderstanding what it means by "fix y", but as it stands this seems false. E.g. say $U(x)=\ln x$; then $f'(x)=x^{-1}-x/x=x^{-1}-1\not=0$.

What am I not understanding?

  • $\begingroup$ y = I(x) = x...? $\endgroup$ – BCLC Dec 5 '14 at 21:19
  • $\begingroup$ @BCLC: yes, $U(x)=\ln x\implies U'(x)=1/x \implies I(x)=x$. $\endgroup$ – Xodarap Dec 5 '14 at 21:30
  • $\begingroup$ Then f'(x) = 1/x - 2x? $\endgroup$ – BCLC Dec 5 '14 at 21:31
  • $\begingroup$ @BCLC: Sorry, the inverse of $1/x$ is $1/x$, so my comment is wrong and the formula in the question is right. $\endgroup$ – Xodarap Dec 5 '14 at 23:42
  • 1
    $\begingroup$ The inverse of 1/x is x :P $\endgroup$ – BCLC Dec 6 '14 at 9:27

From a different edition plus solutions manual.

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