Log Utility Makes Sense
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Under multiplicative wealth dynamics, “equal and opposite” moves are multiplying and dividing, not adding and subtracting. Let’s show that any utility function that respects this fact must be affine in \(\ln w\).
We want a utility function such that that at any wealth level \(w\), the agent is indifferent between keeping \(w\) and taking a fair \(50/50\) bet giving either \(w_{+} = wr\) or \(w_{-} = w/r\) (with \(r>1\)). In expected utility terms we want:
\[u(w)=\frac{u(wr)+u(w/r)}{2}\]Discrete Case
Fix some \(r>1\) and \(w_0>0\), and define the lattice:
\[a_i := u(w_0r^{\,i}), \quad i\in\mathbb{Z}\]Applying the indifference condition at wealth \(w_0r^i\) gives \(a_i=\frac{a_{i+1}+a_{i-1}}{2}\), hence
\[2a_i=a_{i+1}+a_{i-1} \Longleftrightarrow a_{i+1}-a_i = a_i-a_{i-1}\]The increment \(a_i-a_{i-1}\) does not depend on \(i\), so there exists a constant \(A\) such that
\[a_i-a_{i-1}=A, \qquad \forall i\in\mathbb{Z}\]Telescoping, and setting \(B := a_0\):
\[a_i =a_0 + \sum_{k=1}^{i}(a_k-a_{k-1})=Ai + B,\qquad \forall i\in\mathbb{Z}\]Let \(x := w_0 r^{\,i}\). Then
\[i = \frac{\ln(x/w_0)}{\ln r}\]so the previous equation becomes
\[u(x)=A\frac{\ln(x/w_0)}{\ln r}+B=\underbrace{\frac{A}{\ln r}}_{\alpha}\,\ln x+\underbrace{\left(B-\frac{A}{\ln r}\ln w_0\right)}_{\beta}\]Therefore,
\[u(w)=\alpha \ln w + \beta\]Continuous Case
Let \(u\) be twice differentiable, and write \(r=1+\varepsilon\). Then,
\[wr=w(1+\varepsilon), \qquad \frac{w}{r}=\frac{w}{1+\varepsilon}=w\left(1-\varepsilon+\varepsilon^2+O(\varepsilon^3)\right)\]Taylor expanding,
\[u(w(1+\varepsilon)) = u(w) + wu'(w)\varepsilon + \frac12 w^2u''(w)\varepsilon^2 + O(\varepsilon^3)\]and
\[u\!\left(\frac{w}{1+\varepsilon}\right) = u(w) - wu'(w)\varepsilon + \left(wu'(w) + \frac12 w^2u''(w)\right)\varepsilon^2 + O(\varepsilon^3)\]Averaging,
\[\frac{u(w(1+\varepsilon)) + u(w/(1+\varepsilon))}{2} = u(w) + \frac12\left(wu'(w)+w^2u''(w)\right)\varepsilon^2 + O(\varepsilon^3)\]Assuming this average equals \(u(w)\) for all sufficiently small \(\varepsilon\). Then,
\[\frac12\left(wu'(w)+w^2u''(w)\right)\varepsilon^2 + O(\varepsilon^3)=0\]Dividing by \(\varepsilon^2\) gives
\[\frac12\left(wu'(w)+w^2u''(w)\right) + O(\varepsilon)=0\]Letting \(\varepsilon\to0\), we get
\[wu'(w)+w^2u''(w)=0\]Equivalently,
\[(wu'(w))'=0\]Hence,
\[wu'(w)=\alpha\]for some constant \(\alpha\), so
\[u'(w)=\frac{\alpha}{w}\]Integrating,
\[u(w)=\alpha\ln w+\beta\]