0
$\begingroup$

Two utility functions $u,v:\mathbb{R}_{>0}\rightarrow\mathbb{R}$ (giving the utility of, say, an amount of money) are considered equivalent if $u(x)$ is given by $m\,v(x)+c$, for some constants $c$ and $m$ with $m>0$. They then give the same preference ordering on any set of gambles or investments. So we could call a utility function scale invariant if there exist functions $c:\mathbb{R}_{>0}\rightarrow\mathbb{R}$ and $m: \mathbb{R}_{>0}\rightarrow\mathbb{R}_{>0}$ such that $$u(ax)=m(a)\,u(x)+c(a).$$ Which utility functions are scale invariant?

I thought of this question while reading a justification of the Kelly Criterion. It was assumed that a gambler would bet a constant fraction of their wealth, regardless of how much money they currently had.


What I've attempted so far:

There are solutions $u(x)=x^\beta$ along with their equivalents (i.e. of the form $\alpha x^\beta+\gamma$). We also get $u(x)=\log(x)$ which we can think of as a limit of $x^\beta$ as $\beta\rightarrow 0$ (i.e. it's the pointwise limit of $\frac{x^\beta-1}{\beta}$), and its equivalents. So far these are the only solutions I've found.

As far as solving the equation goes, my first step was to sub in $x=1$ and then $x=1/a$ and use the two resulting equations to eliminate $m$ and $c$. This got too messy to write down. Instead I tried first replacing $u$ by $f(x)=u(x)-u(1)$ so that $f(1)=0$. This eventually gives $$f(ax)f(1/a)-f(a)f(1/a)+f(a)f(x)=0$$ or equivalently $$f(ax)=f(a)+\frac{-f(a)}{f(1/a)}f(x)$$ [which is beginning to look like the Cauchy functional equation $$f(ax)=f(a)+f(x)$$ which has as its solutions $\log(x)$ to different bases (and some pathologies which we can safely ignore since no-one would ever pick them as utility functions).]

Can anyone find the complete set of solutions? Feel free to assume continuity, differentiability etc.

$\endgroup$
0
$\begingroup$

Answering my own question:

Take the equation $$f(ax)f(1/a)-f(a)f(1/a)+f(a)f(x)=0$$ and differentiate wrt $x$ to give $$af'(ax)f(1/a)+f(a)f'(x)=0$$ and hence $$af'(ax)f(1/a)=-f(a)f'(x).$$ Differentiating a second time gives $$a^2f''(ax)f(1/a)=-f(a)f''(x).$$ Taking the ratio of the last two equations and setting $x=1$ gives $$\frac{f'(a)}{f''(a)}=a\frac{f'(1)}{f''(1)}.$$ So we have reducced the problem to a differential equation. Solving this gives precisely the functions I mentioned in the question. These are precisely the CRRA utility functions.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.