3
$\begingroup$

I'll ask this question as pure economics, as pure math, and showing the translation.

Economics (micro): Which utilities have the property that whenever $EU(X)>EU(Y)$, the same is true after perturbing both $X$ and $Y$ by $N(0,1)$? Is it just linear and exponential utilities?

Math (integro-differential equations): \begin{align} \text{For }EU(x+N(0,1))&=\int_{-\infty}^\infty \frac{e^{-(t-x)^2/2}}{\sqrt{2\pi}_\phantom{2_2}}\, U(t)\, dt,\\ \text{does }\ \frac{dEU(x+N(0,1))}{dx}&=c\ \frac{dU(x)}{dx} \ \text{ imply }\ U' U''' = U'' U'' \ ? \\ \end{align}

Translation:

Suppose my utility is a function of my wealth $x$, and my wealth may receive a normal shock.

Currently my utility is $U(x)$. My expected utility after the shock is $EU(x+N(0,1))$, as above.

  • For a linear utility function, the shock does not change my expected utility.

  • For a nonlinear polynomial utility function, the shock transforms it nonlinearly.

  • For an exponential utility function, of the form $$U(x)=a-be^{cx}$$ the expected utility becomes $$EU(x+N(0,1))\ =\ a-be^{c^2/2}e^{cx}\ =\ a-e^{c^2/2}(a-EU(x))$$ which is a linear transformation. Then integrating over many possibilities gives $$EU(X+N(0,1))>EU(Y+N(0,1))\ \leftrightarrow\ EU(X)>EU(Y),$$ and conversely we know from Von Neumann that this property holds iff $EU(X+N(0,1))$ and $EU(X)$ are linear transforms of each other.

The question is whether this property characterizes the exponential and linear utility functions, which are also characterized by the differential equation $U' U''' = U'' U''$.

$\endgroup$
  • $\begingroup$ Your question utterly confuses me. 1) Do you consider the initial wealth $X$ and $Y$ to be random variables or deterministic. You are also switching from lower case to capital letters which makes it even less clear. 2) Do you want to perturb them with the same standard normal random variable or with random variables with the same (standard normal) distribution? 3) Are you asking about linear transformation (as in the title) or for monotonicity under transformation (as in the first paragraph)? $\endgroup$ – Stephan Sturm Jun 3 '18 at 17:44
  • $\begingroup$ @StephanSturm, 1) I'm using $X$ and $Y$ for random variables, and $x$ for a single real number. 2) I'm using a standard normal random variable for concreteness. 3) It is a theorem that if $U$ and $V$ are utilities with $\forall X,Y,\ EU(X)>EU(Y) \leftrightarrow EV(X)>EV(Y)$, then $U$ and $V$ must be linear transforms of each other, so the monotonicity and linear transforms are closely related. $\endgroup$ – Matt F. Jun 4 '18 at 2:00
  • $\begingroup$ This clarifies some things, but I still do not understand how you are going from the economic to the mathematic formulation. It seems that the economic formulation depends heavily on the joint distribution of the random variables $X$ and $Y$ and the standard normal perturbation (let's call it $Z$). It seems to be that the desired relation is never true for strictly concave utility functions, save you make some additional assumptions such as independence of the perturbation of $X$ and $Y$. The beauty of utility theory is that it is law invariant - adding a perturbation destroys this. $\endgroup$ – Stephan Sturm Jun 4 '18 at 4:55
  • $\begingroup$ To wit, for $\varepsilon>0$ small (to be determined later) set $X = \varepsilon + Z$ and $Y = -Z$. As $Z$ and $_Z$ have the same law, $X$ dominates $Y$ for every strictly increasing utility function. However, adding now $Z$ on both sides leaves us with $\tilde{X} = 2Z + \varepsilon$ and $\tilde{Y} = 0$ and we have that $\tilde{Y}$ dominates $\tilde{X}$ for each strictly increasing, strictly concave utility function as long as $\varepsilon$ was chosen small enough. [In particular your claim for exponential $U$ seems wrong.] $\endgroup$ – Stephan Sturm Jun 4 '18 at 5:19
  • $\begingroup$ @StephanSturm, I’m assuming that the normal perturbation is independent of both $X$ and $Y$. $\endgroup$ – Matt F. Jun 4 '18 at 7:50
2
$\begingroup$

$\newcommand{\al}{\alpha} \newcommand{\de}{\delta} \newcommand{\De}{\Delta} \newcommand{\ep}{\varepsilon} \newcommand{\ga}{\gamma} \newcommand{\Ga}{\Gamma} \newcommand{\la}{\lambda} \newcommand{\Si}{\Sigma} \newcommand{\thh}{\theta} \newcommand{\om}{\omega} \newcommand{\R}{\mathbb{R}} \newcommand{\Z}{\mathbb{Z}} \newcommand{\F}{\mathcal{F}} \newcommand{\E}{\operatorname{\mathsf E}} \renewcommand{\P}{\operatorname{\mathsf P}} \newcommand{\ii}[1]{\operatorname{\mathsf I}\{#1\}} \newcommand{\tf}{\widetilde{f}}$

A brief answer, at least to the "Math (integro-differential equations)" question, is no. For example, let $U(x)=\cosh ax$ for $a\ne0$. Then \begin{equation} \frac{dEU(x+N(0,1))}{dx}=c\ \frac{dU(x)}{dx} \tag{1} \end{equation} for $c=e^{a^2/2}$, but $U' U''' \ne U'' U''$.


A (much) longer story is to consider a general solution of equation (1), which can be rewritten as \begin{equation*} V*\phi=cV, \end{equation*} where $V:=U'$ and $\phi$ is the standard normal density. Formally, considering the Fourier transform $\hat V(t):=\int_\R e^{itx}V(x)\,dx$, we have \begin{equation} (\sqrt{2\pi}\,\phi-c)\hat V=0 \tag{1.5} \end{equation} and "hence" \begin{equation} \hat V=\sum_j C_j\de_{iu_j}, \tag{2} \end{equation} where the $C_j$'s are constants, the $iu_j$'s are the complex roots of the equation $\sqrt{2\pi}\,\phi-c=0$, and $\de_b$ is the Dirac distribution at point $b$, so that $\langle \de_b,f\rangle=f(b)$ for test functions $f$. "Therefore", $V(x)$ is a linear combination of the $e^{u_jx}$'s and hence $U(x)$ is a linear combination of the $e^{u_jx}$'s and of a constant (and also of $x$ if $c=1$). This is how the solution $U(x)=\cosh ax$ was found.

On the other hand, it is easy to verify by direct calculation that any linear combination of the $e^{u_jx}$'s (with the $u_j$'s as above) is a solution to equation (1). Particular examples of such linear combinations are $U(x)=\cosh ax$ and $U(x)=\sinh ax$.

Remaining problems with the above approach to finding a general solution of equation (1) are as follows:

  1. We need to define an appropriate class of functions $U$ for which equation (1) is considered.

  2. Since $\hat V$ will in general be a distribution, we need to define an appropriate space of test functions. The space of all compactly supported smooth functions will not do, since the Fourier transform does not map this space into itself. The space of all tempered test functions will not do here either, since we should allow $U$ (and hence $V$) to grow at least exponentially fast. Perhaps the space of all functions $f$ such that \begin{equation} |f(x)|\le Ce^{-ax^2}\quad\text{and}\quad |\hat f(x)|\le Ce^{-bx^2} \tag{3} \end{equation} for some positive $C,a,b$ (possibly depending on $f$) and all real $x$ will do as the test functions, but

(i) this will exclude functions such as $V(x)=e^{x^2/4}$, for which $V*\phi$ is well defined and

(ii) I am not aware of a Fourier transform theory for the distributions corresponding to the test functions as in (3); cf. Hardy, A Theorem Concerning Fourier Transforms. Here we also to need to rigorously justify the transition from (1.5) to (2).

  1. We need to attach a rigorous meaning to the linear combination in (2).
$\endgroup$
  • $\begingroup$ I'd be happy to see an answer where $U$ is restricted to being monotonically increasing. $\endgroup$ – Matt F. Jun 4 '18 at 2:10
  • $\begingroup$ @MattF. : If you want $U$ to be monotonically increasing, you can have e.g. $U(x)=\sinh ax$ for $a>0$, still with $U'U'''\ne U''U''$. However, this monotonicity condition does not seem to be of help regarding the general solution of (1) -- all the three listed difficulties seem to be still there. $\endgroup$ – Iosif Pinelis Jun 4 '18 at 2:30
  • $\begingroup$ Interesting. Even better would be monotonically increasing and concave. And it would be better to remove the require of differentiability which I inserted accidentally, rewriting the condition as $EU(x+N(0,1))=a U(x)+b$. $\endgroup$ – Matt F. Jun 4 '18 at 2:42

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.