Utility from Linear Risk Aversion

Article Quant Q&A · Author: Tosh

Summary

The document derives a utility function whose relative risk aversion increases linearly with wealth. Starting from the definition of relative risk aversion, it turns the condition into a first-order differential equation for marginal utility, then integrates that result to obtain utility as an integral involving a power of wealth and an exponential term. The integration variable is a dummy variable used to distinguish the variable of integration from the upper bound.

The posted derivation contains errors in the questioner's integrating-factor work; the accepted answer instead separates variables to get marginal utility proportional to wealth raised to a constant power times an exponential decay. Integrating marginal utility yields the stated form, up to a positive scale constant. The treatment is a compact symbolic derivation rather than an empirical finance result. Its expression assumes a domain and parameter values for which the integral is defined; the document does not discuss boundary behavior or utility normalization.

Key ideas

Linear relative risk aversion specifies a differential equation for marginal utility.
Separating variables solves for marginal utility as a power of wealth times an exponential factor.
Utility is recovered by integrating marginal utility with respect to wealth.
The integration variable is a dummy variable, while the upper limit sets the wealth argument.

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Statistics Risk management

Full text

# Differentiating risk aversion based on utility theory


# Differentiating risk aversion based on utility theory












A utility function $U$ whose corresponding relative risk aversion function is a linear, increasing function satisfies the differential equation

$$-x\frac{U''(x)}{U'(x)}=ax+b$$

for some constants $a>o$ and $b\in \mathbf{R}$

Show that

$$U(x)=c \int _0 ^x t^{-b}e^{-at} dt$$ where $c>0$ is an arbitrary constant.

I massaged the equation so it becomes friendly.

$$-x \frac{d^2U}{dx^2}=(ax+b)\frac{dU}{dx}$$

Letting $\frac{dU}{dx}=v$

$$-x \frac{dv}{dx}=(ax+b)v$$

$$x \frac{dv}{dx}+axv=-bv$$

I.F

$$e^{\int (ax) dx}=e^{a\frac{x^2}{2}}$$

$$\therefore ve^{a\frac{x^2}{2}}=b\int e^{a\frac{x^2}{2}} dx$$

$$\therefore ve^{a\frac{x^2}{2}}=ba^2x e^{a\frac{x^2}{2}}+C$$

$$v=ba^2x+Ce^{-a\frac{x^2}{2}}$$

$$\frac{dU}{dx}=ba^2x+Ce^{-a\frac{x^2}{2}}$$

$$u=ba^2\frac{x^2}{2}+C\int e^{-a\frac{x^2}{2}} dx$$

But the answer is

$$U(x)=c\int^c_0 t^{-b}e^{-at}dt$$

I cannot understand where the $t$ comes to sit in the equation.

## Answer by Tosh (score 2, accepted)

https://quant.stackexchange.com/a/30249

$$-x \frac{d^2U}{dx^2}=(ax+b)\frac{dU}{dx}$$

Letting $\frac{dU}{dx}=v$

$$-x \frac{dv}{dx}=(ax+b)v$$

Rearranging

$$-\frac{1}{v} dv=\left(a+\frac{b}{x}\right)dx$$

$$ -\int(\ln{v}) dv= \int \left(a+\frac{b}{x}\right) dx $$

$$ v=C e^{-ax}x^b$$

Where $C$ is a constant. By taking $t$ as a dummy variable.

$$U(x)= C\int ^x_0e^{-at}t^{-b}\, dt$$

Shown in full with attribution under the source's licence. Licence: CC BY-SA 4.0 (Stack Exchange)

This summary was written by Stratmill's research agent from the original; it is not a copy of the source.

Deriving Utility from Linear Relative Risk Aversion

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