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How Utility and Return Distributions Shape Mean-Variance Allocation

Article Quant Q&A · Author: Machinus

Summary

The document examines when optimizing expected utility leads to an investment fraction proportional to expected return divided by variance. It writes the first-order condition for an investor choosing a fraction of wealth in a risky asset, then identifies assumptions that reduce the condition to a balance between expected return and a second-moment or variance term.

The examples include quadratic utility with finite first and second return moments, a second-order approximation to utility, a directly specified mean-variance objective, and exponential utility with normally distributed returns. The latter yields an allocation linked to expected return-to-variance and relative risk aversion. These are conditional derivations, not a universal risk measure: the result depends on the chosen utility, distribution, and objective. The document also notes uncertainty about whether similar results extend to broader distribution families.

Key ideas

  • A mean-variance allocation follows from specific utility and return-distribution assumptions.
  • Quadratic utility can produce an allocation proportional to expected return divided by a second moment.
  • A second-order utility approximation can reduce a broader problem to a similar form.
  • Exponential utility with normal returns links optimal allocation to relative risk aversion and the return-to-variance ratio.
  • The derivation does not establish that mean-variance form applies universally.

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Full text
# What is the proper way to derive risk definitions from utility functions?


# What is the proper way to derive risk definitions from utility functions?












In typical mean-variance analysis, the risk-adjusted relative value of an individual asset takes the general form

$\frac{\mu}{\sigma^2}$,

with further weighting and normalization depending on the particular optimization conditions.

In economics books, I have also read the suggestion that this structure is a consequence of assuming a quadratic utility function in combination with the assumption of gaussian returns. (Is that sufficient?)

I have seen some examples of expectation value computations, but I have not seen the general connection between utility functions, distributions, and the form of risk derived explicitly. What is the proper procedure to derive the form of risk-adjusted value from an arbitrary utility function and distribution? Or is there a good reference that shows this?

## Answer by Kermittfrog (score 2, accepted)

https://quant.stackexchange.com/a/79362

Note: This is (still) the starting point to a proper answer. If time permits, I'll add some flesh from time to time.

From a mathematical point of view, in order to obtain the optimal investment fraction $\alpha$ to be of the shape $\alpha^*\propto\frac{\mathrm{E}(r)}{\mathrm{E}(r^2)}$ (or $\alpha^*\propto\mu/\sigma^2$), the first order condition of the optimization problem must be of the form

$$ c_0\mathrm{E}(r)+c_1\alpha\mathrm{E}(r^2)=0 $$

where $c_0,c_1$ are some constants specific to the problem. Thus, our problem boils down to finding meaningful combinations of utility functions and return distribution functions that jointly yield

$$ \begin{align} & \max_\alpha\mathrm{E}(u(W_0(1-\alpha+\alpha (1+r))))\\ =&\max_\alpha\mathrm{E}(u(W_0(1+\alpha r)))\\ \Rightarrow 0\stackrel{!}{=}&W_0\mathrm{E}(ru'(W_0(1+\alpha r)))\\ \propto&c_0\mathrm{E}(r)+c_1\alpha\mathrm{E}(r^2) \end{align} $$

I can think of three sets of specific assumptions that result in this form, and maybe there exists a unifying class of $\mu-\sigma-$problems.

- Assume a quadratic utility function $u(w)=c_0w-\frac{1}{2}c_1w^2$ and (any) distribution function with finite first and second moments.

- Taylor approximate the utility function to second order and apply step 1 above,

- Directly impose a utility / goal function of the form $c_0\mathrm{E}(W)-\frac{1}{2}c_1\mathrm{E}\left(W-\mathrm{E}(W)\right)^2$

- Assume an exponential utility function and normally distributed returns (maybe it suffices for the return distribution to be from the exponential family of distributions, but I cannot prove that)

For cases 1 and 2, the solution can be directly read off the problem formulation as $u'$ has the form $c_0+c_1\alpha r$.

For case 3, we find $\mathrm{E}(W)=W_0(1+\alpha\mathrm{E}(r))$ and $\mathrm{E}\left(W-\mathrm{E}(W)\right)^2=W_0^2\alpha^2\mathrm{Var}(r)$, thus

$$ \begin{align} 0&\stackrel{!}{=}\frac{\partial}{\partial \alpha}\left(c_0\mathrm{E}(W)-\frac{1}{2}c_1\mathrm{E}\left(W-\mathrm{E}(W)\right)^2\right)\\ &=\frac{\partial}{\partial \alpha}\left(c_0W_0(1+\alpha\mathrm{E}(r))-\frac{1}{2}c_1W_0^2\alpha^2\mathrm{Var}(r)\right)\\ &=c_0W_0\mathrm{E}(r)-\alpha c_1W_0^2\mathrm{Var}(r)\\ \Rightarrow & \alpha^*=\frac{c_0}{W_0 c_1}\frac{\mathrm{E}(r)}{\mathrm{Var}(r)} \end{align} $$

For case 4, we have the exponential utility of the form $u(w)=\frac{1-e^{-\lambda w}}{\lambda}$, i.e. $u'(w)\propto e^{-\lambda w}$. Thus:

$$ 0\stackrel{!}{=}W_0\mathrm{E}\left(re^{-\lambda W_0(1+\alpha r)}\right) $$

Assuming $r\sim\mathrm{N}\left(\mu,\sigma^2\right)$, the expectation is obtained in close form as

$$ \begin{align} 0&\stackrel{!}{=}W_0e^{-\lambda W_0-\alpha\lambda\mu W_0+\frac{1}{2}\alpha^2\lambda^2\sigma^2 W^2}(\mu-\alpha\lambda\sigma^2 W)\\ \Rightarrow \alpha^*&=\frac{1}{\lambda W}\frac{\mu}{\sigma^2}\\ &=\left(-W_0\frac{u''}{u'}\right)^{-1}\frac{\mu}{\sigma^2}\\ &=\frac{1}{\rho(W_0)}\frac{\mu}{\sigma^2} \end{align} $$

where we have introduced the coefficient of relative risk aversion

$$ \rho(w)\equiv-\frac{wu''(w)}{u'(w)} $$

## Answer by Machinus (score 1)

https://quant.stackexchange.com/a/79368

My reply was too long, so I am posting it as a separate comment. Thank you Kermittfrog for the derivation! (I think you meant to write $\rho(W_0)$ instead of $\rho(w)$ in the value of $w^*$.)

Since the question is specifically about the implications of a quadratic $u$, there are only two terms to take in the Taylor series, and this approximation seems acceptable. But, even if the utility function were of higher order, you could still approximate the value of $u$ by keeping a finite number of terms. For example, I've seen a paper that shows that up to fourth order, the terms in the expectation values are still of the form $<r>$ or $<r^2>$, which I think is part of why those two statistics are so heavily used. That is an interesting outcome, but I think this method is a little primitive, regardless. Is polynomial approximation the canonical way of expressing risk-adjusted value? Are there other ways used, maybe more mathematically difficult and/or only applying with some constraints on $u$ or $r$?

I originally became interested in this question by looking for alternative measures of risk-adjusted value based on a variety of return distributions, but the quantitative application of "utility" was (and still is) somewhat vague to me. This helps clarify that nothing mysterious is going on.

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.