# Why do we assume quadratic utility in portfolio theory?

In my text (Investments by BKM), the investor's mean-variance utility (given as $U = E[R] - \frac12A\sigma^2$) is stated to be the objective function we wish to maximize. Upon further digging, it seems that this stems from the assumption of quadratic utility functions ($U = aW - bW^2$). This kind of bothers me since I see two unrealistic properties for quadratic utility functions. (1) They exhibit increasing absolute risk aversion, and (2) they achieve a satiation point, beyond which money/return begins to have negative value.

So why do we assume quadratic utility? Are there no other simple, more realistic functional forms for utility that would still lead to a reasonably clean portfolio optimization theory? Or are the issues I cited about the quadratic just negligible in practice?

• If utility is quadratic then we can do optimization on it. So it is convenient to have quadratic function. – Naz Apr 30 '16 at 13:10
• It is not $U = E[R] - \frac12A\sigma^2$ but $E(U) = E[R] - \frac12A\sigma^2$. – Richard Hardy Feb 7 '20 at 9:00

## 3 Answers

The assumption of quadratic utility function is very convenient in ptf Theory because it is possible to demonstrate that also if the ptf return are not normally distributed the mean-variance approach is still the best. The best in the sense that any other distributional properties is amenable into mean and variance. For converse, if the return are normally distributed the choice of utility function is irrelevant. More in general, if the ptf return are not known distribution and we use a general utility function, the mean-variance approach is valid yet, but only as approximation.

• More precisely, under normality the choice of utility function is irrelevant up to the constants specifying the relative importance of mean vs. variance reflecting the degree of risk aversion. – Richard Hardy Feb 7 '20 at 9:03

In most settings, utility functions are defined up to an affine transformation: if $u(x)$ defines the preference of an investor, then so does $a*u(x)+b.$ This implies, you can normalize the Taylor expantion of any smooth utility function to $u(x)=x+a*x^2+\ldots$ around 0. So the next step is just to drop off higher order terms. The investor is also usually assumed to be risk-averse, which implies, that $a < 0.$ You can check the details, e.g. here: https://www.empiwifo.uni-freiburg.de/lehre-teaching-1/winter-term-10-11/materialien-portfolio-analysis/utility.pdf

What you learn in school are models, meant to illustrate the concepts and methods of the field. Later you will learn about other forms of utility functions (power utility most prominently). With such families of utility functions the computations aren't as clean as with quadratic utility, but by then you will have understood the concepts and methods, and you will understand the approximate methods that you will need to use at that stage.

With that said, Markowitz tried, in a few papers, to explain why approximating one's utility function by a quadratic utility function makes sense in some cases. Not very convincing in my opinion, but it's out there.

• so you mean this is due to pedagogical purpose rather than pratical one? – SiXUlm May 31 '16 at 12:12
• At this time, pretty much, yes. In real life there are almost no practitioners (people with PhD's in math and physics working in finance) who use quadratic utility functions in their work. Power utility (including logarithmic utility) are the most commonly used family. – mathguy May 31 '16 at 12:17
• I do not know myself what practitioners actually use, but in this answer there is a counterargument to yours. See paragraphs starting with First, most quants and Why do the Nobels matter?. – Richard Hardy Feb 7 '20 at 9:16
• @RichardHardy - I do (or did) know what practitioners actually use. I am a mathematician by training (PhD, taught up to graduate level courses at a top university - including in statistics and in financial math). Math is the avenue by which I was able to enter the actual practice of finance, retiring in 2015 as managing partner of a firm that gives investment advice based on modern portfolio theory and such. It is true that most people use mean-variance models; the Answer you linked to, in some places, seems to confuse between that and "quadratic utility". – mathguy Feb 7 '20 at 20:04
• @RichardHardy - Mean-variance models work equally validly with other classes of utility functions - as long as utility is a two-parameter family (so, it is completely determined by its mean and variance). "Power utility", for example, is in this class. But the math calculations can no longer be done in closed form. A large number of "quants" do use power utility; honestly, I can't recall ever meeting one who assumed quadratic utility. And the bit about the Nobels providing a legal safe harbor is pure BS - that is exactly the business I was in, and there is no such safe harbor. – mathguy Feb 7 '20 at 20:07