Limits of Taylor Approximations in Mean-Variance Portfolio Choice
Summary
The document considers whether a second-order Taylor expansion of expected utility can select a preferred portfolio from the mean-variance efficient frontier. It illustrates the idea using power utility, final wealth tied to returns, and a normal-return assumption. The response says this approximation can be useful in practice when risks are small, but it is not generally equivalent to maximizing exact expected utility.
The explanation expands utility around expected wealth and retains the variance term, while higher-order effects are omitted. A constructed comparison shows that approximations through several orders can rank two normal return distributions differently from the exact utility comparison, with the ranking aligning only when more terms are included. The example is deliberately unusual, and the text offers no general error bound or universal threshold for when the approximation is safe. Portfolio rankings may therefore require checking exact expected utility when approximation error could matter.
Key ideas
- A second-order Taylor expansion approximates expected utility using expected wealth and a variance adjustment.
- The approximation can be practical when risks are small.
- Omitted higher-order terms can change the ranking of investment choices.
- The example shows that several low-order approximations may disagree with exact expected utility.
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# Utility Theory and Mean Variance Analysis
# Utility Theory and Mean Variance Analysis
I was wondering if it's pertinent to use this interpretation of the expected utility function given by the Taylor series expansion,
$${E(U(W)}\approx{U[E(W)}]+\frac{U''[E(W)]\sigma^2_W}{2}\tag{1}$$
to delineate my optimal portfolio from the set of portafolios that lie on the efficient frontier?
So, let's say I were to have the power utility function,
$$\frac{W^{1-\gamma}}{1-\gamma}\tag{2}$$ and defined final wealth as $W=W_0(1+r)$, and $\gamma=.5$, I also assume that returns are normally distributed. Can I just plug this function and its second derivative back into 1, and then find the max utility among the set of efficient portfolios, by plugging each portfolio's $[E(r),\sigma^2]$? Am I allowed to do this? Is this completely wrong? Thanks !
## Answer by Kermittfrog (score 5)
https://quant.stackexchange.com/a/70507
Theoretically: no.
For most practical purposes: yes; given that risks are small risks, see these lecture notes on p76.
Belows's the background and one example showing you why you can run into problems:
Given two portfolio compositions $P_i,P_j$ with $(\mu_i,\sigma_i)$ and $(\mu_j,\sigma_j)$, the agent prefers $i$ over $j$ if
$E(u(P_i))\geq E(u(P_j))$.
In your example, the Taylor approximation reads
$$\begin{align} E\left[u(1+x)\right]&=E\left[u(1+\mu+x-\mu)\right]\\ &=E\left[u(1+\mu)+\sum_{k=1}^{\infty}\left.\frac{\partial^k u}{\partial x^k}\right|_{x=(1+\mu)}\frac{(x-\mu)^k}{k!}\right]\\ &\approx u(1+\mu)+\frac{1}{2}u''(1+\mu)\sigma^2\\ &\equiv \tilde{E}_2[u(1+x)] \end{align} $$
This approximation holds well in most cases, but it will break in corner cases as higher order terms are lost in the approximation. Note that each contribution from the higher order terms is negative; thus we can have $E_2(u)$ induce a different ordering of investment opportunities than $E(u)$, as $E_2(u)$ does not consider all higher order effects.
Here's an (somewhat contrived) example: Given the two normal distributions in the table below, we see that the second order Taylor approximation (E2) yields a preference for distribution 2 over 1, the same holds when increasing the length of the Taylor approximation up to the sixth order (E4,E6). Only after incorporating the eighth order (or more), we see that the Taylor approximation reveals the true preference relation, i.e. that distribution 1 is preferred over 2.
```
i mu sigma E2 E4 E6 E8 .. E(u)
1 0.0795066 0.10 - - - + .. +
2 0.0800000 0.11 + + + - .. -
```
For most practical purposes, this error should be sufficiently small. Eg when comparing investment choices, the error bound should be so tight that you can work with approximate results.
HTH?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.