Deriving the risk-aversion coefficient

Question

By considering the parametrised formulation of the mean-variance criterion by Markowitz, the risk aversion coefficient $\lambda$ can be derived as follow.

1. As suggested by Arrow and Pratt, given the utility function of the investor $U(x)$, $\lambda$ for a specific level of initial wealth $x$ can be approximated by recurring to the absolute $A_a$ and relative $A_r$ Arrow-Prat risk aversion measures.

$$A_a(x)=-\frac{U''(x)}{U'(x)}$$

$$A_r(x)=-x\frac{U''(x)}{U'(x)}$$

2. Deriving the entire efficient frontier, it is possible to obtain $\lambda$ implicitly. It would be the one that leads to the preferred level of risk.

I was wondering if there are other approaches to compute such coefficient without the need of identify the utility function. I have been able find one that compute $\lambda$ as follow but I do not understand the idea behind it with the exception that it vaguely resemble the Safety First Ratio or Sharpe ratio with $r_f=0$. Specifically, if $\mu_B$ and $\sigma^2_B$ are respectively the expected return and variance of a benchmark $B$, then

$$\lambda=\frac{\mu_B}{2\sigma^2_B}.$$

It is peculiar the fact that for the same level of $\mu_B$, when $\sigma^2_B\to +\infty$ the coefficient $\lambda\to0$. Is this result compliant with theory of choice? It appears to be more likely related with prospect theory or since there are no other parameters this formula seems to imply only risk seeking behaviour.

Answer 1

I understand that you want to derive some form of risk preference parameter from portfolios that you can observe 'in the wild', and I will discuss that accordingly. As a side note, there is a whole thread in the literature that discusses elicitability of risk preferences using cleverly designed choice experiments -- and the form of the utility function. The link is one random example.

RE your question

The AP measures are defined locally and can be used (in theory) to compare risk aversion across agents. The AP measures require a functional utility form of the utility function, and require its first and second derivatives to be calculated.

In practice, you would thus need a way to compute a second derivative (numerically: at least three data points).

IF you assume a functional form in the first place, you can find its risk preference parameter under some additional restrictions, I think. Below, I will discuss two cases: One where you can obtain the parameter, and another one where this is not possible (I think).

Assumptions

Our agent is risk averse with CARA utility function $u(x)=1-e^{-\gamma x}$ with risk aversion parameter $\gamma>0$. The agent invests in some portfolio weights $w$ and for simplicity, we assume that the log returns are multivariate normally distributed, $x\sim N(\mu,\Sigma)$. As the agent wants to maximize expected utility, we thus have them:

$$ \begin{align} \max_{w}\mathrm{E(u(w))}&=\max_{w}\left(1-\mathrm{E}(e^{-\gamma w^Tx})\right)\\ &=\max_{w}\left(1-e^{-\gamma w^T\mu+\frac{1}{2}\gamma^2w^T\Sigma w}\right)\\ &\propto\max_{w}\left(w^T\mu-\frac{1}{2}\gamma w^T\Sigma w\right)\\ \end{align} $$ subject to $\sum_i w_i=1$, i.e. $w^Te=1$ with $e$ a vector of ones.

1. Efficient portfolio

In our first example, the agent faces not only the risky investment set $x$ but also a risk free rate $r_f$. Their portfolio optimization decision is hence

$$ \max_{w}\quad w^T\mu-\frac{1}{2}\gamma w^T\Sigma w+\left(1-w^Te\right)r_f $$

with optimality condition

$$ \gamma \Sigma w=\mu-er_f $$

Clearly, once we observe the optimal risky portfolio $w^*$, we can rewrite the optimality condition and find

$$ \gamma (w^*)^T\Sigma w^*=(w^*)^T(\mu-er_f) \Rightarrow \gamma = \frac{(w^*)^T(\mu-er_f)}{(\sigma^*)^2} $$

2. No risk free investment If, on the other hand, there is no risk free investment available, the agent maximizes their expected utility under a full investment restriction, resulting in the FOC:

$$ \begin{align} \gamma\Sigma w -\lambda e &= \mu\\ w^Te&=1 \end{align} $$

Since we will only be able to observe their 'optimal' portfolio $w^*$ and not their optimal Lagrange parameter $\lambda$, we cannot elicit their risk aversion parameter $\gamma$ in this case.

HTH?

Addendum

To get towards an answer to your other comments / post:

Say you want to measure the risk aversion parameter given a utility form (CARA, as above) and an observed fraction of wealth that is invested in the risky asset. Then we should at first note that this is inherently the same as my example 1. from above, but in a univariate setting without a risk free rate. Nevertheless, let me try to sketch the path:

Everything is assumed as above, and the agent decides on a share of wealth $W$ (at this point, not restricted between 0% and 100%) that is invested in the risky asset. Let us simplify and set $W=1$, then the risky consumption is

$$ c=(1-\alpha)+\alpha x $$

and with $x\sim N(\mu,\sigma^2)$, expected utility is

$$ EU(\alpha)=1-e^{-\lambda (1-\alpha)-\alpha\lambda\mu+\frac{1}{2}\alpha^2\lambda^2\sigma^2} $$

Optimization of the expected utility is akin to maximizing the following

$$ \max_{\alpha} \quad 1-\alpha + \alpha\mu-\frac{1}{2}\alpha^2\lambda\sigma^2 $$

with FOC

$$ \mu-1 =\alpha\lambda\sigma^2 $$

and hence you are able to back-out the parameter $\lambda$ from an observed investment fraction $\alpha$ as

$$ \lambda^* = \frac{\mu-1}{\alpha \sigma^2} $$

NB: Don't worry about the $-1$ in the nominator, this stems from the way returns and utility are setup. If done more carefully, you'd indeed arrive at $\lambda^* = \frac{\mu}{\alpha \sigma^2}$

HTH?

Answer 2

EDITED Let $x$ be the available investor's wealth. Given a benchmark $B$ which can be considered as a proxy to the market portfolio, let $x_B$ the amount of invested wealth. Let also $\mu_b$ and $\sigma^2_B$ be the expected return and variance of $B$, respectively.

The investor solves a trade off between investing and not investing by considering the risk-return profile of $B$. Personal preferences are expressed by means of $\lambda$.

$$\max_{x_B}\mu_Bx_B-\lambda x_B^2\sigma_B^2$$

Then the first order condition is

$$\frac{\partial f}{\partial x_B}=\mu_B-2\lambda x_B\sigma_B^2=0$$ which leads to

$$\lambda=\frac{\mu_B}{2\sigma^2_Bx_B}$$