# 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?

• Upvoting because I am interested in an answer as well. May 7 at 4:51
• This would probably be a better fit for Economics Stack Exchange. May 7 at 6:47
• May 7 at 12:50
• Could you please define the term "risk adjusted relative value" or provide a literature reference? Thanks! May 8 at 15:32

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.

1. 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.
2. Taylor approximate the utility function to second order and apply step 1 above,
3. Directly impose a utility / goal function of the form $$c_0\mathrm{E}(W)-\frac{1}{2}c_1\mathrm{E}\left(W-\mathrm{E}(W)\right)^2$$
4. 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

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

• Thank you! I posted my reply as a comment. May 14 at 0:57
• The notation $E[x^2]$ looks like it represents the second moment of the distribution, which is not the variance of the distribution. In the case of gaussian returns, that would mean $E[x^2] = \sigma^2 + \mu^2$. How do you interpret $E[f(x)]$? May 14 at 23:42
• Yes, in order to arrive at a result that incorporates $E((x-E(x))^2)$ we need more structure, see my amended answer. Please explain what you mean by $E(f(x))$? May 15 at 6:58
• Do you mean E is expectation value? If so, the denominator of the weight would be different, giving $w = \frac{\mu}{\mu^2+\sigma^2}$ May 15 at 15:47

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 $$$$ or $$$$, which I think is part of why those two statistics are so heavily used. That is an interesting outcome, but I think this method is a little primitive, regardless. Is polynomial approximation the canonical way of expressing risk-adjusted value? Are there other ways used, maybe more mathematically difficult and/or only applying with some constraints on $$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.

• It would be more appropriate if you posted your "answer", for example, in two parts as comments to the related answer. May 14 at 8:24