# Is DARA utility implying CRRA most of the time?

The Wikipedia page on risk aversion states that a "Constant Relative Risk Aversion implies a Decreasing Absolute Risk Aversion, but the reverse is not always true". Let me decompose this statement in two parts:

1/ "Constant Relative Risk Aversion implies a Decreasing Absolute Risk Aversion."

A simple example is the log utility function, $u(c) = \ln(c)$, with $c>0$ satisfies the DARA because the utility function is positively skewed $\left(u'''=\frac{2}{c^3} >0\right)$ and implies a Relative Risk Aversion equals to $1 \left(=-c\frac {u''(c)}{u'(c)}\right)$.

2/ "but the reverse is not always true".

I am wondering if this is the most frequent case? Or if most of the time DARA utility functions also exhibit CRRA?

I would be grateful if you can illustrate your answer with some utility functions.

• How about functions with decreasing absolute and decreasing relative risk aversion? – HRSE Feb 26 '16 at 8:43
• "Most of the time CRRA utility functions also exhibit DARA": yes, this is exactly your first point, isn't it? Or did you mean the opposite? – Oliv Feb 26 '16 at 10:37
• Thanks @Oliv. I meant the opposite. Sorry for the confusion. – emeryville Feb 26 '16 at 15:49
• – Herr K. Feb 26 '16 at 16:06

Using the results derived in this answer we have the following relations for any utility function:

(Absolute Risk Aversion = $A(c)$, Relative Risk Aversion = $R(c)$) : $$A(c) = -\frac {u''(c)}{u'(c)},\;\;\; R(c) = cA(c), \;\; A(c) = \frac 1c R(c)$$

and so

$$\frac {\partial A(c)}{\partial c} = \frac {\partial [(1/c)R(c)]}{\partial c}= -\frac 1{c^2} R(c) + \frac 1{c}\frac {\partial R(c)}{\partial c} \tag{1}$$

and

$$\frac {\partial R(c)}{\partial c} = \frac {\partial [cA(c)]}{\partial c}=A(c) + c\frac {\partial A(c)}{\partial c} \tag{2}$$

Remembering that these measures are mainly discussed for utility functions where $u''<0$, and so they are seen as algebraically positive, we can deduce which relations hold with certainty and which do not.

Now $ARA$ satisfies as an identity

$$u''+ A(c)u' \equiv 0 \implies u''' + A'u' + A u'' = 0 \implies u''' = -A'u' + A (-u'')$$

The $DARA$ family is characterized by $A' <0$ so we obtain that (sufficient)

$$DARA \implies u'''>0$$

So assuming $DARA$ what we gain in knowledge is that $u'''>0$.

In turn $RRA$ satisfies as an identity

$$cu''+ R(c)u' \equiv 0 \implies u''+ cu''' + R'u' + R u'' = 0$$

$$\implies c u''' = -R'u' + (-u'')(1+R)$$

Assuming $DARA$, we have $u'''>0$ and so what we know is that

$$-R'u' + (-u'')(1+R) > 0 \implies (-u'')(1+R) > R'u'$$

$$\implies R(1+R) > c\cdot R'$$

This is the constraint that $DARA$ imposes on $RRA$. We see that the inequality can be satisfied with $R'$ negative or zero, and even positive, up to a degree.

• I expected a brilliant answer and I got it! Thanks a lot Alecos. – emeryville Feb 27 '16 at 5:09

Let me turn my comment into a quick answer: Using the notation of the article you quoted $A(c)$ is the absolute risk aversion and $c A(c)$ the relative risk aversion. If $A(c)$ is decreasing, the preferences fulfill DARA. If $c A(c)$ is constant, the preferences fulfill CRRA. If CRRA holds, then $A(c)$ must be decreasing in $c$.

If we take any $A(c)$ such that $c A(c)$ is decreasing and $A(c)$ is positive, then $A(c)$ will also be decreasing. In this case, the preferences are both DARA and DRRA. You asked whether DARA implies CRRA in almost all cases. From this analysis, it seems to be rather the case that CRRA is the exceptional case.

• Thanks @HRSE. So it would have been more precise to say "but the reverse is generally not true", right? – emeryville Feb 26 '16 at 16:02