In its Wikipedia article, absolute risk aversion is defined as $ARA = -\frac{u''(c)}{u'(c)}$. However, I have alternatively seen absolute risk aversion defined as half the decrease in consumption that an investor is willing to accept to avoid a gamble $\varepsilon$, where $E[\varepsilon] = 0$, $E[\varepsilon^2] = 1$, and $\varepsilon$ is independent of consumption: $$ U(C - ARA/2) \equiv E[U(C + \varepsilon) \mid C]. $$ Using the latter definition, how can I show that $ARA = -U''(C) / U'(C)$?
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$\begingroup$ why is there a lowercase $c$ and conditional expectation on the right? and don't we need to specify that this is an infinitesimal gamble, rather than having a constant variance 1? $\endgroup$– nominally rigidDec 30, 2014 at 17:30
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2 Answers
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@Alecos's answer is great. For pedagogical purposes, I'm just going to rephrase some of the steps.
We want to show that $ARA = -u''(c)/u'(c)$ given that ARA is defined such that $u(c - ARA/2) = E[u(c + \varepsilon) \mid c]$. So, following Alecos' answer, take a 2nd-order Taylor expansion to get \begin{equation} E[u(c + \varepsilon)\mid c] \approx u(c) + \frac 12 u''(c). \end{equation} Then by definition, $u(c - ARA/2) \approx u(c) + \frac 12 u''(c)$. Now, taking of 1st order Taylor series expansion of the left-hand side of this expression, we see that \begin{equation} u(c - ARA/2) \approx u(c) - u'(c) \cdot \frac{ARA}{2}, \end{equation} which implies that \begin{align} u(c) - u'(c) \cdot \frac{ARA}{2} &\approx u(c) + \frac 12 u''(c) \\ ARA &\approx -\frac{u''(c)}{u'(c)}. \end{align}
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1$\begingroup$ this is certainly right and well-written, but I still don't think it's a good idea to do this with $E[\epsilon^2]=1$; in that case, the relation holds only as an approximation, and it may be a very poor one depending on the relative levels of $c$ and $\epsilon$. In the limit $E[\epsilon^2]\rightarrow 0$ it holds exactly, and that's the clearest way to define risk aversion here (in the limit, as a coefficient that relates the variance of an infinitesimal gamble to its consumption equivalent). $\endgroup$ Dec 30, 2014 at 23:58
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$\begingroup$ Thanks for the comment. I was wondering about this. So, just to clarify, are you saying that for this reason, it's usually better to just define ARA as -u''/u' ? $\endgroup$– jmbejaraDec 31, 2014 at 0:18
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$\begingroup$ well, that's the more convenient way, but it's equally valid to define ARA as the limit, as you take $\sigma^2=E[\epsilon^2]\rightarrow 0$, of twice the ratio of the consumption equivalent to $\sigma^2$. For it to be a precise definition it needs to be explicit about this limit, though, rather than taking a finite $\sigma^2$ as given. $\endgroup$ Dec 31, 2014 at 0:34
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1$\begingroup$ I'm sorry, I'm just not understanding. So, you would define ARA such that $\lim_{\sigma^2 \rightarrow 0} E[u(c + \varepsilon) \mid c] = \lim_{\sigma^2 \rightarrow 0} u(c + \frac{\sigma^2}{2} ARA)$? But in this case, doesn't $\lim_{\sigma^2 \rightarrow 0} u(c + \frac{\sigma^2}{2} ARA) = u(c)$ regardless of the value of ARA? $\endgroup$– jmbejaraDec 31, 2014 at 20:08
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$\begingroup$ yes, the limits are the same, so it's about the derivative. Suppose $\epsilon$ has mean 0 and variance 1, and define the certainty equivalent $z(c,\gamma)$ implicitly as the $z$ satisfying the equation $E[u(c+\sqrt{\gamma}\epsilon)] = u(c-z)$. Then the derivative of $z$ with respect to the variance $\gamma$, at $\gamma=0$, is $ARA(c)/2$. In principle, this can be used as a definition of $ARA$. (Btw, if we want to write $z$ in terms of sd $\sigma$ rather than variance $\gamma=\sigma^2$, then we need to talk about the second derivative. Also see MWG problem 6.C.20, which is related.) $\endgroup$ Jan 5, 2015 at 6:38
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Set $y \equiv c+\varepsilon$. So $y$ represents changed consumption around and "close" to a given level $c$. Take a 2nd-order Taylor expansion of the function $E[u(y)\mid c]$ around $c$, which is treated as fixed since we condition on it :
$$E[u(y)\mid c] \approx E[u(c)\mid c] + E[u'(c)(y-c)\mid c] + E[\frac12u''(c)(y-c)^2\mid c]$$
But $y-c = \varepsilon$
so
$$E[u(y)\mid c] \approx E[u(c)\mid c] + E[u'(c)\varepsilon\mid c] + E[\frac12u''(c)\varepsilon^2\mid c]$$
Due to the conditioning, and the independence of $\varepsilon$ from $c$ the expected value distributes and applies only to $\varepsilon$:
$$E[u(y)] \approx u(c) + u'(c)E[\varepsilon] + \frac12u''(c)E[\varepsilon^2]$$
Since we assume
$$E[\varepsilon] = 0 \Rightarrow {\rm Var}(\varepsilon)=E[\varepsilon^2] =1$$ we obtain
$$E[u(c+\varepsilon)] \approx u(c) + \frac12u''(c) \tag{1}$$
Now consider $u(c-\frac 12 ARA)$, with $ARA \equiv -\frac{u''(c)}{u'(c)}$, and take a 1st-order Taylor expansion in this case, again around $c$:
$$u\left(c-\frac 12 ARA\right) = u(c) + u'(c)\cdot (c- \frac 12 ARA - c) = u(c) - \frac 12 u'(c)\cdot ARA $$
Using the definition of $ARA$ to replace it we obtain
$$u\left(c-\frac 12 ARA\right) \approx u(c) - \frac 12 u'(c)\cdot\left (-\frac{u''(c)}{u'(c)}\right)$$
$$\Rightarrow u\left(c-\frac 12 ARA\right) \approx u(c) + \frac 12 \cdot u''(c) \tag{2}$$
The right-hand sides of equations $(1)$ and $(2)$ are equal therefore, approximately, so are their left-hand sides or,
$$E[u(c+\varepsilon) \mid c] \approx u\left(c-\frac 12 ARA\right)$$
which is valid as long as $ARA$ is defined the way it is. QED.
If the variance of the gamble is not unity but $\sigma^2 \neq 1$, then the more general equation is
$$E[u(c+\varepsilon) \mid c] \approx u\left(c-\frac {\sigma^2}2 ARA\right)$$
answered Dec 30, 2014 at 13:03