<p>@Alecos's answer is great. For pedagogical purposes, I'm just going to rephrase some of the steps.</p>

<p>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} &amp;\approx u(c) + \frac 12 u''(c) \\
ARA &amp;\approx -\frac{u''(c)}{u'(c)}.
\end{align}</p>

CC BY-SA 3.0
2014-12-30T23:58:25.350Z
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).
nominally rigid
CC BY-SA 3.0
2014-12-31T00:18:08.123Z
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' ?
jmbejara
CC BY-SA 3.0
2014-12-31T00:34:34.400Z
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.
nominally rigid
CC BY-SA 3.0
2014-12-31T20:08:31.710Z
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?
jmbejara
CC BY-SA 3.0
2015-01-05T06:38:44.320Z
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.)
nominally rigid
CC BY-SA 3.0
2014-12-30T23:46:46.207Z
2014-12-30T23:46:46.207Z
jmbejara
<p>Set $y \equiv c+\varepsilon$. So $y$ represents <em>changed</em> 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 :</p>

<p>$$E[u(y)\mid c] \approx E[u(c)\mid c] + E[u'(c)(y-c)\mid c] + E[\frac12u''(c)(y-c)^2\mid c]$$</p>

<p>But $y-c = \varepsilon$<br>
so</p>

<p>$$E[u(y)\mid c] \approx E[u(c)\mid c] + E[u'(c)\varepsilon\mid c] + E[\frac12u''(c)\varepsilon^2\mid c]$$</p>

<p>Due to the conditioning, and the independence of $\varepsilon$ from $c$ the expected value distributes and applies only to $\varepsilon$:</p>

<p>$$E[u(y)] \approx u(c) + u'(c)E[\varepsilon] + \frac12u''(c)E[\varepsilon^2]$$</p>

<p>Since we assume </p>

<p>$$E[\varepsilon] = 0 \Rightarrow {\rm Var}(\varepsilon)=E[\varepsilon^2] =1$$
we obtain</p>

<p>$$E[u(c+\varepsilon)] \approx u(c) + \frac12u''(c)  \tag{1}$$</p>

<p>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 $c$:</p>

<p>$$u\left(c-\frac 12 ARA\right) = u(c) + u'(c)\cdot (c- \frac 12 ARA - c) = u(c) - \frac 12 u'(c)\cdot ARA $$</p>

<p>Using the definition of $ARA$ to replace it we obtain</p>

<p>$$u\left(c-\frac 12 ARA\right) \approx u(c) - \frac 12 u'(c)\cdot\left (-\frac{u''(c)}{u'(c)}\right)$$</p>

<p>$$\Rightarrow u\left(c-\frac 12 ARA\right) \approx u(c) + \frac 12 \cdot u''(c) \tag{2}$$</p>

<p>The right-hand sides of equations $(1)$ and $(2)$ are equal therefore, approximately, so  are their left-hand sides or,</p>

<p>$$E[u(c+\varepsilon) \mid c] \approx u\left(c-\frac 12 ARA\right)$$</p>

<p>which is valid as long as $ARA$ is defined the way it is. QED.  </p>

<p>If the variance of the gamble is not unity but $\sigma^2 \neq 1$, then the more general equation is</p>

<p>$$E[u(c+\varepsilon) \mid c] \approx u\left(c-\frac {\sigma^2}2 ARA\right)$$</p>

CC BY-SA 3.0
2014-12-30T13:03:49.227Z
2014-12-31T02:49:11.113Z
2014-12-31T02:49:11.113Z
Alecos Papadopoulos
<p>In its <a href="https://en.wikipedia.org/wiki/Risk_aversion" rel="nofollow">Wikipedia article</a>, 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 $ARA = -U''(C) / U'(C)$?</p>

CC BY-SA 3.0
2014-12-30T17:30:17.813Z
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?
nominally rigid
<decision-theory><risk>
Definition of Absolute Risk Aversion
jmbejara
# Definition of Absolute Risk Aversion

(Asked by: username_0 on 2014-12-30)

In its [Wikipedia article](https://en.wikipedia.org/wiki/Risk_aversion), 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 $ARA = -U''(C) / U'(C)$?

**Tags:** decision-theory, risk

> **Comments:**
> **username_1**: 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?  

---

## 2 answers

### Answer by username_0 on 2014-12-30. Score: 4

@username_2'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 username_2' 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}

> **Comments:**
> **username_1**: 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).  
> **username_0**: 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' ?  
> **username_1**: 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.  
> **username_0**: 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?  
> **username_1**: 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.)  

---

### Answer by username_2 on 2014-12-30. Score: 4

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

$$E[u(c+\varepsilon) \mid c] \approx u\left(c-\frac {\sigma^2}2 ARA\right)$$


---

