<p>It is worth noting that the "risk premium" you are talking about is in fact more accurately referred to as the Arrow-Pratt approximation of the cost of a small additive risk. You can approximate it however you want (and a Taylor series approximation is an unbiased way of doing so), but ultimately you only need to do that to provide a simple explanation of other, more interesting concepts on the economics of risk.</p>

<p>My understanding for the different levels of approximation was the different levels of variance and movement in the functions you are approximating. Expected utility will vary in a concave way with the added riskiness of x, whereas the utility of the difference between the certainty equivalent and the risk premium will be much closer to a linear function.</p>

CC BY-SA 3.0
2015-11-10T06:55:12.053Z
Thanks! This answer helps as well. I share your understanding of the different levels of approximation.
emeryville
CC BY-SA 3.0
2015-11-09T16:18:37.660Z
2016-04-06T06:38:13.897Z
2016-04-06T06:38:13.897Z
globule
<blockquote>
  <p><em>Is there any (economic) rational for the first-order expansion of the RHS? And for its different neighborhood evaluation?</em></p>
</blockquote>

<p>As for your first question:<br>
This is a purely mathematical tactic in order to obtain an (approximate) equation for $R$. The expansion of first order on the RHS is motivated by this fact, i.e. to bring $R$ alone "in the surface".  The reason why the LHS is subjected to a second-order expansion is in order for something to be left (the variance term). Higher-order expansions of the LHS can certainly be applied.</p>

<p>As for your second question, you are making a mistake. The center of expansion for the RHS expansion is the same as that for the LHS expansion, namely $w-E(x)$ (or equivalently, around $R=0$). It is meaningless (and it fails) to expand a function around its exact argument. Specifically we have</p>

<p>$$u\left(w+E(x)-R\right) \approx u\left(w+E(x)\right) + u'\cdot [(w+E(x)-R)-(w+E(x))] = u\left(w+E(x)\right) - u'\cdot R$$</p>

<p>Finally, why not consider a second-order expansion on the RHS? We would then get</p>

<p>$$u\left(w+E(x)-R\right) \approx u\left(w+E(x)\right) + u'\cdot [(w+E(x)-R)-(w+E(x))] \\+\frac 12 u''\cdot [(w+E(x)-R)-(w+E(x))]^2 $$</p>

<p>$$= u\left(w+E(x)\right) - u'\cdot R + \frac 12 u''\cdot R^2$$ </p>

<p>Then we would obtain a quadratic polynomial in $R$,</p>

<p>$$\frac 12 u''\cdot R^2 - u'\cdot R - \frac 12 u''\sigma^2_x = 0$$</p>

<p>$$\ R^2 - \frac {2u'}{u''}\cdot R - \sigma^2_x = 0$$</p>

<p>This has roots</p>

<p>$$R_1,R_2 = \frac {(2u'/u'') \pm \sqrt{(2u'/u'')^2+4\sigma^2_x}}{2}$$</p>

<p>$$\implies R = \frac{u'}{u''} + \sqrt{\left( \frac{u'}{u''}\right)^2+\sigma^2_x}$$</p>

<p>You can totally validly use this expression for $R$, but I guess you understand why the simpler one is used instead.</p>

CC BY-SA 3.0
2015-11-10T06:53:39.657Z
Thanks a bunch for this Alecos. Without doing the math I understood that only a first-order approximation will lead to a much simpler expression! I figured out my mistake in a previous comment but thanks for confirming.
emeryville
CC BY-SA 4.0
2024-02-27T16:59:59.587Z
Could you help me solve this question? Thank you! https://economics.stackexchange.com/questions/57774/how-to-show-that-it-is-possible-to-approximate-this-risk-premium
Jonathen
CC BY-SA 3.0
2015-11-08T23:37:11.743Z
2015-11-09T02:39:13.780Z
2015-11-09T02:39:13.780Z
Alecos Papadopoulos
<p>Consider an agent with utility function $u$, initial wealth $\omega$, and a random variable $x$. By definition of the risk premium $R$, we have </p>

<p>$$ Eu(w+x) = u(w+E(x)-R). $$</p>

<p>The classical derivation of the risk premium is as follows:</p>

<p>A Taylor series expansion of <strong>order 2</strong> in the neighborhood of $(\omega + E(x))$ of the left-hand side (LHS) gives</p>

<p>$$u(\omega+x) \approx u[\omega+E(x)] + u'[x-E(x)] + \frac{1}{2} u''[x-E(x)]^2,$$</p>

<p>A Taylor series expansion of <strong>order 1</strong> in the neighborhood of  $(\omega + E(x) - R)$ of the right hand side (RHS) gives</p>

<p>$$u(\omega + E(x)-R)  \approx  u(\omega+E(x)) - u'R.$$</p>

<p>Taking expectation of the first series expansion and combining the results of the two series with the definition of the risk premium yields</p>

<p>$$u(\omega + E(x)) + u'E[x-E(x)] + \frac{1}{2} u''E[x-E(x)]^2  \approx   u(\omega+E(x)) - u'R.$$</p>

<p>This implies</p>

<p>$$R \approx - \frac{1}{2} \frac{u''}{u'} E[x-E(x)]^2.$$</p>

<p>My understanding of this derivation is that we can take a 2- or higher-order expansion of the LHS if we want the risk premium to be related not only to the variance of $R$ but also to higher moments of $R$. </p>

<p>However, is there any (economic) rational for the first-order expansion of the RHS? And for its different neighborhood evaluation?</p>

CC BY-SA 3.0
2015-11-06T05:18:02.517Z
Notice the order of $E[\cdot]$ and $u$ are different on the RHS and LHS, They don't commute.
Michael
CC BY-SA 3.0
2015-11-06T17:14:03.667Z
I noticed this. I also noticed that for the RHS expansion evaluated at $(\omega+E(x)−R)$, $R=0$ because by definition of the risk premium $(\omega+E(x)−R)$ is the certainty equivalent. But I don't see why we stop at the first-order level.
emeryville
CC BY-SA 3.0
2015-11-07T01:56:29.800Z
@Michael may ask you what you had in mind?
emeryville
CC BY-SA 4.0
2022-03-25T09:51:29.300Z
u(ω+E(x))+u′E[x−E(x)]+12u′′E[x−E(x)]2≈u(ω+E(x))−u′R. I odont understand how we add the E[x-E(x)]
user40792
<microeconomics><preferences><expected-utility><risk-aversion>
Risk Premium in the Expected Utility Theory
emeryville
# Risk Premium in the Expected Utility Theory

(Asked by: username_0 on 2015-11-06)

Consider an agent with utility function $u$, initial wealth $\omega$, and a random variable $x$. By definition of the risk premium $R$, we have

$$ Eu(w+x) = u(w+E(x)-R). $$

The classical derivation of the risk premium is as follows:

A Taylor series expansion of **order 2** in the neighborhood of $(\omega + E(x))$ of the left-hand side (LHS) gives

$$u(\omega+x) \approx u[\omega+E(x)] + u'[x-E(x)] + \frac{1}{2} u''[x-E(x)]^2,$$

A Taylor series expansion of **order 1** in the neighborhood of $(\omega + E(x) - R)$ of the right hand side (RHS) gives

$$u(\omega + E(x)-R) \approx u(\omega+E(x)) - u'R.$$

Taking expectation of the first series expansion and combining the results of the two series with the definition of the risk premium yields

$$u(\omega + E(x)) + u'E[x-E(x)] + \frac{1}{2} u''E[x-E(x)]^2 \approx u(\omega+E(x)) - u'R.$$

This implies

$$R \approx - \frac{1}{2} \frac{u''}{u'} E[x-E(x)]^2.$$

My understanding of this derivation is that we can take a 2- or higher-order expansion of the LHS if we want the risk premium to be related not only to the variance of $R$ but also to higher moments of $R$.

However, is there any (economic) rational for the first-order expansion of the RHS? And for its different neighborhood evaluation?

**Tags:** microeconomics, preferences, expected-utility, risk-aversion
> **Comments:**
> **username_1**: Notice the order of $E[\cdot]$ and $u$ are different on the RHS and LHS, They don't commute.  
> **username_0**: I noticed this. I also noticed that for the RHS expansion evaluated at $(\omega+E(x)−R)$, $R=0$ because by definition of the risk premium $(\omega+E(x)−R)$ is the certainty equivalent. But I don't see why we stop at the first-order level.  
> **username_0**: @Michael may ask you what you had in mind?  
> **username_2**: u(ω+E(x))+u′E[x−E(x)]+12u′′E[x−E(x)]2≈u(ω+E(x))−u′R. I odont understand how we add the E[x-E(x)]  

---

## 2 answers

### Answer by username_4 on 2015-11-08. Score: 9 | Accepted answer

> *Is there any (economic) rational for the first-order expansion of the RHS? And for its different neighborhood evaluation?*

As for your first question:  
This is a purely mathematical tactic in order to obtain an (approximate) equation for $R$. The expansion of first order on the RHS is motivated by this fact, i.e. to bring $R$ alone "in the surface". The reason why the LHS is subjected to a second-order expansion is in order for something to be left (the variance term). Higher-order expansions of the LHS can certainly be applied.

As for your second question, you are making a mistake. The center of expansion for the RHS expansion is the same as that for the LHS expansion, namely $w-E(x)$ (or equivalently, around $R=0$). It is meaningless (and it fails) to expand a function around its exact argument. Specifically we have

$$u\left(w+E(x)-R\right) \approx u\left(w+E(x)\right) + u'\cdot [(w+E(x)-R)-(w+E(x))] = u\left(w+E(x)\right) - u'\cdot R$$

Finally, why not consider a second-order expansion on the RHS? We would then get

$$u\left(w+E(x)-R\right) \approx u\left(w+E(x)\right) + u'\cdot [(w+E(x)-R)-(w+E(x))] \\+\frac 12 u''\cdot [(w+E(x)-R)-(w+E(x))]^2 $$

$$= u\left(w+E(x)\right) - u'\cdot R + \frac 12 u''\cdot R^2$$

Then we would obtain a quadratic polynomial in $R$,

$$\frac 12 u''\cdot R^2 - u'\cdot R - \frac 12 u''\sigma^2\_x = 0$$

$$\ R^2 - \frac {2u'}{u''}\cdot R - \sigma^2\_x = 0$$

This has roots

$$R\_1,R\_2 = \frac {(2u'/u'') \pm \sqrt{(2u'/u'')^2+4\sigma^2\_x}}{2}$$

$$\implies R = \frac{u'}{u''} + \sqrt{\left( \frac{u'}{u''}\right)^2+\sigma^2\_x}$$

You can totally validly use this expression for $R$, but I guess you understand why the simpler one is used instead.

> **Comments:**
> **username_0**: Thanks a bunch for this Alecos. Without doing the math I understood that only a first-order approximation will lead to a much simpler expression! I figured out my mistake in a previous comment but thanks for confirming.  
> **username_5**: Could you help me solve this question? Thank you! https://economics.stackexchange.com/questions/57774/how-to-show-that-it-is-possible-to-approximate-this-risk-premium  

---

### Answer by username_3 on 2015-11-09. Score: 4

It is worth noting that the "risk premium" you are talking about is in fact more accurately referred to as the Arrow-Pratt approximation of the cost of a small additive risk. You can approximate it however you want (and a Taylor series approximation is an unbiased way of doing so), but ultimately you only need to do that to provide a simple explanation of other, more interesting concepts on the economics of risk.

My understanding for the different levels of approximation was the different levels of variance and movement in the functions you are approximating. Expected utility will vary in a concave way with the added riskiness of x, whereas the utility of the difference between the certainty equivalent and the risk premium will be much closer to a linear function.

> **Comments:**
> **username_0**: Thanks! This answer helps as well. I share your understanding of the different levels of approximation.  

---

