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?

  • $\begingroup$ Notice the order of $E[\cdot]$ and $u$ are different on the RHS and LHS, They don't commute. $\endgroup$
    – Michael
    Nov 6, 2015 at 5:18
  • $\begingroup$ 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. $\endgroup$
    – emeryville
    Nov 6, 2015 at 17:14
  • $\begingroup$ @Michael may ask you what you had in mind? $\endgroup$
    – emeryville
    Nov 7, 2015 at 1:56
  • $\begingroup$ 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)] $\endgroup$
    – user40792
    Mar 25, 2022 at 9:51

2 Answers 2


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.

  • $\begingroup$ 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. $\endgroup$
    – emeryville
    Nov 10, 2015 at 6:53

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.

  • $\begingroup$ Thanks! This answer helps as well. I share your understanding of the different levels of approximation. $\endgroup$
    – emeryville
    Nov 10, 2015 at 6:55

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