Max R. P. Grossmann

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Certainty equivalent of a normally distributed lottery

Sometimes, the following claim—important for the analysis of an incentive scheme known as the "LEN model"—is made:

Given a normally distributed lottery $ X $, an agent with constant absolute risk aversion $r$ is indifferent to the following certainty equivalent: $$ x^{ce} = E[X] - \frac{1}{2}\cdot r\cdot V[X].$$

We will attempt to prove this statement. First of all, let me clarify the terms that are involved.

Normally distributed lottery

A lottery is a scheme that has payments that occur with a certain probability. For example, a lottery could be: Win $ 10 with probability 60 % or lose 5 dollars with probability 40 %.

You should know the normal distribution, and I will not repeat its tenets here.

A normally distributed lottery is therefore simply a lottery whose payments occur according to a normal distribution with mean $\mu$ and variance $\sigma^2$. Here, the payments are the possible values of the normal distribution and their probabilities are defined through the cumulative distribution function. For example, a lottery with $\mu = 5$ and $\sigma = 10$ has a 15 % chance of paying you 10 to 15 dollars. However, there's a 31 % chance that you make a loss (if we assume that participating in the lottery is free)!

Agent

That's someone just like you!

Just kidding, it's actually someone like me.

We assume that the agent has an utility function that relates the money earned ($x$) to the utility derived from that money ($u(x)$). The agent strives to maximize his (expected) utility function. You should already know about agents and utility functions from your introductory microeconomics course. I also assume basic knowledge of expected utility.

Constant absolute risk aversion

Arrow and Pratt defined the following measure of absolute risk aversion: $$ A(x) = -\frac{u''(x)}{u'(x)} $$ where $u(x)$ is the utility function introduced above. If you want to find out why this definition makes sense, I recommend that you read chapter 11 of Varian's Microeconomic Analysis. It's a great book.

However, our proposition demands constant absolute risk aversion. This means that $A(x)$ must be independent of $x$ and just some numeric value. We call that numeric value $r$. We say that the higher $r$ is, the more risk averse is the agent. If $r = 0$, we say that the agent is risk neutral. If $r < 0$, the agent is risk loving.

But note that our proposition does not tell us the utility function of the agent! We only know that his or her utility function has constant absolute risk aversion $r$. This means that we have to infer the agent's utility function based on that fact alone. Oh snap! We have to solve the following differential equation:

$$ -\frac{u''}{u'} = r. $$

First of all, substitute $v = u'$ and use differential notation.

$$ -\frac{\frac{dv}{dx}}{v} = r. $$

Integrate with respect to $x$ on both sides.

$$ -\int \frac{\frac{dv}{dx}}{v}\,dx = \int r\,dx. $$

Notice that $dx$ cancels out; rewrite the leftmost integral.

$$ -\int \frac{1}{v}\,dv = \int r\,dx. $$

Compute both integrals (we add the constant on the right-hand side).

$$ -\log(v) = rx+c. $$

Solve for $v$:

$$ v = \exp(-rx+c). $$

Use the multiplicative property of exponentiation ($c$ is not the same as before, we simplified the constant):

$$ v = c\,\exp(-rx). $$

Substitute back and use differential notation:

$$ \frac{du}{dx} = c\,\exp(-rx). $$

Integrate with respect to $x$, let $dx$ cancel out and evaluate the integral:

$$ \int \frac{du}{dx}\,dx = c\,\exp(-rx), $$

$$ \int 1\,du = c\exp(-rx), $$

$$ u = c\exp(-rx) + d. $$

Finally, reapply functional notation:

$$ u(x) = c\exp(-rx) + d .$$

This is how the utility function is derived. This $u(x)$ has constant absolute risk aversion $r$. To simplify, we choose $c = 1$, $d = 0$. Therefore, our utility function takes the following form:

$$ u(x) = \exp(-rx) $$

Certainty equivalent

The certainty equivalent is a fixed payment that makes the agent indifferent to the lottery. In other words: Which certain payment has the same utility to the agent as the lottery?

Therefore, the certainty equivalent $x^{ce}$ is implicitly defined by the following relation:

$$ u(x^{ce}) = EU(X). $$

This is a crucial equation that we will apply later. After all, we are looking for the certainty equivalent. But wait a second... what is $EU(X)$?

Expected utility

$EU(X)$ is the expected utility of lottery $X$. If our agent is risk averse (or risk loving), the lottery will be worth less (or more) to them than the expected lottery payout $E[X]$. Once again, I refer you to chapter 11 of Varian's Microeconomic Analysis if you want to know why expected utility makes sense.
In essence, $EU(X)$ is the expected utility of partaking in the lottery: It is defined as $E[u(X)]$.

The basic idea is that we multiply the utility of each possible payment by its probability of occuring and add them up. If our lottery is discrete, our expected utility is therefore defined as

$$ EU(X) = \sum_{x} P(X = x)\cdot u(x). $$

If our lottery is continuous, as in our case with a normally distributed lottery, expected utility takes the following form:

$$ EU(X) = \int_{-\infty}^{\infty} f(x)u(x)\,dx. $$

Putting it all together

We will now solve for the certainty equivalent. The certainty equivalent is defined by $u(x^{ce}) = EU(X)$ and expected utility is $EU(X) = \int_{-\infty}^{\infty} f(x)u(x)\,dx$. The utility function is $u(x) = \exp(-rx)$ and since our lottery $X$ is normally distributed, we know the p.d.f. as well: It is $f(x) = \frac{\exp\left(-\frac{(x-\mu)^2}{2\sigma^2}\right)}{\sqrt{2\pi\sigma^2}}$.

Knowing all this, our ansatz to the problem is

$$ \exp(-rx^{ce}) = \int_{-\infty}^{\infty} \frac{\exp\left(-\frac{(x-\mu)^2}{2\sigma^2}\right)}{\sqrt{2\pi\sigma^2}}\,\exp(-rx)\,dx. $$

First of all, use the multiplicative property of exponentiation and bring the constant in front of the second integral:

$$ \exp(-rx^{ce}) = \frac{1}{\sqrt{2\pi\sigma^2}}\int_{-\infty}^{\infty} \exp\left(-\frac{(x-\mu)^2}{2\sigma^2} - rx\right)\,dx. $$

Let us consider the integral without limits first. We then apply the limits.

Using SAGE, I found the following solution to the integral:

$$\int \exp\left(-\frac{(x-\mu)^2}{2\sigma^2} - rx\right)\,dx $$ $$ = \frac{1}{2} \sqrt{2} \sqrt{\pi} \sigma\,\text{erf}\left(\frac{\sqrt{2} r \sigma^{2} - \sqrt{2} \mu + \sqrt{2} x}{2 \sigma}\right) \exp\left(\frac{1}{2} r^{2} \sigma^{2} - \mu r\right). $$

To compute the integral with limits, we now take the limits of that expression with respect to $x$. Notice that in our solution to the integral, only $\text{erf}\left(\frac{\sqrt{2} r \sigma^{2} - \sqrt{2} \mu + \sqrt{2} x}{2 \sigma}\right)$ depends on $x$. Therefore, if we find a limit to this term we can simply plug in the limit as a multiplicative factor. Also notice that in this special case, as the argument of the function is simply a positive, monotone transformation of $x$, we can simply take the limit of $\text{erf}(x)$ instead. This makes our life a whole lot simpler:

$$ \lim_{x \rightarrow -\infty} \text{erf}\left(\frac{\sqrt{2} r \sigma^{2} - \sqrt{2} \mu + \sqrt{2} x}{2 \sigma}\right) = \lim_{x \rightarrow -\infty} \text{erf}(x) = -1. $$

This result can be obtained by looking it up. Similarly, we take the limit as $x$ goes to positive infinity:

$$ \lim_{x \rightarrow \infty} \text{erf}\left(\frac{\sqrt{2} r \sigma^{2} - \sqrt{2} \mu + \sqrt{2} x}{2 \sigma}\right) = \lim_{x \rightarrow \infty} \text{erf}(x) = 1. $$

We apply our knowledge:

$$ \left[\frac{1}{2} \sqrt{2} \sqrt{\pi} \sigma\,\text{erf}\left(\frac{\sqrt{2} r \sigma^{2} - \sqrt{2} \mu + \sqrt{2} x}{2 \sigma}\right) \exp\left(\frac{1}{2} r^{2} \sigma^{2} - \mu r\right)\right]_{-\infty}^{\infty} $$ $$ = \frac{1}{2} \sqrt{2} \sqrt{\pi} \sigma \underbrace{(1)}_{\text{second limit}} \exp\left(\frac{1}{2} r^{2} \sigma^{2} - \mu r\right) $$ $$ - \left(\frac{1}{2} \sqrt{2} \sqrt{\pi} \sigma \underbrace{(-1)}_{\text{first limit}} \exp\left(\frac{1}{2} r^{2} \sigma^{2} - \mu r\right)\right)$$ $$ = \frac{1}{2} \sqrt{2} \sqrt{\pi} \sigma (1) \exp\left(\frac{1}{2} r^{2} \sigma^{2} - \mu r\right) + \frac{1}{2} \sqrt{2} \sqrt{\pi} \sigma (1) \exp\left(\frac{1}{2} r^{2} \sigma^{2} - \mu r\right)$$ $$ = \sqrt{2} \sqrt{\pi} \sigma \exp\left(\frac{1}{2} r^{2} \sigma^{2} - \mu r\right)$$

We use this result:

$$ \exp(-rx^{ce}) = \frac{1}{\sqrt{2\pi\sigma^2}}\,\sqrt{2} \sqrt{\pi} \sigma \exp\left(\frac{1}{2} r^{2} \sigma^{2} - \mu r\right). $$

We simplify tremendously:

$$ \exp(-rx^{ce}) = \exp\left(\frac{1}{2} r^{2} \sigma^{2} - \mu r\right). $$

We take the natural logarithm on both sides:

$$ -rx^{ce} = \frac{1}{2} r^{2} \sigma^{2} - \mu r. $$

We divide both sides by $(-r)$ and we rearrange:

$$ x^{ce} = \mu - \frac{1}{2} r \sigma^{2}. $$

Since $E[X] = \mu$ and $V[X] = \sigma^2$, it does indeed follow that

$$ x^{ce} = E[X] - \frac{1}{2}\cdot r\cdot V[X]. $$

Pretty amazing, isn't it?