<p>Implementing the "affine transformation", let $$u(x) = A-B\exp\{-\lambda x\}$$</p>

<p>Then we want to solve $(x_1 = 2400, x_2 = 5000)$</p>

<p>$$A-B\exp\{-\lambda x_1\} = \frac 12 \Big[A-B\exp\{-\lambda x_2\}\Big]+\frac 12 \Big[A-B\exp\{-\lambda \cdot 0\}\Big]$$</p>

<p>$$\implies A-B\exp\{-\lambda x_1\} = A - \frac 12B - \frac 12B\exp\{-\lambda x_2\}$$</p>

<p>$$\implies \frac 12B -B\exp\{-\lambda x_1\} + B\frac 12\exp\{-\lambda x_2\} =0$$</p>

<p>$$\implies -\frac 12 +\exp\{-\lambda x_1\} - \frac 12\exp\{-\lambda x_2\} =0$$</p>

<p>and we verify that $A,B$ do not matter.</p>

<p>This is a non-linear function in one variable, and a computer algorithm should be used to find the possible solutions. In this particular case, trial-and-error by hand (or naive graphing in a spreadsheet) may mislead, because, as the posted solution by the OP in a comment shows, the non-zero value for $\lambda$ that satisfies the equation is very small, $\lambda = 0.0032$.</p>

<p>So if one tries to graph the function in a spreadsheet, it must specify very small increments for $\lambda$ otherwise it will miss the small interval for which the function becomes positive for strictly positive values of $\lambda$, and then falls to zero and negative values again.</p>

<p>Making matters worse,the values of the actual prizes are too large numerically when inserted into the exponential function, and so the increments for lambda should be even smaller, if one would want to "see" the solution in a spreadsheet.. For example to use the actual levels of the prices as coefficients of the equation, one needs to specify increments of the order $10^{-5}$. </p>

<p>One could input the prizes in hundreds or in thousands... (the solution in the link posted in the comment uses hundreds), which allows for larger increments.</p>

CC BY-SA 3.0
2018-01-21T15:02:19.723Z
Many thanks! But it seems that the equation actually has 2 solutions. The thing is that, as you say, the actual prizes are too large so that the 2 solutions are numerically indistinguishable. One can see through the trick by dealing it with a unit of $100. Here's an answer provided by the instructor: http://nbviewer.jupyter.org/gist/oyamad/e51f2363416a92b59a8d4ec4f5b7aacb
ZLIU
CC BY-SA 3.0
2018-01-21T14:04:32.737Z
2018-01-21T15:32:29.290Z
2018-01-21T15:32:29.290Z
Alecos Papadopoulos
<p>Consider the following scenario:</p>

<p>A consumer with CARA (constant absolute risk aversion) claims that she is indifferent between "getting ＄2400 for sure" and "getting ＄5000 or ＄0, each with 50% possibility". </p>

<p>I know that a consumer has CARA if and only if the vNM utility function is an affine transformation of $-e^{-\lambda x}$, where $x$ is the prize of the gamble. I was wondering how I could solve for the coefficient $\lambda$, given the above scenario.</p>

<p>To do so I wrote down the following equation:</p>

<p>$e^{-2400\lambda}=\frac{1}{2}e^{-5000\lambda}+\frac{1}{2}e^{-0\lambda}$</p>

<p>It is clear that $\lambda=0$ should be a solution to it. However, I found that this could not be the case, as $\lambda=0$ means risk neutral, which is controversial with the consumer's claim. </p>

<p>What should be the correct way of solving it? I think I must have made some stupid mistakes here but I couldn't figure it out.</p>

<p>(For reference, this is actually Problem 6.2 in Kreps' textbook - any suggestion or hint would be much appreciated.)</p>

<consumer-theory><risk-aversion><computation>
CARA Coefficient Calculation
ZLIU
# CARA Coefficient Calculation

(Asked by: username_0 on 2017-06-13)

Consider the following scenario:

A consumer with CARA (constant absolute risk aversion) claims that she is indifferent between "getting ＄2400 for sure" and "getting ＄5000 or ＄0, each with 50% possibility".

I know that a consumer has CARA if and only if the vNM utility function is an affine transformation of $-e^{-\lambda x}$, where $x$ is the prize of the gamble. I was wondering how I could solve for the coefficient $\lambda$, given the above scenario.

To do so I wrote down the following equation:

$e^{-2400\lambda}=\frac{1}{2}e^{-5000\lambda}+\frac{1}{2}e^{-0\lambda}$

It is clear that $\lambda=0$ should be a solution to it. However, I found that this could not be the case, as $\lambda=0$ means risk neutral, which is controversial with the consumer's claim.

What should be the correct way of solving it? I think I must have made some stupid mistakes here but I couldn't figure it out.

(For reference, this is actually Problem 6.2 in Kreps' textbook - any suggestion or hint would be much appreciated.)

**Tags:** consumer-theory, risk-aversion, computation

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## 1 answers

### Answer by username_1 on 2018-01-21. Score: 2

Implementing the "affine transformation", let $$u(x) = A-B\exp\{-\lambda x\}$$

Then we want to solve $(x\_1 = 2400, x\_2 = 5000)$

$$A-B\exp\{-\lambda x\_1\} = \frac 12 \Big[A-B\exp\{-\lambda x\_2\}\Big]+\frac 12 \Big[A-B\exp\{-\lambda \cdot 0\}\Big]$$

$$\implies A-B\exp\{-\lambda x\_1\} = A - \frac 12B - \frac 12B\exp\{-\lambda x\_2\}$$

$$\implies \frac 12B -B\exp\{-\lambda x\_1\} + B\frac 12\exp\{-\lambda x\_2\} =0$$

$$\implies -\frac 12 +\exp\{-\lambda x\_1\} - \frac 12\exp\{-\lambda x\_2\} =0$$

and we verify that $A,B$ do not matter.

This is a non-linear function in one variable, and a computer algorithm should be used to find the possible solutions. In this particular case, trial-and-error by hand (or naive graphing in a spreadsheet) may mislead, because, as the posted solution by the OP in a comment shows, the non-zero value for $\lambda$ that satisfies the equation is very small, $\lambda = 0.0032$.

So if one tries to graph the function in a spreadsheet, it must specify very small increments for $\lambda$ otherwise it will miss the small interval for which the function becomes positive for strictly positive values of $\lambda$, and then falls to zero and negative values again.

Making matters worse,the values of the actual prizes are too large numerically when inserted into the exponential function, and so the increments for lambda should be even smaller, if one would want to "see" the solution in a spreadsheet.. For example to use the actual levels of the prices as coefficients of the equation, one needs to specify increments of the order $10^{-5}$.

One could input the prizes in hundreds or in thousands... (the solution in the link posted in the comment uses hundreds), which allows for larger increments.

> **Comments:**
> **username_0**: Many thanks! But it seems that the equation actually has 2 solutions. The thing is that, as you say, the actual prizes are too large so that the 2 solutions are numerically indistinguishable. One can see through the trick by dealing it with a unit of $100. Here's an answer provided by the instructor: http://nbviewer.jupyter.org/gist/oyamad/e51f2363416a92b59a8d4ec4f5b7aacb  

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