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Let $U(\cdot)$ be a twice differentiable (increasing) utility function and $x$ the agent's initial wealth. For $\varepsilon\in[0, x]$ define a bet that pays either $+\varepsilon$, or $-\varepsilon$. Define $p(x, \varepsilon)$ by $$U(x) = \left(\dfrac12 + p(x, \varepsilon)\right)U(x + \varepsilon) + \left(\dfrac12 - p(x, \varepsilon)\right)U(x - \varepsilon).$$

a) Prove that if the agent is risk-averse, then $p(x, \varepsilon)\ge 0$.

b) Prove that $A(x) = 4 \lim_{\varepsilon\to 0}\frac{\partial p(x, \varepsilon)}{\partial\varepsilon}$, where $A(x)$ is the agent's coefficient of absolute risk aversion.

c) Consider now a multiplicative bet. Define $q(x, \varepsilon)$ by $$U(x) = \left(\dfrac12 + q(x, \varepsilon)\right)U(x(1 + \varepsilon)) + \left(\dfrac12 - q(x, \varepsilon)\right)U(x(1 - \varepsilon)).$$ Derive a relation between $\lim_{\varepsilon\to 0}\frac{\partial q(x, \varepsilon)}{\partial\varepsilon}$ and the coefficient of relative risk-aversion $R(x)$.

Hi, I am trying to answer this financial mathematics problem but I am unable to come up with the solution so looking for some help. The agent being risk-averse is equivalent to the concavity of the function U(x), so I think the solution might use that but not sure about it. On the other hand, for the other parts of the exercise the definitions of A(x) and R(x) are A(x) = -U''(x)/U'(x) and R(x) = -xU''(x)/U'(x). Hope someone can help me. Thank you in advance.

In my approach for the part a) i have tried to check that if the second derivative is negative then $p(x,\epsilon)$ must be positive. However, when i calculate $U(x)''$ it depends on $\frac{\partial p}{\partial x}$ and I dont have any information about it so I dont know if this is the approach I should follow to solve it.

On the other hand, on my approach to part b), I find the explicit expression for $p(x,\epsilon)$ and then derive, getting to the following expression:

$$\frac{\partial p}{\partial \epsilon} = \frac{U'(x+\epsilon)[U(x-\epsilon) - U(x)] + U'(x-\epsilon)[U(x+\epsilon)-U(x)]}{[U(x+\epsilon) - U(x-\epsilon)]^2}$$

However, after trying to solve the limit by L'Hopital I cant find the desire expression which should be $-\frac{U''(x)}{4U'(x)}$.

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Sergio
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1 Answers1

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For a) restructure your definition of $p(x,\epsilon)$ as follows:

$$-\frac{1}{2}\left[(U(x+\epsilon)-U(x) - (U(x)-U(x-\epsilon))\right] = p(x,\epsilon)\left[ U(x+\epsilon) - U(x-\epsilon)\right] \; .$$

First, the utility is an increasing function, so the factor of $p(x,\epsilon)$ should be positive. Second, the person is risk-averse, therefore utility should have decreasing slope and therefore the term on the left hand side should be positive. Hence $p(x,\epsilon)$ is positive as well.

For b) take the derivative of the expression with respect to $\epsilon$ and you'll get

$$-\frac{1}{2}\left[(U'(x+\epsilon)-U'(x-\epsilon))\right] = \frac{\partial p(x,\epsilon)}{\partial \epsilon}\left[ U(x+\epsilon) - U(x-\epsilon)\right] + p(x,\epsilon)\left[ U'(x+\epsilon) + U'(x-\epsilon)\right]\; .$$

Taking the limit for $\epsilon \to 0$, with $\lim_{\epsilon \to 0} p(x,\epsilon) = 0$ which you can derive from taking the same limit on the expression before derivation, you get

$$\lim_{\epsilon \to 0} \frac{\partial p(x,\epsilon)}{\partial \epsilon} = -\frac{U''(x)}{2U'(x)} $$

which seems to be a factor 2 off from the usual definition of $A$. I don't see where I made the mistake here.

Try to work c) out for yourself.

Raskolnikov
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