<p>We have
<span class="math-container">$$-\frac{f''}{f'}-\frac 1 x - a = 0 \implies f'' = (-\frac 1x -a ) f'.$$</span></p>
<p>Set <span class="math-container">$y\equiv f'$</span> to arrive at
<span class="math-container">$$y' = (-\frac 1x -a ) y.$$</span></p>
<p>This has solution
<span class="math-container">$$f' = y : \int \frac 1y dy = \int (-\frac 1x -a ) dx$$</span>
<span class="math-container">$$\implies \ln y = -\ln x -ax \implies y = \frac 1x e^{-ax}.$$</span></p>
<p>So
<span class="math-container">$$f' = \frac 1x e^{-ax} \implies f(x) = \int^{ax}_{-\infty}\frac 1t e^{-t}dt.$$</span></p>
<p>This is related to the <a href="https://en.wikipedia.org/wiki/Exponential_integral" rel="nofollow noreferrer">Exponential Integral</a>. It satisfies the required inequality for any <span class="math-container">$a&gt;0$</span>. The lower bound need not be minus infinity, for your purposes.
Explore how you can insert the exogenous parameter <span class="math-container">$w$</span> in there.</p>

CC BY-SA 4.0
2023-10-28T02:49:53.543Z
2023-10-28T02:49:53.543Z
Alecos Papadopoulos
<p>Ok, I think the answer is no. There does not exist a utility function satisfying DARA such that <span class="math-container">$A(x)&gt;\frac{1}{x}$</span> for all <span class="math-container">$x\in R_+$</span>.</p>
<p>To prove that, let's assume such a function exists. Then, there must exists <span class="math-container">$a\in R_+$</span> such that:
<span class="math-container">$$\frac{-f''}{f'}&gt;\frac{1}{x}\Leftrightarrow\frac{-f''}{f'}-\frac{1}{x}-a=0$$</span>
If we rearrange this function a bit by taking <span class="math-container">$y=x+w$</span>, we have:
<span class="math-container">$$-\frac{f''(y)}{f'(y)}=\frac{1+a(y-w)}{y-w}\qquad\text{(1)}$$</span>
Expression (1) is a second-order nonlinear ordinary differential equation. Solving it (here pardon my laziness, I used <a href="https://www.wolframalpha.com/input?i2d=true&amp;i=Divide%5By%27%27%5C%2840%29x%5C%2841%29%2Cy%27%5C%2840%29x%5C%2841%29%5D%3DDivide%5B%5C%2840%291-ax%5C%2841%29%2Cx%5D" rel="nofollow noreferrer">Wolfram</a>) gives:
<span class="math-container">$$f(y)=\int_1^{y}exp\bigg(\int_{1}^{\phi}\frac{a(\rho-w)+1}{w-\rho}d\rho\bigg)c_1d\phi+c_2$$</span>
Taking the derivative of <span class="math-container">$f$</span> we get:
<span class="math-container">$$f'(y)=exp\bigg(\int_1^{y}\frac{a(\rho-w)+1}{w-\rho}d\rho\bigg)c_1$$</span>
So we do have the condition <span class="math-container">$f'&gt;0$</span> fulfilled as long as <span class="math-container">$c_1&gt;0$</span>. Taking the second-order derivative of <span class="math-container">$f$</span>, we get:
<span class="math-container">$$f''(y)=\frac{a(y-w)+1}{w-y}exp\bigg(\int_1^{y}\frac{a(\rho-w)+1}{w-\rho}d\rho\bigg)c_1$$</span>
This expression is negative if and only if:
<span class="math-container">$$\frac{a(y-w)+1}{w-y}&gt;0$$</span>
Using the fact that <span class="math-container">$x=y-w$</span>:
<span class="math-container">$$-\frac{1+ax}{x}&gt;0\Leftrightarrow x&lt;-\frac{1}{a}$$</span>
We have a contradiction -- if I did not make any mistake in the computations -- there does not exist a DARA utility function where <span class="math-container">$A(x)&gt;\frac{1}{x}$</span>, for all <span class="math-container">$x\in R_+$</span>.</p>

CC BY-SA 4.0
2023-03-23T13:45:24.447Z
2023-03-23T15:15:12.990Z
2023-03-23T15:15:12.990Z
ju_pi_car
<p>Say I have the following utility function:</p>
<p><span class="math-container">$$u(x,w)=f(w+x)$$</span></p>
<p>This utility exhibits decreasing absolute risk aversion (<span class="math-container">$f'&gt;0$</span>, <span class="math-container">$f''&lt;0$</span> and <span class="math-container">$f'''&gt;0$</span>). <span class="math-container">$x\in R_+$</span> is the control variable and <span class="math-container">$w\in R_+$</span> is an exogenous parameter.</p>
<p>Is there a parametric form of <span class="math-container">$f$</span> for which <span class="math-container">$A(x)&gt;\frac{1}{x}$</span>?</p>
<p>Where <span class="math-container">$A(x)=\frac{-f''}{f'}$</span> is the Arrow-Pratt measure of absolute risk-aversion. I cannot manage to find such a function, which makes me believe that maybe the Arrow-Pratt measure has a lower bound when the utility is DARA.</p>
<p>Many thanks!</p>

<utility><risk-aversion>
Does the Arrow-Pratt measure of absolute risk aversion have a lower bound with DARA utlity?
ju_pi_car
# Does the Arrow-Pratt measure of absolute risk aversion have a lower bound with DARA utlity?

(Asked by: username_0 on 2023-03-23)

Say I have the following utility function:

$$u(x,w)=f(w+x)$$

This utility exhibits decreasing absolute risk aversion ($f'>0$, $f''<0$ and $f'''>0$). $x\in R\_+$ is the control variable and $w\in R\_+$ is an exogenous parameter.

Is there a parametric form of $f$ for which $A(x)>\frac{1}{x}$?

Where $A(x)=\frac{-f''}{f'}$ is the Arrow-Pratt measure of absolute risk-aversion. I cannot manage to find such a function, which makes me believe that maybe the Arrow-Pratt measure has a lower bound when the utility is DARA.

Many thanks!

**Tags:** utility, risk-aversion

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## 2 answers

### Answer by username_1 on 2023-10-28. Score: 2

We have
$$-\frac{f''}{f'}-\frac 1 x - a = 0 \implies f'' = (-\frac 1x -a ) f'.$$

Set $y\equiv f'$ to arrive at
$$y' = (-\frac 1x -a ) y.$$

This has solution
$$f' = y : \int \frac 1y dy = \int (-\frac 1x -a ) dx$$
$$\implies \ln y = -\ln x -ax \implies y = \frac 1x e^{-ax}.$$

So
$$f' = \frac 1x e^{-ax} \implies f(x) = \int^{ax}\_{-\infty}\frac 1t e^{-t}dt.$$

This is related to the [Exponential Integral](https://en.wikipedia.org/wiki/Exponential_integral). It satisfies the required inequality for any $a>0$. The lower bound need not be minus infinity, for your purposes.
Explore how you can insert the exogenous parameter $w$ in there.


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### Answer by username_0 on 2023-03-23. Score: 0

Ok, I think the answer is no. There does not exist a utility function satisfying DARA such that $A(x)>\frac{1}{x}$ for all $x\in R\_+$.

To prove that, let's assume such a function exists. Then, there must exists $a\in R\_+$ such that:
$$\frac{-f''}{f'}>\frac{1}{x}\Leftrightarrow\frac{-f''}{f'}-\frac{1}{x}-a=0$$
If we rearrange this function a bit by taking $y=x+w$, we have:
$$-\frac{f''(y)}{f'(y)}=\frac{1+a(y-w)}{y-w}\qquad\text{(1)}$$
Expression (1) is a second-order nonlinear ordinary differential equation. Solving it (here pardon my laziness, I used [Wolfram](https://www.wolframalpha.com/input?i2d=true&i=Divide%5By%27%27%5C%2840%29x%5C%2841%29%2Cy%27%5C%2840%29x%5C%2841%29%5D%3DDivide%5B%5C%2840%291-ax%5C%2841%29%2Cx%5D)) gives:
$$f(y)=\int\_1^{y}exp\bigg(\int\_{1}^{\phi}\frac{a(\rho-w)+1}{w-\rho}d\rho\bigg)c\_1d\phi+c\_2$$
Taking the derivative of $f$ we get:
$$f'(y)=exp\bigg(\int\_1^{y}\frac{a(\rho-w)+1}{w-\rho}d\rho\bigg)c\_1$$
So we do have the condition $f'>0$ fulfilled as long as $c\_1>0$. Taking the second-order derivative of $f$, we get:
$$f''(y)=\frac{a(y-w)+1}{w-y}exp\bigg(\int\_1^{y}\frac{a(\rho-w)+1}{w-\rho}d\rho\bigg)c\_1$$
This expression is negative if and only if:
$$\frac{a(y-w)+1}{w-y}>0$$
Using the fact that $x=y-w$:
$$-\frac{1+ax}{x}>0\Leftrightarrow x<-\frac{1}{a}$$
We have a contradiction -- if I did not make any mistake in the computations -- there does not exist a DARA utility function where $A(x)>\frac{1}{x}$, for all $x\in R\_+$.


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