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Some conditions for the equivalence between risk aversion, prudence and temperance

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Abstract

We study relationships between different aspects of risk preferences. We show that, under the assumptions of non-satiation and bounded marginal utility, some additional conditions on the asymptotic behaviour of the indices of relative prudence and relative temperance ensure that risk aversion, prudence and temperance are equivalent. Similar conclusions are derived for higher-degree risk aversion. Moreover, some links between indices of relative risk aversion of different degrees are derived. The implications of these results for several economic problems which involve risk changes are discussed.

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Notes

  1. De Donno et al. (2019) derived, in the context of a portfolio problem, some restrictive conditions which ensure that the threshold condition on the index of partial relative risk aversion of some order implies the threshold conditions on the same indices of lower orders. We also recall that Maggi et al. (2006) studied the linkages between the monotonicity of the index of relative risk aversion and the index of relative prudence.

  2. These are just some of the possible applications of our results, which can generally be used in all problems where changes in risks are relevant, such as production localization problems (as in Eeckhoudt et al. 2009) labour supply choice (as in Chiu and Eeckhoudt 2010), optimal self protection (as in Crainich et al. 2016).

  3. Throughout the paper, we say that the limit of a function f as x approaches infinity exists when \(\lim _{x\rightarrow +\infty } f(x) = l \in {\mathbb {R}}\) or \(\lim _{x\rightarrow +\infty } f(x) = \pm \infty\)

  4. Consider as an example for the risk averse case the classical CRRA utility and for the risk lover case the utility function \(U(x)=x\;\arctan (x)-\log (1+x^2)\) on the domain \([0, +\infty )\).

  5. Many economic models suggest that this additional utility should decrease when wealth increases. This is what is called the “Law of diminishing marginal utility” (e.g. Marshall 1920, p.79). This idea, however, implies risk aversion, which is clearly an assumption we do not introduce in our paper since risk aversion is one of the aspects of preferences involved in our results.

  6. All rational functions composed of exponentials, logarithmic, polynomial functions or the inverse tangent have a limit at \(+\infty\) (which can be infinite).

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Acknowledgements

We would like to thank the Associate Editor and two anonymous referees, whose comments and suggestions helped to improve the presentation of our results.

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Appendix: Lemmas and proofs

Appendix: Lemmas and proofs

Lemma 1

Under non-satiation, if there exists\(x_0 \ge a\)such that\(U^{\prime \prime }(x) < 0\)for all\(x\in [x_0, +\infty )\), then

  1. (i)

    \(\lim _{x \rightarrow +\infty } U^{\prime }(x) = l\ge 0\)

  2. (ii)

    \(\lim _{x \rightarrow +\infty } U^{\prime \prime }(x)=0.\)

  3. (iii)

    If \(\lim _{x \rightarrow +\infty } xU^{\prime \prime }(x)\)exists, then it must be 0.

Proof

To prove (i), we observe that since \(U^{\prime \prime }(x) <0\), the marginal utility \(U^{\prime }(x)\) is strictly decreasing, it has a limit as \(x \rightarrow +\infty\). Being \(U^{\prime }(x) > 0\), the limit must be greater than or equal to 0.

Since \(U^{\prime }(x) = U^{\prime }(x_0) + \int _{x_0}^x U^{\prime \prime }(t)\mathrm{d}t,\) and \(\lim _{x \rightarrow +\infty } U^{\prime }(x) = l < + \infty\), the integral \(\int _{x_0}^{+\infty } U^{\prime \prime }(t)\mathrm{d}t\) converges. Since \(U^{\prime \prime }(x)\) is constant in sign, we have necessarily \(\displaystyle \lim _{x \rightarrow +\infty } U^{\prime \prime }(x)=0,\) so (ii) is proved.

To show (iii), note that the function \(xU^{\prime \prime }(x)\) is strictly negative, so the limit, if it exists, is some \(k \le 0\) (k can possibly take the value \(-\infty\)). Assume by contradiction that \(k<0\). Thus, there exist some \(M < 0\) and \(z \ge \max (x_1, 0)\) such that \(xU^{\prime \prime }(x) \le M\) on \([z, +\infty )\), or equivalently \(U^{\prime \prime }(x) \le \frac{M}{x}\). Integrating between z and x for \(x > z\), we obtain the following inequalities:

$$\begin{aligned} 0 \le l - U^{\prime }(z) \le U^{\prime }(x)-U^{\prime }(z) = \int _z^x U^{\prime \prime }(t) \mathrm{d}t \le \int _z^x \frac{M}{t} \mathrm{d}t \le M \ln \left( \frac{x}{z}\right) < 0, \end{aligned}$$

where the first inequality follows from the fact that \(U^{\prime }(x)\) is decreasing and converges to l when \(x \rightarrow + \infty\). We thus obtain a contradiction. \(\square\)

Lemma 2

Under non-satiation and marginal utility bounded at\(+\infty\), if there exists\(x_0 \ge a\)such that\(U^{\prime \prime \prime }(x) > 0\, \forall x\in [x_0, +\infty )\), then, in addition to\(U^{\prime \prime }(x) < 0\)and conditions (i) and(ii) of Lemma 1we have that

  1. (iii)

    \(\displaystyle \lim _{x \rightarrow +\infty } xU^{\prime \prime }(x)=0.\)

  2. (iv)

    \(\displaystyle \lim _{x \rightarrow +\infty } U^{\prime \prime \prime }(x)=0.\)

  3. (v)

    \(\displaystyle \lim _{x \rightarrow +\infty } xU^{\prime \prime \prime }(x)=0.\)

Proof

We have that for \(x> y > 0\),

$$\begin{aligned} |xU^{\prime \prime }(x)| = |U^{\prime \prime }(x)(x-y) + y U^{\prime \prime }(x)| \le |U^{\prime \prime }(x)(x-y)| + |y U^{\prime \prime }(x)| \end{aligned}$$

Exploiting the mean value theorem and the fact that \(U^{\prime \prime }\) is increasing, we obtain the following relations, with \(\xi \in (y,x)\):

$$\begin{aligned} \frac{U^{\prime }(x) - U^{\prime }(y)}{x-y} = U^{\prime \prime }(\xi ) \le U^{\prime \prime }(x). \end{aligned}$$

All the quantities above are negative, because \(U^{\prime \prime }\) is negative (and \(U^{\prime }\) decreasing). Taking the absolute values, the inequality becomes \(|U^{\prime \prime }(x)(x-y)| \le | U^{\prime }(x) - U^{\prime }(y) |\); hence we obtain

$$\begin{aligned} |xU^{\prime \prime }(x)| \le | U^{\prime }(x) - U^{\prime }(y) |+ y | U^{\prime \prime }(x)|. \end{aligned}$$

Claim (iii) thus follows from (i) and (ii) in Lemma 1.

The convergence of \(U^{\prime \prime }(x) = U^{\prime \prime }(x_0) + \int _{x_0}^{x} U^{\prime \prime \prime }(t) \mathrm{d}t\) as \(x \rightarrow + \infty\) implies that the integral \(\int _{x_0}^{+\infty } U^{\prime \prime \prime }(t) \mathrm{d}t\) is finite; hence \(\lim _{x \rightarrow +\infty } U^{\prime \prime \prime }(x) = 0\) (since \(U^{\prime \prime \prime }(x) > 0\)) and (iv) is proved.

To prove claim (v), we first observe that by the integration by parts formula we have

$$\begin{aligned} \int _{x_0}^x U^{\prime \prime }(t) \mathrm{d}t = xU^{\prime \prime }(x) - x_0 U^{\prime \prime }(x_0) - \int _{x_0}^x tU^{\prime \prime \prime }(t)\mathrm{d}t. \end{aligned}$$

On the other hand \(\int _{x_0}^x U^{\prime \prime }(t) \mathrm{d}t = U^{\prime }(x) - U^{\prime }(x_0)\). Combining these two equalities, we obtain

$$\begin{aligned} xU^{\prime \prime }(x) = x_0U^{\prime \prime }(x_0) + U^{\prime }(x)-U^{\prime }(x_0) + \int _{x_0}^x tU^{\prime \prime \prime }(t)\mathrm{d}t. \end{aligned}$$

We then apply the same argument as in the proof of claim (iv). \(\square\)

Lemma 3

Under non-satiation and marginal utility bounded at\(+\infty\), if there exists\(x_0 \ge a\)such that\((-1)^{n-1} U^{(n)} (x) > 0\, \forall x\in [x_0, +\infty )\), then we have that

  1. (vi)

    \(\displaystyle \lim _{x \rightarrow +\infty } U^{(j)}(x) =0\) for \(j=2, \ldots , n\)

  2. (vii)

    \(\displaystyle \lim _{x \rightarrow +\infty } xU^{(j)}(x)= 0\) for \(j=2, \ldots , n\).

Proof

Proposition 3 in Menegatti (2015) implies that \((-1)^{j-1} U^{(j)} (x) > 0\) for \(j=1, \ldots , n\) for all \(x\in [x_0, +\infty )\). Then (vi) and (vii) can be proved by induction, exploiting the same reasoning as in the Proof of (iv) and (v) in Lemma 2. \(\square\)

Proof of Theorem 1

  1. (a)

    The implication \(\mathrm{(ii)}\, \Rightarrow \, \mathrm{(i)}\) follows from Menegatti (2014), Proposition 1(b), with \(x_1=x_2\).

    To prove the opposite implication, we first show that \(\mathrm{ARP}(U) \ge 0\). Assume by contradiction that \(\mathrm{ARP}(U)< 0\). Then there should exist a value \({\bar{x}} > 0\) such that \(-\frac{x U^{\prime \prime \prime }(x)}{U^{\prime \prime }(x)} < 0\) for all \(x\ge {\bar{x}}\). It follows that \(U^{\prime \prime \prime }(x) < 0\) for all \(x\ge \max ({\bar{x}}, x_1)\). But, in this case, Proposition 1(a) in Menegatti (2014) implies \(U^{\prime \prime }(x) > 0\) for all \(x\ge \max ({\bar{x}}, x_1)\); hence we obtain a contradiction.

    Assume now that \(\mathrm{ARP}(U)= k,\) with \(0\le k < + \infty\). Then \(\left| \frac{x U^{\prime \prime \prime }(x)}{U^{\prime \prime }(x)} \right|\) is definitely bounded by some constant \(C>0\). In addition, we know that \(U^{\prime }(x)\) tends to a finite limit l (Lemma 1(i)). If we fix \(\varepsilon > 0\), the two previous conditions allow us to find \(z \ge x_1\) such that \(\left| \frac{x U^{\prime \prime \prime }(x)}{U^{\prime \prime }(x)} \right| \le C\) and \(|U^{\prime }(x)-U^{\prime }(y)| \le \varepsilon\) for all \(x,y \in [z, + \infty )\). Moreover, as in the proof of Lemma 2, we can write \(\int _x^y U^{\prime \prime }(t) \mathrm{d}t = xU^{\prime \prime }(x)-yU^{\prime \prime }(y) - \int _y^x t U^{\prime \prime \prime }(t) \mathrm{d}t\). Then, for \(x\ge y \ge z\)

    $$\begin{aligned} |xU^{\prime \prime }(x)-yU^{\prime \prime }(y)|= & {} \left| \int _y^x \left( U^{\prime \prime }(t) + t U^{\prime \prime \prime }(t) \right) \mathrm{d}t \right| \le \int _y^x\left| U^{\prime \prime }(t) \left( 1+\frac{tU^{\prime \prime \prime }(t)}{U^{\prime \prime }(t)}\right) \right| \mathrm{d}t \\\le & {} (1 + C) \int _y^x\left| U^{\prime \prime }(t)\right| \mathrm{d}t = (1+C) |U^{\prime }(x)-U^{\prime }(y)| \le (1+C)\varepsilon . \end{aligned}$$

    This implies that \(\lim _{x \rightarrow +\infty } xU^{\prime \prime }(x)\) exists, and by Lemma 1 (iii), it is equal to 0. We can thus apply L’Hôpital’s rule to \(-\frac{xU^{\prime \prime }(x)}{U^{\prime }(x)-l}\) and obtain

    $$\begin{aligned} 0 \le \lim _{x\rightarrow +\infty } -\frac{xU^{\prime \prime }(x)}{U^{\prime }(x)-l} = \lim _{x\rightarrow +\infty } \left( -1 -\frac{x U^{\prime \prime \prime }(x)}{U^{\prime \prime }(x)} \right) = -1+k. \end{aligned}$$

    Hence, either \(1 \le \mathrm{ARP}(U) < + \infty\) or \(\mathrm{ARP}(U) = +\infty .\) In both cases, for a fixed positive constant c, one can find some \(x_2 \ge x_1\) such that \(\displaystyle -\frac{x U^{\prime \prime \prime }(x)}{U^{\prime \prime }(x)} \ge c\), that is

    $$\begin{aligned} U^{\prime \prime \prime }(x) \ge c \, \frac{-U^{\prime \prime }(x)}{x} > 0 \quad \text{ for } \text{ all } x\ge x_2. \end{aligned}$$
  2. (b)

    The implication \(\mathrm{(ii)}\, \Rightarrow \,\mathrm{(i)}\) is proved in Menegatti (2014), Proposition 2 (a). To show the opposite implication, we use a similar argument as in the previous proof. First of all we exclude the case where \(\mathrm{ART}(U) < 0\), since by part (a) of this theorem, \(U^{\prime \prime }(x) < 0\) for \(x\ge x_1\). Moreover, from Lemma 2, we know that \(xU^{\prime \prime \prime }(x)\) and \(U^{\prime \prime }(x)\) tend to 0 as \(x \rightarrow +\infty\), hence \(\frac{-xU^{\prime \prime \prime }(x)}{U^{\prime \prime }(x)}\) is an indeterminate form at \(+\infty\). By virtue of L’Hôpital’s rule, it has a finite limit if and only if \(\mathrm{ART}(U)\) is finite and, in this case,

    $$\begin{aligned} 0 \le \lim _{x \rightarrow +\infty } \frac{-xU^{\prime \prime \prime }(x)}{U^{\prime \prime }(x)} = -1 + \mathrm{ART}(U) \end{aligned}$$

    This implies that \(\mathrm{ART}(U)\) is strictly positive (possibly infinite) and the claim follows as above.

  3. (c)

    The implication \(\mathrm{(ii)}\,\Rightarrow \,\mathrm{(i)}\) follows from Menegatti (2015), Proposition 3. To prove that \(\mathrm{(i)}\, \Rightarrow \, \mathrm{(ii)}\), we proceed as above to show that \(\mathrm{ARRA}_n(U) \ge 0\). Lemma 3 implies that the ratio \(\frac{-xU^{(n)}(x)}{U^{(n-1)}(x)}\) is an indeterminate form at \(+\infty\) and by L’Hôpital’s rule, the limit exists and is finite if and only if \(\mathrm{ARRA}_n(U)\) is finite. In this case,

    $$\begin{aligned} 0 \le \lim _{x \rightarrow +\infty } \frac{-xU^{(n)}(x)}{U^{(n-1)}(x)} = -1 + \mathrm{ARRA}_n(U); \end{aligned}$$

    hence \(\mathrm{ARRA}_n(U)\) is strictly positive (possibly infinite). The result follows as above.

\(\square\)

Proof of Theorem 2

  1. (a)

    Menegatti (2014), Proposition 1 (b) implies that \(U^{\prime \prime }(x) < 0\) for all x greater than \(x_0\). Theorem 1(b) implies that \(U^{(4)} < 0\) for all x greater than some \(x^*\). Exploiting Lemma 2, we can show, as in the proof of Theorem 1, that

    $$\begin{aligned} 0 \le \lim _{x\rightarrow +\infty } -\frac{xU^{\prime \prime \prime }(x)}{U^{\prime \prime }(x)} = -1 +\mathrm{ART}(U). \end{aligned}$$

    Since \(\mathrm{ART} (U)\) is either strictly greater or smaller than 3, the claim follows.

    Assume now that \(\mathrm{ART}(U) = 3\) and \(-\frac{xU^{(4)}(x)}{U^{\prime \prime \prime }(x)}< 3\). The function \(g(x) = -xU^{\prime \prime \prime }(x) - 2U^{\prime \prime }(x)\) is definitely strictly decreasing since its derivative \(g'(x) = -x U^{(4)}(x) - 3U^{\prime \prime \prime }(x)\) is by assumption definitely strictly negative. Moreover, by Lemma 2, it tends to 0 as x goes to \(+\infty\). This implies that \(g(x) > 0\), which is equivalent to the inequality \(-\frac{x U^{\prime \prime \prime }(x)}{U^{\prime \prime }(x)} < 2\), since \(U^{\prime \prime }(x) < 0\). The other case (\(-\frac{xU^{(4)}(x)}{U^{\prime \prime \prime }(x)}>3 \Rightarrow -\frac{xU^{\prime \prime \prime }(x)}{U^{\prime \prime }(x)}> 2\)) can be similarly proved.

  2. (b)

    Menegatti (2015), Proposition 3 implies that \((-1)^{n-2} U^{(n-1)}(x) > 0\) for all x greater than \(x_0\), while Theorem 1(c) implies that \((-1)^n U^{(n+1)} > 0\) for all x greater than some \(x^*\). Using Lemma 3 and L’Hôpital’s rule, we obtain

    $$\begin{aligned} 0 \le \lim _{x\rightarrow +\infty } -\frac{xU^{(n)} (x)}{U^{(n-1)}(x)} = -1 +\mathrm{ARRA}_n(U). \end{aligned}$$

    Since \(\mathrm{ARRA}_n (U)\) is either strictly greater or smaller than n, the claim follows.

    Let now \(\mathrm{ARRA}_n(U) = n\) and \(-\frac{xU^{(n+1)}(x)}{U^{(n)}(x)}< n\) and consider the function \(g(x) = -xU^{(n)}(x) - (n-1) U^{(n-1)}(x)\). We have that \(\lim _{x\rightarrow +\infty } g(x) = 0\) by Lemma 3. Moreover, the derivative \(g'(x) = -x U^{(n+1)}(x) - nU^{(n)}(x)\) is by assumption definitely strictly negative if n is odd (\(U^{(n)}(x) > 0\)) and definitely strictly positive if n is even (\(U^{(n)}(x) < 0\)). Hence g(x) is definitely decreasing if n is odd and definitely increasing if n is even, and, as a consequence, \(g(x) > 0\) for n odd and \(g(x) > 0\) for n even. In both cases, this condition is equivalent to the inequality \(-\frac{x U^{(n)}(x)}{U^{(n-1)} (x)} < n-1\), since \((-1)^{n-2} U^{(n-1)}(x) > 0\). The other case follows in a similar way.

\(\square\)

Proof of Corollary 2

Menegatti (2015), Proposition 3 implies that \((-1)^{j-1} U^{(j)}(x) > 0\) for all x greater than \(x_0\), \(2\le j \le n\). Lemma 3 implies that the ratio \(\frac{-xU^{(j+1)}(x)}{U^{(j)}(x)}\) is an indeterminate form at \(+\infty\) for all \(2\le j\le n\). Then, a recursive application of L’Hôpital’s rule shows that \(\mathrm{ARRA}_j(U)\) exists. Moreover, it is different from j if and only if \(\mathrm{ARRA}_n(U) \not = n\), and it is equal to j otherwise. A recursive application of Theorem 2 yields the result. \(\square\)

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De Donno, M., Menegatti, M. Some conditions for the equivalence between risk aversion, prudence and temperance. Theory Decis 89, 39–60 (2020). https://doi.org/10.1007/s11238-020-09745-5

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