Probabilistic risk attitudes and local risk aversion: a paradox

Abstract

Prominent theories of decision under risk that challenge expected utility theory model risk attitudes at least partly with transformation of probabilities. This paper shows how attributing local risk aversion to attitudes towards probabilities can produce extreme probability distortions that imply paradoxical risk aversion.

Appendix: Proof of the Theorem

We first prove one proposition and one Lemma. Then we use these two results to prove the main theorem.

Let set $\Psi (k_*,k^{*})=\{k_*/2n,(k_*+1)/2n,..., k^{*}/2n\}$ be defined for any given two integers $k_*,k^{*}\in N$ such that $1\le k_*<k^{*}\le 2n-1.$ For any given non-negative prizes $h>m>\ell$ let $q$ denote $(v(h)-v(m))/(v(m)-v(l))$.

Proposition 5.1

Let $n\in N$ and non-negative prizes $h>m>\ell$ be given. Suppose that $q>1$ and $S_i \;{\succeq }\;R_i ,$ for all $p_i \in \Psi (k_*,k^{*})$. Then $\forall \hbox {k}\in \Psi (k_*,k^{*})$

$$\begin{aligned} f\left( \frac{k}{2n}\right) -f\left( \frac{k_*-1}{2n}\right) \le \frac{1}{\kappa }\left[ {f\left( \frac{k^{*}+1}{2n}\right) -f\left( \frac{k_*-1}{2n}\right) } \right] , \end{aligned}$$

(5.1)

where $\kappa =\kappa (q,k^{*}-k,k-k_*)$.

Proof

If $S_i \;{\succeq }\;R_i$ for some $i\in \{k_*,\ldots , k^{*}\}$ then according to statement (2), the agent has revealed

$$\begin{aligned}&v(h)f\left( \frac{i-1}{2n}\right) +v(m)\left( {f\left( \frac{i+1}{2n}\right) -f\left( \frac{i-1}{2n}\right) } \right) +v(\ell )\left( {1-f\left( \frac{i+1}{2n}\right) } \right) \nonumber \\&\quad \ge v(h)f\left( \frac{i}{2n}\right) +v(\ell )\left( {1-f\left( \frac{i}{2n}\right) } \right) , \end{aligned}$$

(5.2)

for all$\;\;i=k_*,\ldots , k^{*}$. Adding and subtracting $v(m)f(i/2n)$ on the left-hand-side of the above inequality, rearranging terms and using notation $q=(v(h)-v(m))/(v(m)-v(\ell ))$, one has

$$\begin{aligned} f\left( \frac{i+1}{2n}\right) -f\left( \frac{i}{2n}\right) \ge q\left( {f\left( \frac{i}{2n}\right) -f\left( \frac{i-1}{2n}\right) } \right) ,\quad i=k_*,\ldots , k^{*}.\quad \quad \end{aligned}$$

(5.3)

If $S_j \;{\succeq }\;R_j$ for all $j=i,\ldots , i+t$ where $i,i+t\in \{k_*,\ldots , k^{*}\},$ then application of inequality (5.3) $t$ times gives

$$\begin{aligned}&f\left( \frac{i+t+1}{2n}\right) -f\left( \frac{i+t}{2n}\right) \ge q\left( {f\left( \frac{i+t}{2n}\right) -f\left( \frac{i+t-1}{2n}\right) } \right) \nonumber \\&\quad \ge ... \ge q^{t+1}\left( {f\left( \frac{i}{2n}\right) -f\left( \frac{i-1}{2n}\right) } \right) . \end{aligned}$$

(5.4)

To show statement (5.1) and complete the proof it suffices to show that $\forall k\in \{k_*,..., k^{*}\},$

$$\begin{aligned} f\left( \frac{k}{2n}\right) -f\left( \frac{k_*-1}{2n}\right) \le \left( {f\left( \frac{k}{2n}\right) -f\left( \frac{k-1}{2n}\right) } \right) \sum _{j=0}^{k-k_*} {q^{-j}}, \end{aligned}$$

(5.5)

and

$$\begin{aligned} f\left( \frac{k^{*}+1}{2n}\right) -f\left( \frac{k}{2n}\right) \ge \left( {f\left( \frac{k}{2n}\right) -f\left( \frac{k-1}{2n}\right) } \right) \sum _{j=0}^{k^{*}-k} {q^{j+1}} \end{aligned}$$

(5.6)

because the last two inequalities imply that

$$\begin{aligned} \frac{1}{\sum _{j=0}^{k-k_*} {q^{-j}} }\left( {f\left( \frac{k}{2n}\right) \!-\!f\left( \frac{k_*\!-\!1}{2n}\right) } \right) \le \frac{1}{\sum _{j=0}^{k^{*}-k} {q^{j+1}} }\left( {f\left( \frac{k^{*}+1}{2n}\right) \!-\!f\left( \frac{k}{2n}\right) } \right) ; \end{aligned}$$

rearrange terms and use notation $\kappa =1+\sum _{j=0}^{k^{*}-k} {q^{j+1}} /\sum _{j=0}^{k-k_*} {q^{-j}}$ to obtain statement (5.1)

$$\begin{aligned} f\left( \frac{k}{2n}\right) -f\left( \frac{k_*-1}{2n}\right) \le \frac{1}{\kappa }\left[ {f\left( \frac{k^{*}+1}{2n}\right) -f\left( \frac{k_*-1}{2n}\right) } \right] . \end{aligned}$$

Inequality (5.5) follows directly from inequality (5.4) and some rearrangement of terms

$$\begin{aligned} f\left( \frac{k}{2n}\right) -f\left( \frac{k_*-1}{2n}\right)&= \sum _{i=k_*}^k {\left( {f\left( \frac{i}{2n}\right) -f\left( \frac{i-1}{2n}\right) } \right) }\\&\le \left( {f\left( \frac{k}{2n}\right) -f\left( \frac{k-1}{2n}\right) } \right) \sum _{j=0}^{k-k_*} {q^{-j}}. \end{aligned}$$

Similarly, for inequality (5.6) verify that

$$\begin{aligned} f\left( \frac{k^{*}+1}{2n}\right) -f\left( \frac{k}{2n}\right)&= \sum _{i=k+1}^{k^{*}+1} {\left( {f\left( \frac{i}{2n}\right) -f\left( \frac{i-1}{2n}\right) } \right) }\\&\ge \left( {f\left( \frac{k}{2n}\right) -f\left( \frac{k-1}{2n}\right) } \right) \sum _{j=0}^{k^{*}-k} {q^{j+1}}. \end{aligned}$$

$\square$

Lemma 5.1

If $q>1$ then $\mathop {\hbox {lim}}\limits _{t\rightarrow \infty } \kappa \left( {q,t,s} \right) =\infty$

Proof

It follows from $\kappa (q,t,s)=1+\sum _{j=0}^t {q^{j+1}} /\sum _{i=0}^s {q^{-i}} =\frac{q^{2+t+s}-1}{q^{1+s}-1}>q^{1+t}$

$\square$

Proof of the Theorem

Part a. Suppose that $S_i \;{\succeq }\;R_i ,$ for all $p_i =i/2n\in (p_*,p^{*})$ for some $n\in N.$ Let $p\in (p_*,p^{*})$ be given. Without any loss of generality let $p_*$ be written as $p_*=(k_*-1)/2n$ for some $k_*\in N.$ Let $k$ denote the smallest integer larger than 2np, that is $k=\left\lceil {2np} \right\rceil$ and $k^{*}$ denote the largest integer smaller than 2np $^{*}-1$, that is $k^{*}=\left\lfloor {2np^{*}-1} \right\rfloor .$ By the supposition in the theorem and the construction of $k,k_*,k^{*}$ one has $S_i \;{\succeq }\;R_i ,$ for all $p_i \in \Psi (k_*,k^{*})$. By Proposition 5.1 and construction one has

$$\begin{aligned} f(p)-f(p_*)\le f\left( \frac{k}{2n}\right) -f\left( \frac{k_*-1}{2n}\right)&\le \frac{1}{\kappa }\left[ {f\left( \frac{k^{*}+1}{2n}\right) -f\left( \frac{k_*-1}{2n}\right) } \right] \\&\le \frac{1}{\kappa }\left[ {f(p^{*})-f(p_*)} \right] , \end{aligned}$$

where $\kappa =\kappa (q,k^{*}-k,k-k_*),$ which completes the proof.

Part b. For any given $\varepsilon \in (0,p^{*}-p_*),G\in N$ and $q>1$ take $n^{*}\ge \left( {1+\ln G/\ln q^{2}} \right) /\varepsilon$. Let $n\ge n^{*}$ be given and suppose that $S_i \;{\succeq }\;R_i$, for all $p_i =i/2n\in [p_*,p^{*}].$ Construct $k=\left\lceil {2np} \right\rceil ,p=p^{*}-\varepsilon$ and $k^{*}=\left\lfloor {2np^{*}-1} \right\rfloor ,$and apply Lemma 5.1 and $n\ge n^{*}$ to get

$$\begin{aligned} \kappa (q,k^{*}-k,k-1)\ge q^{k^{*}-k+1}\ge q^{2(n\varepsilon -1)}>G. \end{aligned}$$

Next apply subadditivity of $v(.)$, the last inequality and inequality (*) in part (a) of the theorem to get

$$\begin{aligned} v(zG)(f(p^{*}\!-\!\varepsilon )\!-\!f(p_*))&\le Gv(z)(f(p^{*}\!-\!\varepsilon )\!-\!f(p_*))\!\le \! \kappa v(z)(f(p^{*}\!-\!\varepsilon )\!-\!f(p_*)) \\ \!&\le \! v(z)\left[ {f(p^{*})\!-\!f(p_*)} \right] . \end{aligned}$$

Hence, $v(zG)f(p^{*}-\varepsilon )\le v(zG)f(p_*)+v(z)\left[ {f(p^{*})-f(p_*)} \right] ,$ which completes the proof for it reveals that

$$\begin{aligned} {\{zG},z,0;p_*,p^{*}-p_*\}\;{\succeq }\;{\{zG,0};p^{*}-\varepsilon \}. \end{aligned}$$