Menu-dependent risk attitudes: Theory and evidence

Abstract

We test for a novel pattern of menu-dependent risk attitudes that forms the basis of recent theories of risky choice: Does expanding the range of potential prizes from lotteries in a choice set lead people to overweight those prizes and make riskier choices? Contrary to our hypothesis, we find no evidence of such a menu effect. Varying the potential prize offered by an actuarially unfavorable, high-risk lottery does not affect the likelihood of choosing a different, moderate-risk gamble in favor of a safer alternative. Our well-powered null results cast doubt on prominent theories of menu-dependent risk preferences.

Appendices

Appendix A: Proofs

1.1 Proof of prediction 1

Proof: In the choice problem presented in Section 2.2, the upside-range is $(\frac{4}{3\alpha }-1)z$ when $\alpha < 2/(3\beta )$, and the downside range is either 0 ($p>0$) or z ($p=0$), both of which are independent of $\alpha$ and $\beta$. Thus, for convenience we denote $g^+(\alpha )=g(\bar{\Delta }(\mathcal {C}(\alpha ,\beta )))$ when $\alpha < 2/(3\beta )$ and $g^-\equiv g(\underline{\Delta }(\mathcal {C}))$. Also, we normalize $u(0)=0$. Then $x_1\succeq x_2(\beta )$ in menu $\mathcal {C}(\alpha ,\beta )$ if and only if

$$\begin{aligned} \begin{aligned}&\frac{pg^+(\alpha )u(z)}{pg^+(\alpha )+(1-p)g^-}\ge \frac{\frac{p}{2}g^+(\alpha )u(2\beta z)}{\frac{p}{2}g^+(\alpha )+(1-\frac{p}{2})g^-}\\&\Leftrightarrow \frac{u(2\beta z)}{u(z)}\le \frac{pg^+(\alpha )+(2-p)g^-}{pg^+(\alpha )+(1-p)g^-}=1+\frac{g^-}{pg^+(\alpha )+(2-p)g^-}. \end{aligned} \end{aligned}$$

(4)

Function $\tau (\alpha ):=1+\frac{g^-}{pg^+(\alpha )+(2-p)g^-}$ is a increasing function of $\alpha$ since function $4/(3\alpha )-1$ is decreasing in $\alpha$ but g is increasing. Thus, a smaller $\alpha$, i.e., a larger upside range, lowers the minimum requirement of $\beta$ to make $x_2$ the preferred option.

1.2 The properties of the expected-value premium

By Eq. (4), the expected-value premium $\beta ^*(\alpha )$ is given by

$$\begin{aligned} \beta ^*(\alpha )=\frac{1}{2z}u^{-1}[\tau (\alpha )u(z)] \end{aligned}.$$

(5)

This function is an increasing function of $\alpha$ since both $\tau (\alpha )$ and u are increasing.

Appendix B: Focusing and behavioral anomalies

This section demonstrates how the model we developed in Section 2 predicts classic behavioral anomalies in risky choices. We show how our simple model of focusing in risky choice can predict the certainty effect, preference reversals, and the common ratio effect.

1.1 Certainty effect

Numerous empirical studies have shown that there is a discontinuous change in utility when a small probability is reduced to zero (Kahneman & Tversky, 1979; Cohen & Jaffray, 1988; Conlisk, 1989). Such discontinuities can arise in the model proposed in Section 2. To see this, consider two lotteries $x=(z_1,z_2;p)$ and $x'(\varepsilon )=(z'_1,z'_2;1-\varepsilon )$ for some $\varepsilon >0$. Denote by $\mathcal {C}(\varepsilon )=\{x_1,x_2(\varepsilon )\}$, and then $\bar{\Delta }(\mathcal {C}(\varepsilon ))=|z_1-z_1'|$ and

$$\begin{aligned} \underline{\Delta }(\mathcal {C}(\varepsilon ))=\left\{ \begin{array}{cc} |z_2-z_1'| &{} \varepsilon =0, \\ |z_2-z_2'| &{} \varepsilon >0, \end{array} \right. \end{aligned}$$

which is invariant to the value of $\varepsilon$. Assume that (i) $|z_2-z_2'| < |z_2-z_1'|$, and (ii)

$$\begin{aligned} V(x;\mathcal\;{C}(\varepsilon ))=\frac{pg(|z_1-z'_1|)u(z_1)+(1-p)g(|z_2-z'_2|)u(z_2)}{pg(|z_1-z'_1|)+(1-p)g(|z_2-z'_2|)}=u(z_1'). \end{aligned}$$

Since $u(z_1')>u(z_2')$, function

$$\begin{aligned} V(x';\mathcal{C}(\varepsilon ))=\frac{g(|z_1-z_1'|)(1-\varepsilon )u(z_1')+g(|z_2-z_2'|)\varepsilon u(z_2')}{g(|z_1-z_1'|)(1-\varepsilon )+g(|z_2-z_2'|)\varepsilon } \end{aligned}$$

is decreasing in $\varepsilon$. When $\varepsilon$ approaches to 0, it gets infinitely close to, but still smaller than, $u(z_1)=V(x;\mathcal{C}(\varepsilon ))$. Therefore, the agent chooses x in $\mathcal {C}(\varepsilon )$ when $\varepsilon >0$.

However, when $\varepsilon =0$, the utility of $x'$ approaches continuously to $V(x';\mathcal{C}(0))=u(z_1)$, but the utility of x becomes

$$\begin{aligned} V(x;\mathcal{C}(\varepsilon ))=\frac{pg(|z_1-z'_1|)u(z_1)+(1-p)g(|z_2-z'_1|)u(z_2)}{pg(|z_1-z'_1|)+(1-p)g(|z_2-z'_1|)}, \end{aligned}$$

which is strictly smaller than $u(z_1)$ because $|z_2-z_2'| < |z_2-z_1'|$. Therefore, there is a discontinuous change in preference when prize $z_2'$ shifts from an improbable prize to an impossible prize.

1.2 Preference reversal

Prior work has also demonstrated that different elicitation methods can lead to inconsistent choices (Lichtenstein & Slovic, 1971; Slovic, 1975; Grether & Plott, 1979). The model introduced in Section 2 can predict such preference reversal because different methods of elicitation led to different comparison menus.

To see how preference reversals can arise, consider the following choice problem $\mathcal {C}=\{x_{\$},x_{P}\}$, where $x=(z_1,z_2;p)$ and $x_{P}=(z'_1,z'_2;p')$. $x_{\$}$ is the “dollar bet” and $x_P$ is the “probability bet” because we assume $z_1>z_1'$, $z_2>z_2'$ and $p < p'$. When compared directly, the agent makes a choice in menu $\mathcal {C}$ and then a preference relation $\succeq _{\mathcal {C}}$ is revealed such that $x\succeq _{\mathcal {C}}x'$ if x is selected in menu $\mathcal {C}$. Alternatively, the two lotteries can also be ranked according to their certainty equivalences. Here, the certainty equivalence of lottery x is defined as a prize $z^*(x)$ such that x and $\delta _{z^*(x)}$ are equally likely to be chosen in menu $\{x,z^*(x)\}$. By deriving the certainty equivalence, another preference $\succeq _{\text {CE}}$ can be identified such that $x\succeq _{\text {CE}}x'$ if and only if $z^*(x)\ge z^*(x')$.Therefore, a preference reversal emerges when $x\succeq _{\mathcal {C}}x'$ but $x'\succ _{\text {CE}}x$.

Consider an example in which the following restrictions are imposed to $x_{\$}$ and $x_{P}$:

• (a) $z_1-z_1'=z_2-z_2'$;

• (b) $pu(z_1)+(1-p)u(z_2)=p'u(z'_1)+(1-p')u(z'_2)=0$;

• (c) $p'=0.5$.

where a) and b) is sufficient to guarantee that $x_{\$}=x_P$ are equally attractive in menu $\mathcal {C}$, because $\bar{\Delta }(\mathcal {C})=z_1-z_1'=z_2-z_2'=\underline{\Delta }(\mathcal {C})$. Even when the two options are indifferent when compared directly, empirical studies typically find that the minimum acceptable selling price of the dollar bet is larger than that of the probability bet; that is, $z^*(x_{\$})$ > $z^*(x_P)$. The following proposition outlines when this prediction arises in our framework.

Proposition 1

Under assumptions a), b), and c), $z^*(x_{\$})$ > $z^*(x_P)$.

Proof

Consider the certainty equivalence of lottery $x=(z_1,z_2;p)$, where $pu(z_1)+(1-p)u(z_2)=0$. Define:

$$\begin{aligned} h_x(z):=\frac{g(z_1-z)-g(z-z_2)}{g(z-z_2)}, \end{aligned}$$

which is a decreasing function in range $[z_2,z_1]$. Thus, $h_x(z) < 0$ if and only if z > $\frac{z_1+z_2}{2}$; that is, the prize is larger than the mean of the two prizes. Since $pu(z_1)=-(1-p)u(z_2)$, x, and $z^*(x)$ are equally likely to be chosen if and only if

$$\begin{aligned} \frac{pg(z_1-z^*(x))u(z_1)+(1-p)g(z^*(x)-z_2)u(z_2)}{pg(z_1-z^*(x))+(1-p)g(z^*(x)-z_2)}=\frac{pu(z_1)h_x(z^*(x))}{ph_x(z^*(x))+1}=u(z^*(x)). \end{aligned}$$

Let $H(z,x):=\frac{pu(z_1)h_x(z)}{ph_x(z)+1}$, and $H(\cdot ,x)$ is decreasing since $h_x(\cdot )$ is decreasing. Hence, $z^*(x)$ is the intersection of decreasing function $H(\cdot ,x)$ and increasing function $u(\cdot )$.

Claim 1:

The certainty equivalence of $x_P$ is smaller than its expectation, $\frac{z_1'+z_2'}{2}$.

Proof

To compare lotteries $x_{\$}$ and $x_P$, note that since $u(\cdot )$ is concave and $p'=0.5$, by Jensen’s inequality we have $u(\frac{z'_1+z'_2}{2})$ > $0=p'u(z_1')+(1-p')u(z_2')$. Meanwhile, $H(\frac{z'_1+z'_2}{2},x_P)=0$ since $h_{x_P}(\frac{z'_1+z'_2}{2})=0$. Thus, the certainty equivalence of $x_P$ is smaller than $\frac{z'_1+z'_2}{2}$ because u is increasing but $H(\cdot ,x)$ is decreasing.

Claim 2:

$H(z,x_{\$})$ > $H(z,x_P)$ for all $z\in (\frac{z_1'+z_2'}{2},z_1')$.

Proof

It is trivial that $h_{x_{\$}}(z)$ > $h_{x_{P}}(z)$ for all $z\in (z_2,z_1')$. By assumption b) we have $p'u(z_1')$ > $pu(z_1)$. Thus, for any $z > \frac{z_1'+z_2'}{2}$ we have $h_{x_P}(z) < 0$, and then

$$\begin{aligned}&(p'u(z_1')-pu(z_1))h_{x_P}(z) < 0 < pp'(u(z_1)-u(z_1'))h^2_{x_P}(z)\\ \Leftrightarrow&pp'u(z'_1)h^2_{x_P}(z)+p'u(z'_1)h_{x_P}(z) < pp'u(z_1)h^2_{x_P}(z)+pu(z_1)h_{x_P}(z)\\ \Leftrightarrow&\frac{p'u(z'_1)h_{x_P}(z)}{p'h_{x_P}(z)+1}=H(z,x_P) < \frac{pu(z_1)h_{x_P}(z)}{ph_{x_P}(z)+1} < \frac{pu(z_1)h_{x_{\$}}(z)}{ph_{x_{\$}}(z)+1}=H(z,x_{\$}), \end{aligned}$$

which completes the proof.

By Claim 1, the certainty equivalence of $x_P$, $z^*(x_P)$, is greater than $\frac{z_1'+z_2'}{2}$. Thus, by Claim 2,

$$\begin{aligned} u(z^*(x_P))=H(z^*(x_P),x_P) < H(z^*,x_{\$}). \end{aligned}$$

Therefore, since u is increasing and $H(\cdot ,x_{\$})$ is decreasing, $z^*(x_P)$ < $z^*(x_{\$})$.

1.3 Common ratio effect

In the problem 3 and 4 in Kahneman and Tversky (1979), the menus are $\mathcal {C}=(x_1,x_2)$, where $x_1=(\$4000,0;0.8)$ and $x_2=(\$3000,\$3000;1)$, and $\mathcal {C}'=(x_1',x_2')$, where $x_1'=(\$4000,0;0.2)$ and $x_2'=(\$3000,0;0.25)$. Then in menu $\mathcal {C}$, $\bar{\Delta }(\mathcal {C})=1000$ and $\underline{\Delta }(\mathcal {C})=3000$. However, in menu $\mathcal {C}'$, $\bar{\Delta }(\mathcal {C}')=1000$ but $\underline{\Delta }(\mathcal {C}')=0$. Thus, the DM is more downside-focused in menu $\mathcal {C}$ than in menu $\mathcal {C'}$. To simplify, normalize $u(0)=0$. Then we have the following observation:

Observation 1

There exists a parameter environment such that the DM chooses $x_2$ in $\mathcal {C}$ and chooses $x_1$ in $\mathcal {C}'$.

Proof

By function (1) and (2), $x_1$ is chosen in menu $\mathcal {C}$ if

$$\begin{aligned} \frac{u(3000)}{u(4000)}\le \frac{0.8 g(1000)}{0.8 g(1000)+0.2g(3000)}. \end{aligned}$$

(6)

Similarly, $x_1'$ is chosen in menu $\mathcal {C}'$ if

$$\begin{aligned} \frac{u(3000)}{u(4000)}\le \frac{0.2 g(1000)}{0.2 g(1000)+0.8g(0)}\cdot \frac{0.25 g(1000)+0.75g(0)}{0.25g(1000)}. \end{aligned}$$

(7)

It is sufficient to show the threshold in Eq. (6) is smaller than that in Eq. (7). Observe that

$$\begin{aligned} \begin{aligned}&\frac{0.8 g(1000)}{0.8 g(1000)+0.2g(3000)}\le \frac{0.2 g(1000)}{0.2 g(1000)+0.8g(0)}\cdot \frac{0.25 g(1000)+0.75g(0)}{0.25g(1000)}\\ \Leftrightarrow&\frac{4g(1000)}{4g(1000)+g(3000)}\le \frac{g(1000)+3g(0)}{g(1000)+4g(0)}\\ \Leftrightarrow&4g(1000)g(0)\le g(1000)g(3000)+g(0)g(3000), \end{aligned} \end{aligned}$$

which holds automatically since g is increasing.