On the sensitivity of the value of information to risk aversion in two-action decision problems

Abstract

This paper discusses the sensitivity of the value of information to the risk aversion in two-action decision problems when the initial wealth is uncertain. We demonstrate that there is no general monotonicity between information value and the Arrow–Pratt risk aversion in this setting. We then show that monotonicity exists in the sense of Rubinstein’s measure of risk aversion when the lottery is independent of the initial wealth. Finally, we show that if the lottery is dependent on the initial wealth, then Ross’s measure of risk aversion is needed to characterize this monotonic relation. Our results explain the shape of the sensitivity analysis curve of the value of information to risk aversion and interpret various measures of risk aversion based on their monotonicity with information value.

Appendix

To simplify the notation, we denote A and B as the utility functions $U_{A}$ and $U_{B}$. We also denote $a_{A}$ and $a_{B}$ as ${\text{VOIA}}_{A}$ and ${\text{VOIA}}_{B}$, and $r_{A}$ and $r_{B}$ as ${\text{VOIR}}_{A}$ and ${\text{VOIR}}_{B}$.

Proof of Theorem 1

(i) $\rightarrow$ (ii): We prove this direction in three steps.

Step 1

Define a strictly increasing and concave function G relating two utility functions.

Since B is strictly increasing, we can uniquely define a function G such that

$$E_{{\tilde{w}}} [A(x + \tilde{w})] = G(E_{{\tilde{w}}} [B(x + \tilde{w})]),\quad \forall x.$$

Taking the first derivative gives

$$E_{{\tilde{w}}} [A^{\prime } (x + \tilde{w})] = G^{\prime } (E_{{\tilde{w}}} [B(x + \tilde{w})]) \cdot E_{{\tilde{w}}} [B^{\prime } (x + \tilde{w})].$$

(10)

Since $A^{\prime } ,B^{\prime } > 0$, we know that $G^{\prime } > 0$and that G is strictly increasing.

Taking the derivative of (10) gives

$$\begin{aligned} E_{{\tilde{w}}} [A^{\prime \prime } (x + \tilde{w})] & = G^{\prime \prime } (E_{{\tilde{w}}} [B(x + \tilde{w})]) \cdot \{ E_{{\tilde{w}}} [B^{\prime } (x + \tilde{w})]\}^{2} \\ \quad + G^{\prime } (E_{{\tilde{w}}} [B(x + \tilde{w})]) \cdot E_{{\tilde{w}}} [B^{\prime \prime } (x + \tilde{w})]. \\ \end{aligned}$$

(11)

The negative ratio of (11) to (10) gives

$$R_{A} (x;\tilde{w}) = - \frac{{E_{{\tilde{w}}} [A^{\prime \prime } (x + \tilde{w})]}}{{E_{{\tilde{w}}} [A^{\prime } (x + \tilde{w})]}} = - \frac{{G^{\prime \prime } (E_{{\tilde{w}}} [B(x + \tilde{w})])}}{{G^{\prime } (E_{{\tilde{w}}} [B(x + \tilde{w})])}}\quad E_{{\tilde{w}}} [B^{\prime } (x + \tilde{w})] - R_{A} (x;\tilde{w}).$$

(12)

Since $R_{A} (x;\tilde{w}) \ge R_{B} (x;\tilde{w})$, we know that $G^{\prime \prime } \le 0$ and that G is concave.

Step 2

Derive an inequality on the value of information for B from the concavity of G.

If $x \ge 0$, then $E_{{\tilde{w}}} [B(\tilde{w} + x)] \ge E_{{\tilde{w}}} [B(\tilde{w} + x - a_{B} )] \ge E_{{\tilde{w}}} [B(\tilde{w} - a_{B} )]$ and then

$$\begin{gathered} E_{{\tilde{w}}} [A(\tilde{w} + x)] - E_{{\tilde{w}}} [A(\tilde{w} + x - a_{B} )] \le G^{\prime } (E_{{\tilde{w}}} [B(\tilde{w} + x - a_{B} )])(E_{{\tilde{w}}} [B(\tilde{w} + x)] - E_{{\tilde{w}}} [B(\tilde{w} + x - a_{B} )]) \hfill \\ \le G^{\prime } (E_{{\tilde{w}}} [B(\tilde{w} - a_{B} )])(E_{{\tilde{w}}} [B(\tilde{w} + x)] - E_{{\tilde{w}}} [B(\tilde{w} + x - a_{B} )]), \hfill \\ \end{gathered}$$

(13)

since $G^{\prime }$ is positive and monotonically decreasing. If $x < 0$, then

$$E_{{\tilde{w}}} [A(\tilde{w} + x)] - E_{{\tilde{w}}} [A(\tilde{w} - a_{B} )] \le G^{\prime } (E_{{\tilde{w}}} [B(\tilde{w} - a_{B} )])(E_{{\tilde{w}}} [B(\tilde{w} + x)] - E_{{\tilde{w}}} [B(\tilde{w} - a_{B} )]).$$

(14)

From (13) and (14), we know that $\forall x,\forall \tilde{w}$

$$E_{{\tilde{w}}} [A(\tilde{w} + x)] - E_{{\tilde{w}}} [A(\tilde{w} + x^{ + } - a_{B} )] \le G^{\prime } (E_{{\tilde{w}}} [B(\tilde{w} - a_{B} )])(E_{{\tilde{w}}} [B(\tilde{w} + x)] - E_{{\tilde{w}}} [B(\tilde{w} + x^{ + } - a_{B} )]),$$

(15)

where $x^{ + } = \hbox{max} \{ x,0\}$.

Step 3

Prove monotonicity using (15) and the independence assumption.

Since the lottery X is independent of the initial wealth, $\tilde{w}$, taking expectations over X in both hand sides of (15) gives

$$E[A(\tilde{w} + X)] - E[A(\tilde{w} + X^{ + } - a_{B} )] \le G^{\prime } (E_{{\tilde{w}}} [B(\tilde{w} - a_{B} )])(E[B(\tilde{w} + X)] - E[B(\tilde{w} + X^{ + } - a_{B} )]) = 0.$$

The last equality is from the definition of $a_{B} \triangleq {\text{VOIA}}_{B}$. Hence, ${\text{VOIA}}_{A} \triangleq a_{A} \ge a_{B} \triangleq {\text{VOIA}}_{B}$, since $E[A(\tilde{w} + X)] = E[A(\tilde{w} + X^{ + } - a_{A} )]$ and A is strictly increasing.

(ii) $\rightarrow$ (i): We show that (ii) implies (i) by comparing the Taylor’s expansion of the value of information on some small risk lotteries for the two decision makers. For any given x and uncertain initial wealth, $\tilde{w}$, consider the new uncertain initial wealth, $W = x + \tilde{w}$and the independent lottery X with some small risk $\epsilon > 0$ as following,

If p is sufficiently close to 1, then decision makers A and B will accept such lottery without the information and then $E[A(W + X)] = E[A(W + X^{ + } - a_{A} )]$, i.e.,

$$pE[A(W + \epsilon )] + (1 - p)E[A(W - \epsilon )] = pE[A(W + \epsilon - a_{A} )] + (1 - p)E[A(W - a_{A} )].$$

The derivative of $a_{A}$with respect to $\epsilon$,

$$\frac{{{\text{d}}a_{A} }}{{{\text{d}}\epsilon }} = \frac{{pE[A^{\prime } (W + \epsilon - a_{A} )] - pE[A^{\prime } (W + \epsilon )] + (1 - p)E[A^{\prime } (W - \epsilon )]}}{{pE[A^{\prime } (W + \epsilon - a_{A} )] + (1 - p)E[A^{\prime } (W - a_{A} )]}},$$

(16)

and $a_{A} = 0$ when $\epsilon = 0$. Therefore, its value at $\epsilon = 0$,

$$\left. {\frac{{{\text{d}}a_{A} }}{{{\text{d}}\epsilon }}} \right|_{\epsilon = 0} = \frac{{pE[A^{\prime } (W)] - pE[A^{\prime } (W)] + (1 - p)E[A^{\prime } (W)]}}{{pE[A^{\prime } (W)] + (1 - p)E[A^{\prime } (W)]}} = 1 - p.$$

Denote $a_{A}^{\prime } \triangleq \frac{{{\text{d}}a_{A} }}{{{\text{d}}\epsilon }}$, the second derivative of $a_{A}$with respect to $\epsilon$,

$$\begin{gathered} \frac{{{\text{d}}^{2} a_{A} }}{{{\text{d}}\epsilon^{2} }} = \frac{{pE[A^{\prime \prime } (W + \epsilon - a_{A} )](1 - a_{A}^{\prime } ) - pE[A^{\prime \prime } (W + \epsilon )] - (1 - p)E[A^{\prime \prime } (W - \epsilon )]}}{{pE[A^{\prime } (W + \epsilon - a_{A} )] + (1 - p)E[A^{\prime } (W - a_{A} )]}} \hfill \\ \quad - \frac{{pE[A^{\prime } (W + \epsilon - a_{A} )] - pE[A^{\prime } (W + \epsilon )] + (1 - p)E[A^{\prime } (W - \epsilon )]}}{{(pE[A^{\prime } (W + \epsilon - a_{A} )] + (1 - p)E[A^{\prime } (W - a_{A} )])^{2} }} \hfill \\ \quad \cdot (p(1 - a_{A}^{\prime } )E[A^{\prime \prime } (W + \epsilon - a_{A} )] - (1 - p)a_{A}^{\prime } E[A^{\prime \prime } (W - a_{A} )]), \hfill \\ \end{gathered}$$

Since $a_{A}^{\prime } (0) = \left. {\frac{{{\text{d}}a_{A} }}{{{\text{d}}\epsilon }}} \right|_{\epsilon = 0} = 1 - p$, the second derivative of $a_{A}$ with respect to $\epsilon$ at $\epsilon = 0$,

$$\left. {\frac{{{\text{d}}^{2} a_{A} }}{{{\text{d}}\epsilon^{2} }}} \right|_{\epsilon = 0} = \{ p^{2} - p - (1 - p) - (1 - p)[p^{2} - (1 - p)^{2} ]\} \frac{{E[A^{\prime \prime } (W)]}}{{E[A^{\prime } (W)]}} = 3p(1 - p)R_{A} (W).$$

Hence, the Taylor’s expansion of $a_{A}$ at $\epsilon = 0$,

$$a_{A} = (1 - p)\epsilon + 3p(1 - p)R_{A} (W)\epsilon^{2} + o(\epsilon^{2} ).$$

(17)

Similarly, the Taylor’s expansion of $a_{B}$ at $\epsilon = 0$ is

$$a_{B} = (1 - p)\epsilon + 3p(1 - p)R_{B} (W)\epsilon^{2} + o(\epsilon^{2} ).$$

(18)

We know that $a_{A} \ge a_{B}$ for any small $\epsilon$ since decision maker A values the information higher than decision maker B when they both accept the lottery without the information. Hence, $R_{A} (W) \ge R_{B} (W)$by comparing their Taylor’s expansions in (17) and (18) since the residuals, $o(\epsilon^{2} )$, are negligible when $\epsilon$ is small.

Proof of Theorem 2

(i) $\rightarrow$ (ii): Since X is independent of $\tilde{w}$, then applying Jensen’s inequality on the concave function G gives

$$\begin{aligned} E[A(\tilde{w} + X^{ + } - {\text{VOIR}}_{B} )] & = E_{X} [G(E_{{\tilde{w}}} [B(\tilde{w} + X^{ + } - {\text{VOIR}}_{B} )])] \\ & \le G(E_{X} [E_{{\tilde{w}}} [B(\tilde{w} + X^{ + } - {\text{VOIR}}_{B} )]]) \\ & = G(E_{{\tilde{w}}} [B(\tilde{w})]) = E_{{\tilde{w}}} [A(\tilde{w})] = E[A(\tilde{w} + X^{ + } - {\text{VOIR}}_{A} )], \\ \end{aligned}$$

where $E_{X} [ \cdot ]$ is the expectation operator taken over the lottery X. Since A is strictly increasing, then ${\text{VOIR}}_{A} \le {\text{VOIR}}_{B}$.

(ii) $\rightarrow$ (i): We prove it by comparing the Taylor’s expansion of the value of information on a lottery with sufficient small risk. For any given x and uncertain initial wealth, $\tilde{w}$, consider the new uncertain initial wealth, $W = x + \tilde{w}$, and the independent lottery X in the proof of Theorem 1 (see Fig. 6). Decision makers A and B will both reject it without the information, if p is sufficiently small. From the definition of value of information, we know that

$$E[A(W)] = E[A(W + X^{ + } - r_{A} )] = pE[A(W + \epsilon - r_{A} )] + (1 - p)E[A(W - r_{A} )].$$

Fig. 6

Binary lottery with small risk

The derivative of $r_{A}$with respect to $\epsilon$,

$$\frac{{{\text{d}}r_{A} }}{{{\text{d}} \epsilon}} = \frac{{pE[A^{\prime } (W + \epsilon - r_{A} )]}}{{pE[A^{\prime } (W + \epsilon - r_{A} )] + (1 - p)E[A^{\prime } (W - r_{A} )]}},$$

(19)

and $r_{A} = 0$ when $\epsilon = 0$. Therefore, its value at $\epsilon = 0$ is

$$\left. {\frac{{{\text{d}}r_{A} }}{{{\text{d}}\epsilon }}} \right|_{\epsilon = 0} = \frac{{pE[A^{\prime } (W)]}}{{pE[A^{\prime } (W)] + (1 - p)E[A^{\prime } (W)]}} = p.$$

(20)

Denote $r_{A}^{\prime } \triangleq \frac{{{\text{d}}r_{A} }}{{{\text{d}}\epsilon }}$, the second derivative of $r_{A}$ with respect to $\epsilon$,

$$\begin{aligned} \frac{{{\text{d}}^{2} r_{A} }}{{{\text{d}}\epsilon^{2} }} & = \frac{{p(1 - r_{A}^{\prime } )E[A^{\prime \prime } (W + \epsilon - r_{A} )]}}{{pE[A^{\prime } (W + \epsilon - r_{A} )] + (1 - p)E[A^{\prime } (W - r_{A} )]}} \\ \quad - \frac{{pE[A^{\prime } (W + \epsilon - r_{A} )](p(1 - r_{A}^{\prime } )E[A^{\prime \prime } (W + \epsilon - r_{A} )] - (1 - p)r_{A}^{\prime } E[A^{\prime \prime } (W - r_{A} )])}}{{(pE[A^{\prime } (W + \epsilon - r_{A} )] + (1 - p)E[A^{\prime } (W - r_{A} )])^{2} }}. \\ \end{aligned}$$

(21)

Since $r_{A}^{\prime } (0) = \left. {\frac{{{\text{d}}r_{A} }}{{{\text{d}}\epsilon }}} \right|_{\epsilon = 0} = p$, the second derivative of $r_{A}$ with respect to $\epsilon$ at $\epsilon = 0$,

$$\left. {\frac{{{\text{d}}^{2} r_{A} }}{{{\text{d}}\epsilon^{2} }}} \right|_{\epsilon = 0} = p(1 - p)\frac{{E[A^{\prime \prime } (w)]}}{{E[A^{\prime } (W)]}} = - p(1 - p)R_{A} (W).$$

Hence, the Taylor’s expansion of $r_{A}$ at $\epsilon = 0$ is

$$r_{A} = (1 - p)\epsilon - p(1 - p)R_{A} (W)\epsilon^{2} + o(\epsilon^{2} ) .$$

(22)

Similarly, the Taylor’s expansion of $r_{B}$ at $\epsilon = 0$ is

$$r_{B} = (1 - p)\epsilon - p(1 - p)R_{B} (W)\epsilon^{2} + o(\epsilon^{2} ).$$

(23)

We know that $r_{A} \le r_{B}$ for any small $\epsilon$ since decision maker A values the information lower than decision maker B when they both reject the lottery without the information. Hence, $R_{A} (W) \ge R_{B} (W)$by comparing their Taylor’s expansion in (22) and (23) since the residuals, $o(\epsilon^{2} )$, are negligible when $\epsilon$ is small.

Proof of Theorem 3

(i) $\rightarrow$ (ii): We denote $F_{{\tilde{w}}} (w)$ as the marginal distribution of the initial wealth $\tilde{w}$ and $F_{{X|\tilde{w}}} (x|w)$ as the conditional distribution of the lottery X on the initial wealth $\tilde{w}$, then the joint distribution of X and $\tilde{w}$, $F_{{X,\tilde{w}}} (x,w)$, satisfies $F_{{X,\tilde{w}}} (x,w) = \int_{0}^{w} {F_{{X|\tilde{w}}} (x|t){\text{d}}F_{{\tilde{w}}} (t)}$. The proof is in three steps.

Step 1

Represent the value of information as the value of a continuous function $\theta (\tau )$ at $\tau = 1$.

For each $\tau \in [0,1]$, define $\theta (\tau )$ as the unique solution to

$$V_{A} (\tau ) = E[\tau A(\tilde{w} + X^{ + } - \theta (\tau )) + (1 - \tau )A(\tilde{w} + X + E[X^{ - } ] - \theta (\tau ))] = E[A(\tilde{w} + X)].$$

Direct calculation shows that $\theta (0) = E[X^{ - } ]$ and that$\theta (1) = a_{A}$. In addition,

$$\begin{aligned} 0 & = \frac{{{\text{d}}V_{A} (\tau )}}{{{\text{d}}\tau }} = \frac{\text{d}}{{{\text{d}}\tau }}\left\{ {E[\tau A(\tilde{w} + X^{ + } - \theta (\tau )) + (1 - \tau )A(\tilde{w} + X + E[X^{ - } ] - \theta (\tau ))]} \right\} \\ & = E[A(\tilde{w} + X^{ + } - \theta (\tau ))] - E[A(\tilde{w} + X + E[X^{ - } ] - \theta (\tau ))] \\ & \quad - \theta^{\prime } (\tau )\left\{ {\tau E[A^{\prime } (\tilde{w} + X^{ + } - \theta (\tau ))] + (1 - \tau )E[A^{\prime } (\tilde{w} + X + E[X^{ - } ] - \theta (\tau ))]} \right\}. \\ \end{aligned}$$

therefore,

$$\theta^{\prime } (\tau ) = - \frac{{E[A(\tilde{w} + X + E[X^{ - } ] - \theta (\tau ))] - E[A(\tilde{w} + X^{ + } - \theta (\tau ))]}}{{\tau E[A^{\prime } (\tilde{w} + X^{ + } - \theta (\tau ))] + (1 - \tau )E[A^{\prime } (\tilde{w} + X + E[X^{ - } ] - \theta (\tau ))]}}.$$

(24)

Step 2

Transform $\theta^{\prime } (\tau )$ into the ratio of a summation of positive weighted integrals of $A^{\prime \prime }$ to another summation of positive weighted integrals of $A^{\prime }$.

Using the joint distribution $F_{{X,\tilde{w}}} (x,w)$, we can rewrite the numerator of (24) as

$$E[A(\tilde{w} + X + E [X^{ - } ] - \theta (\tau ))] - E[A(\tilde{w} + X^{ + } - \theta (\tau ))] = \int\limits_{0}^{ + \infty } {{\text{I}}(w) - {\text{II}}(w) + {\text{III}}(w){\text{d}}F_{{\tilde{w}}} (w)} ,$$

(25)

where $I \triangleq \int_{{ - E[X^{ - } ]}}^{0} {\int_{0}^{{x + E[X^{ - } ]}} {A^{\prime } (w + s - \theta (\tau )){\text{d}}s} {\text{d}}F_{{X|\tilde{w}}} (x|w)}$, $II \triangleq \int_{ - \infty }^{{ - E[X^{ - } ]}} {\int_{{x + E[X^{ - } ]}}^{0} {A^{\prime}(w + s - \theta (\tau )){\text{d}}s} {\text{d}}F_{{X|\tilde{w}}} (x|w)}$,

$$III \triangleq \int\limits_{0}^{ + \infty } {\int\limits_{x}^{{x + E[X^{ - } ]}} {A^{\prime } (w + s - \theta (\tau )){\text{d}}s} {\text{d}}F_{{X|\tilde{w}}} (x|w)} .$$

Since $E[X^{ - } |\tilde{w}] = E[X^{ - } ]$, then $E[E[X^{ - } ] - X^{ - } |w] = 0$ and

$$0 = A^{\prime } (w - \theta (\tau ))E[E[X^{ - } ] - X^{ - } |w] = \int\limits_{0}^{ + \infty } {{\text{IV}}(w) - {\text{V}}(w) + {\text{VI}}(w){\text{d}}F_{{\tilde{w}}} (w)} ,$$

(26)

where ${\text{IV}} \triangleq \int_{{ - E[X^{ - } ]}}^{0} {\int_{0}^{{x + E[X^{ - } ]}} {A^{\prime } (w - \theta (\tau )){\text{d}}s} {\text{d}}F_{{X|\tilde{w}}} (x|w)}$, $V \triangleq \int_{ - \infty }^{{ - E[X^{ - } ]}} {\int_{{x + E[X^{ - } ]}}^{0} {A^{\prime } (w - \theta (\tau )){\text{d}}s} {\text{d}}F_{{X|\tilde{w}}} (x|w)}$,

$$VI \triangleq \int\limits_{0}^{ + \infty } {\int\limits_{x}^{{x + E[X^{ - } ]}} {A^{\prime } (w - \theta (\tau )){\text{d}}s} {\text{d}}F_{{X|\tilde{w}}} (x|w)} .$$

Since $\int_{a}^{b} {\left( {A^{\prime } (w + s - \theta (\tau )) - A^{\prime } (w - \theta (\tau ))} \right){\text{d}}s} = \int_{a}^{b} {\int_{0}^{s} {A^{\prime \prime } (w + t - \theta (\tau )){\text{d}}t} {\text{d}}s}$, then

$$I - IV = \int\limits_{{ - E[X^{ - } ]}}^{0} {\int\limits_{0}^{{x + E[X^{ - } ]}} {\int\limits_{0}^{s} {A^{\prime \prime } (w + t - \theta (\tau )){\text{d}}t} } {\text{d}}s{\text{d}}F_{{X|\tilde{w}}} (x|w)} ,$$

(27)

$$V - II = \int\limits_{ - \infty }^{{ - E[X^{ - } ]}} {\int\limits_{{x + E[X^{ - } ]}}^{0} {\int\limits_{s}^{0} {A^{\prime \prime } (w + t - \theta (\tau )){\text{d}}t} } {\text{d}}s{\text{d}}F_{{X|\tilde{w}}} (x|w)} ,$$

(28)

and

$$III - VI = \int\limits_{0}^{ + \infty } {\int\limits_{x}^{{x + E[X^{ - } ]}} {\int\limits_{0}^{s} {A^{\prime \prime } (w + t - \theta (\tau )){\text{d}}t} } {\text{d}}s{\text{d}}F_{{X|\tilde{w}}} (x|w)} .$$

(29)

Substituting from (26), (27), (28), and (29) into (25) gives a positive additive form of the integrals of $A^{\prime \prime }$ for the numerator of (24) as

$$\begin{aligned} E[A(\tilde{w} + X + E[X^{ - } ] - \theta (\tau ))] - E[A(\tilde{w} + X^{ + } - \theta (\tau ))] \, & = \int\limits_{0}^{ + \infty } {\int\limits_{{ - E[X^{ - } ]}}^{0} {\int\limits_{0}^{{x + E[X^{ - } ]}} {\int\limits_{0}^{s} {A^{\prime \prime } (w + t - \theta (\tau )){\text{d}}t} } {\text{d}}s{\text{d}}F_{{X|\tilde{w}}} (x|w)} } {\text{d}}F_{{\tilde{w}}} (w) \\ & \quad + \int\limits_{0}^{ + \infty } {\int\limits_{ - \infty }^{{ - E[X^{ - } ]}} {\int\limits_{{x + E[X^{ - } ]}}^{0} {\int\limits_{s}^{0} {A^{\prime \prime } (w + t - \theta (\tau )){\text{d}}t} } {\text{d}}s{\text{d}}F_{{X|\tilde{w}}} (x|w)} } {\text{d}}F_{{\tilde{w}}} (w) \\ & \quad + \int\limits_{0}^{ + \infty } {\int\limits_{0}^{ + \infty } {\int\limits_{x}^{{x + E[X^{ - } ]}} {\int\limits_{0}^{s} {A^{\prime \prime } (w + t - \theta (\tau )){\text{d}}t} } {\text{d}}s{\text{d}}F_{{X|\tilde{w}}} (x|w)} } {\text{d}}F_{{\tilde{w}}} (w) \triangleq N(A^{\prime \prime } ) \\ \end{aligned}$$

(30)

The denominator of (24) can be expressed as positive summation of the integrals of $A^{\prime }$, $M(A^{\prime } )$,

$$\begin{aligned} \tau E[A^{\prime } (\tilde{w} + X^{ + } - \theta (\tau ))] + (1 - \tau )E[A^{\prime } (\tilde{w} + X + E[X^{ - } ] - \theta (\tau ))] & = \tau \int\limits_{0}^{ + \infty } {\int\limits_{ - \infty }^{0} {A^{\prime } (w - \theta (\tau ))F_{{X|\tilde{w}}} (x|w)} } {\text{d}}F_{{\tilde{w}}} (w) + \tau \int\limits_{0}^{ + \infty } {\int\limits_{0}^{ + \infty } {A^{\prime } (w + x - \theta (\tau ))F_{{X|\tilde{w}}} (x|w)} } {\text{d}}F_{{\tilde{w}}} (w) \\ & \quad + (1 - \tau )\int\limits_{0}^{ + \infty } {\int\limits_{ - \infty }^{ + \infty } {A^{\prime } (w + x + E[X^{ - } ] - \theta (\tau ))F_{{X|\tilde{w}}} (x|w)} } {\text{d}}F_{{\tilde{w}}} (w) \triangleq M(A^{\prime } ). \\ \end{aligned}$$

(31)

Step 3

Prove that (ii) implies (i) by applying Machina and Neilson’s result.

For each $\tau \in [0,1]$, define $\bar{\theta }(\tau )$ as the unique solution to

$$\begin{aligned} V_{B} (\tau ) & = E[\tau B(\tilde{w} + X^{ + } - \bar{\theta} (\tau )) + (1 - \tau )B(\tilde{w} + X + E[X] - \bar{\theta} (\tau ))] \\ & = E[B(\tilde{w} + X^{ + } - a_{B} )] = E[B(\tilde{w} + X)]. \\ \end{aligned}$$

Similar to the argument for $V_{A} (\tau )$, we know that $\bar{\theta }(0) = E[X^{ - } ],\bar{\theta }(1) = a_{B}$, and that

$$\theta^{\prime } (\tau ) = - \frac{{N(A^{\prime \prime } )}}{{M(A^{\prime } )}} \ge - \frac{{N(B^{\prime\prime})}}{{M(B^{\prime})}} = \bar{\theta }^{\prime } (\tau ),\quad \forall \tau \in [0,1],$$

The inequality is due to Machina and Neilson (1987) since $\gamma_{A} (x) \ge_{S} \gamma_{B} (x)$. Since $\bar{\theta }(0) = \theta (0) = E[X^{ - } ]$, then $a_{A} = \theta (1) \ge \bar{\theta }(1) = a_{B}$.

(ii) $\rightarrow$ (i): We prove that (ii) implies (i) by comparing Taylor’s expansion of the value of information on such lottery for the two decision makers.

Step 1

Calculate Taylor’s expansion of the value of information for a small risk lottery.

Consider the uncertain initial wealth, $\tilde{w}$, and the lottery, X, in the Fig. 7:

Fig. 7

Binary initial wealth and conditional binary distributed small risk lottery

Note that the condition expectation of the potential loss on the uncertain initial wealth, $E[X^{ - } |\tilde{w}]$, is constant whatever the initial wealth is. More precisely, $E[X^{ - } |\tilde{w}] = E[X^{ - } ] = \tfrac{1}{2}\epsilon$. Moreover, if k is large enough, then two decision makers will both accept the lottery without the information. Statement (ii) indicates $a_{A} \ge a_{B} ,\forall x,y,p,k,\epsilon$ in this case. The definition of $a_{A}$ gives

$$\begin{gathered} \tfrac{p}{2}\left\{ {\left. {A(x + k\epsilon ) + A(x - \epsilon )} \right\}} \right. + \tfrac{1 - p}{2}\left\{ {A(y + \epsilon ) + A(y - \epsilon )} \right\} \hfill \\ = 0.5p\left\{ {\left. {A(x + k\epsilon - a_{A} ) + A(x - a_{A} )} \right\}} \right. + 0.5(1 - p)\left\{ {A(y + \epsilon - a_{A} ) + A(y - a_{A} )} \right\}. \hfill \\ \end{gathered}$$

Straight forward calculation shows that $a_{A} = 0$ when $\epsilon = 0$. Differentiating both hand sides with respect to $\epsilon$ gives

$$\begin{aligned} \frac{{{\text{d}}a_{A} }}{{{\text{d}}\epsilon }} & = \frac{{p\left\{ {kA^{\prime } (x + k\epsilon - a_{A} ) - kA^{\prime } (x + k\epsilon ) + A^{\prime } (x - \epsilon )} \right\}}}{{pA^{\prime } (x + k\epsilon - a_{A} ) + pA^{\prime } (x - a_{A} ) + (1 - p)\{ A^{\prime } (y + \epsilon - a_{A} ) + A^{\prime } (y - a_{A} )\} }} \\ & \quad + \frac{{(1 - p)\left\{ {\left. {A^{\prime } (y + \epsilon - a_{A} ) - A^{\prime } (y + \epsilon ) + A^{\prime } (y - \epsilon )} \right\}} \right.}}{{pA^{\prime } (x + k\epsilon - a_{A} ) + pA^{\prime } (x - a_{A} ) + (1 - p)\{ A^{\prime } (y + \epsilon - a_{A} ) + A^{\prime } (y - a_{A} )\} }}, \\ \end{aligned}$$

and $\left. {\frac{{{\text{d}}a_{A} }}{{{\text{d}}\epsilon }}} \right|_{\epsilon = 0} = 0.5$.

Differentiating $\frac{{{\text{d}}a_{A} }}{{{\text{d}}\epsilon }}$ with respect to $\epsilon$ at $\epsilon = 0$ gives

$$\left. {\frac{{{\text{d}}^{2} a_{A} }}{{{\text{d}}\epsilon^{2} }}} \right|_{\epsilon = 0} = - \frac{{p(2k + 1)A^{\prime \prime } (x) + 3(1 - p)A^{\prime \prime } (y)}}{{4(pA^{\prime } (x) + (1 - p)A^{\prime } (y))}}.$$

Since $a_{A} = 0$ when $\epsilon = 0$, the Taylor’s expansion of $a_{A}$ at $\epsilon = 0$ is

$$a_{A} = \frac{\epsilon }{2} - \frac{{p(2k + 1)A^{\prime \prime } (x) + 3(1 - p)A^{\prime \prime } (y)}}{{4(pA^{\prime } (x) + (1 - p)A^{\prime } (y))}}\epsilon^{2} + o(\epsilon^{2} ),$$

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Similarly, the Taylor’s expansion of $a_{A}$ at $\epsilon = 0$ is

$$a_{B} = \frac{\epsilon }{2} - \frac{{p(2k + 1)B^{\prime \prime } (x) + 3(1 - p)B^{\prime \prime } (y)}}{{4(pB^{\prime } (x) + (1 - p)B^{\prime } (y))}}\epsilon^{2} + o(\epsilon^{2} ).$$

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Step 2

Compare Taylor’s expansions for particular parameters.

It follows that $a_{A} \ge a_{B}$ for all such lotteries and initial wealth only if ($\forall x,y,p,k$)

$$- \frac{{p(2k + 1)A^{\prime \prime } (x) + 3(1 - p)A^{\prime \prime } (y)}}{{pA^{\prime } (x) + (1 - p)A^{\prime } (y)}} \ge - \frac{{p(2k + 1)B^{\prime \prime } (x) + 3(1 - p)B^{\prime \prime } (y)}}{{pB^{\prime } (x) + (1 - p)B^{\prime } (y)}}.$$

If $- \frac{{A^{\prime \prime } (x)}}{{pA^{\prime } (x) + (1 - p)A^{\prime } (y)}} < - \frac{{B^{\prime \prime } (x)}}{{pB^{\prime } (x) + (1 - p)B^{\prime } (y)}}$ for some x, y, and p, then we have a contradiction for k sufficiently large. Hence ($\forall x,y,p$),

$$- \frac{{A^{\prime \prime } (x)}}{{pA^{\prime } (x) + (1 - p)A^{\prime } (y)}} \ge - \frac{{B^{\prime \prime } (x)}}{{pB^{\prime } (x) + (1 - p)B^{\prime } (y)}}.$$

If for some x and y, $- \frac{{A^{\prime \prime } (x)}}{{A^{\prime } (y)}} < - \frac{{B^{\prime \prime } (x)}}{{B^{\prime } (y)}},$ then for p sufficiently small we have a contradiction. Hence ($\forall x,y$), $- \frac{{A^{\prime \prime } (x)}}{{A^{\prime } (y)}} \ge - \frac{{B^{\prime \prime } (x)}}{{B^{\prime } (y)}}$, i.e., $\gamma_{A} \ge_{S} \gamma_{B}$.

Proof of Theorem 4

(i) $\rightarrow$ (ii): Denote $CE(A(x + \tilde{w}),X)$ and $CE(B(x + \tilde{w}),X)$ as the certain equivalents of the lottery, X, for A and B with initial wealth, $\tilde{w}$, respectively. Then

$$E[A(\tilde{w} + X)] = E[A(\tilde{w} + CE(A(x + \tilde{w}),X))],$$

and

$$E[B(\tilde{w} + X)] = E[B(\tilde{w} + CE(B(x + \tilde{w}),X))].$$

If $E[X^{ + } |\tilde{w}] = E[X^{ + } ]$, then $E[X^{ + } |(\tilde{w} - {\text{VOIR}}_{A} )] = E[X^{ + } ]$. Machina and Neilson (1987) showed that if $\gamma_{A} \ge_{S} \gamma_{B}$ and if $E[X^{ + } |(\tilde{w} - {\text{VOIR}}_{A} )] = E[X^{ + } ]$, then

$$CE(A(x + \tilde{w} - {\text{VOIR}}_{A} ),X^{ + } ) \le CE(B(x + \tilde{w} - {\text{VOIR}}_{A} ),X^{ + } ).$$

By directly comparing the definitions of certain equivalent and the value of information in (5), we know that ${\text{VOIR}}_{A} = CE(A(x + \tilde{w} - {\text{VOIR}}_{A} ),X^{ + } )$. Since B is monotonically increasing, we have

$$\begin{aligned} E[B(\tilde{w} - {\text{VOIR}}_{A} + X^{ + } )] & = E[B(\tilde{w} - {\text{VOIR}}_{A} + CE(B(x + \tilde{w} - {\text{VOIR}}_{A} ),X^{ + } ))] \\ & \ge E[B(\tilde{w} - {\text{VOIR}}_{A} + CE(A(x + \tilde{w} - {\text{VOIR}}_{A} ),X^{ + } ))] \\ & = E[B(\tilde{w} - {\text{VOIR}}_{A} + {\text{VOIR}}_{A} )] = E[B(\tilde{w})] = E[B(\tilde{w} - {\text{VOIR}}_{B} + X^{ + } )]. \\ \end{aligned}$$

Therefore, ${\text{VOIR}}_{A} \le {\text{VOIR}}_{B}$, Since B is monotonically increasing.

(ii) $\rightarrow$ (i): We prove that (ii) implies (i) by comparing Taylor’s expansion of the value of information on such lottery for the two decision makers.

Step 1

Calculate Taylor’s expansion of the value of information for a small risk lottery.

Again consider the uncertain initial wealth $\tilde{w}$ and the lottery X in the Fig. 8.

Fig. 8

Binary initial wealth and conditional binary lottery with small risk

Then, the condition expectation of potential gain on the uncertain initial wealth, $E[X|\tilde{w}]$ is constant. More precisely, $E[X^{ + } |\tilde{w}] = 0.5\epsilon$, either $\tilde{w} = x$ or $\tilde{w} = y$. Moreover, two decision makers will both reject the lottery without the information when k is large enough. Hence, statement (ii) indicates that $r_{A} \le r_{B} ,\forall x,y,p,k,\epsilon$ in this case. From the definition of $r_{A}$, we have

$$pA(x) + (1 - p)A(y) = \frac{p}{2k}\left\{ {\left. {A(x + k\epsilon - r_{A} ) + (2k - 1)A(x - r_{A} )} \right\}} \right. + \frac{1 - p}{2}\left\{ {A(y + \epsilon - r_{A} ) + A(y - r_{A} )} \right\}.$$

Differentiating both hand sides with respect to $\epsilon$ gives

$$\frac{{{\text{d}}r_{A} }}{{{\text{d}}\epsilon }} = \frac{{kpA^{\prime } (x + k\epsilon - r_{A} ) + k(1 - p)A^{\prime } (y + \epsilon - r_{A} )}}{{pA^{\prime } (x + k\epsilon - r_{A} ) + (2k - 1)pA^{\prime } (x - r_{A} ) + k(1 - p)\{ A^{\prime } (y + \epsilon - r_{A} ) + A^{\prime } (y - r_{A} )\} }}$$

and $\left. {\frac{{{\text{d}}r_{A} }}{{{\text{d}}\epsilon }}} \right|_{\epsilon = 0} = 0.5$.

Differentiating $\frac{{{\text{d}}r_{A} }}{{{\text{d}}\epsilon }}$with respect to $\epsilon$again at $\epsilon = 0$ gives

$$\left. {\frac{{{\text{d}}^{2} r_{A} }}{{{\text{d}}\epsilon ^{2} }}} \right|_{\epsilon = 0} = \frac{{p(2k - 1)A^{\prime \prime } (x) + (1 - p)A^{\prime \prime } (y)}}{{4(pA^{\prime } (x) + (1 - p)A^{\prime } (y))}}.$$

Since $a_{A} = 0$ when $\epsilon = 0$, the Taylor’s expansion of $a_{A}$ at $\epsilon = 0$is

$$r_{A} = \frac{\epsilon }{2} + \frac{{p(2k - 1)A^{\prime \prime } (x) + (1 - p)A^{\prime \prime } (y)}}{{4(pA^{\prime } (x) + (1 - p)A^{\prime } (y))}}\epsilon^{2} + o(\epsilon^{2} ),$$

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Similarly, the Taylor’s expansion of $a_{A}$at $\epsilon = 0$ is

$$r_{B} = \frac{\epsilon }{2} + \frac{{p(2k - 1)B^{\prime \prime } (x) + (1 - p)B^{\prime \prime } (y)}}{{4(pB^{\prime } (x) + (1 - p)B^{\prime } (y))}}\epsilon^{2} + o(v^{2} ).$$

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Step 2

Compare Taylor’s expansions for the particular chosen parameters.

It follows that $r_{A} \le r_{B}$ for all such lotteries and initial wealth only if ($\forall x,y,p,k$)

$$\frac{{p(2k-1)A^{\prime \prime } (x) + (1 - p)A^{\prime \prime } (y)}}{{pA^{\prime } (x) + (1 - p)A^{\prime } (y)}} \le \frac{{p(2k - 1)B^{\prime \prime } (x) + (1 - p)B^{\prime \prime } (y)}}{{pB^{\prime } (x) + (1 - p)B^{\prime } (y)}}.$$

If $\frac{{A^{\prime \prime } (x)}}{{pA^{\prime } (x) + (1 - p)A^{\prime } (y)}} > \frac{{B^{\prime \prime } (x)}}{{pB^{\prime } (x) + (1 - p)B^{\prime } (y)}}$ for some x, y, and p, then we have a contradiction for k sufficiently large. Hence ($\forall x,y,p$),

$$\frac{{A^{\prime \prime } (x)}}{{pA^{\prime } (x) + (1 - p)A^{\prime } (y)}} \le \frac{{B^{\prime \prime } (x)}}{{pB^{\prime } (x) + (1 - p)B^{\prime } (y)}}.$$

If for some x and y, $\frac{{A^{\prime \prime } (x)}}{{A^{\prime } (y)}} > \frac{{B^{\prime \prime } (x)}}{{B^{\prime } (y)}},$ then for p sufficiently small we have a contradiction. Hence ($\forall x,y$), $- \frac{{A^{\prime \prime } (x)}}{{A^{\prime } (y)}} \ge - \frac{{B^{\prime \prime } (x)}}{{B^{\prime } (y)}}$, i.e., $\gamma_{A} \ge_{S} \gamma_{B}$.