Persuasion under ambiguity

Abstract

This paper introduces a receiver who perceives ambiguity in a binary model of Bayesian persuasion. The sender has a well-defined prior, while the receiver considers an interval of priors and maximizes a convex combination of worst and best expected payoffs ($\alpha$-maxmin preferences). We characterize the sender’s optimal signal and find that the receiver’s payoff differences across states given each action (sensitivities), play a fundamental role in the characterization and the comparative statics. If the sender’s preferred action is the least (most) sensitive one, then the sender’s equilibrium payoff, as well as the sender’s preferred degree of receiver ambiguity, is increasing (decreasing) in the receiver’s pessimism. We document a tendency for ambiguity-sensitive receivers to be more difficult to persuade.

Appendices

Appendix A. Explicit form of optimal signal

Here we provide explicit formulae for the sender’s optimal signal for cases (A) and (B) in Proposition 1. These are given by the solutions to Eqs. (*) and (**).

$$\begin{aligned} \pi _{1L} =&\frac{(m-\varepsilon )((1-\alpha )\Delta _{0}+\alpha \Delta _{1}-\Delta _{L})}{(1-(m-\varepsilon ))2\Delta _{L}}+\frac{(m+\varepsilon )(\alpha \Delta _{0}+(1-\alpha )\Delta _{1}-\Delta _{L})}{(1-(m+\varepsilon ))2\Delta _{L}} \\&+\sqrt{ {\scriptstyle \left[ \frac{(m-\varepsilon )((1-\alpha )\Delta _{0}+\alpha \Delta _{1}-\Delta _{L})}{(1-(m-\varepsilon ))2\Delta _{L}}+\frac{(m+\varepsilon )(\alpha \Delta _{0}+(1-\alpha )\Delta _{1}-\Delta _{L})}{(1-(m+\varepsilon ))2\Delta _{L}}\right] ^{2}+\frac{(m-\varepsilon )(m+\varepsilon )\Delta _{H} }{(1-(m-\varepsilon ))(1-(m+\varepsilon ))\Delta _{L}}}} \\ \end{aligned}$$

(A)

$$\begin{aligned} \pi _{1L} =&\frac{(m-\varepsilon )((1-\alpha )\Delta _{H}-\alpha \Delta _{L})}{(1-(m-\varepsilon ))2\Delta _{L}}+\frac{(m+\varepsilon )(\alpha \Delta _{H}-(1-\alpha )\Delta _{L})}{(1-(m+\varepsilon ))2\Delta _{L}}\\&+\sqrt{{\scriptstyle \left[ \frac{(m-\varepsilon )((1-\alpha )\Delta _{H}-\alpha \Delta _{L})}{(1-(m-\varepsilon ))2\Delta _{L}}+\frac{(m+\varepsilon )(\alpha \Delta _{H}-(1-\alpha )\Delta _{L})}{(1-(m+\varepsilon ))2\Delta _{L}}\right] ^{2}+\frac{(m-\varepsilon )(m+\varepsilon )\Delta _{H}}{(1-(m-\varepsilon ))(1-(m+\varepsilon ))\Delta _{L}}}}. \end{aligned}$$

(B)

The expression for $\pi _{1L}$ for case (B) confirms the fact that the optimal signal is characterized by the ratio $\Delta _{H}/\Delta _{L}$ in the case of comonotonic actions (as discussed in the main text). This, however, is not the case when actions are not comonotonic, as reflected in the expression for $\pi _{1L}$ for case (A) above.

Appendix B. Proofs

Proof of proposition 1

Suppose $G(m-\varepsilon ,m+\varepsilon )<0$. The formulas in the statement follow from the arguments in the main text. Existence of an optimal signal follows from the intermediate value theorem, because $G(\beta _{1}(\pi ,m-\varepsilon ),\beta _{1}(\pi ,m+\varepsilon ))$ is continuous in $\pi _{1L}$, and because $G(\beta _{1}(\pi ,m-\varepsilon ),\beta _{1}(\pi ,m+\varepsilon ))=G(1,1)=\Delta _{H}>0$ for $\pi _{1H}=1$ and $\pi _{1L}=0$, and $G(\beta _{1}(\pi ,m-\varepsilon ),\beta _{1}(\pi ,m+\varepsilon ))=G(m-\varepsilon ,m+\varepsilon )<0$ for $\pi _{1H}=1$ and $\pi _{1L}=1$. Uniqueness follows because $G(\beta _{1}(\pi ,m-\varepsilon ),\beta _{1}(\pi ,m+\varepsilon ))$ is strictly decreasing in $\pi _{1L}$. $\square$

Proof of proposition 2

Continuity of $\tilde{\pi }_{1L}(\varepsilon ,\alpha )$ in $\alpha$ follows from the maximum theorem, as $\tilde{\pi }_{1L}(\varepsilon ,\alpha )=\max \{\pi _{1L}\in [ 0,1]:G(\beta (\pi _{1L},m-\varepsilon ),\beta (\pi _{1L},m+\varepsilon ))\ge 0\}$ and G is continuous.

Suppose $\tilde{\pi }_{1L}(\varepsilon ,\alpha )<1$. Since $\beta (\pi _{1L},m-\varepsilon )$ and $\beta (\pi _{1L},m+\varepsilon )$ are strictly decreasing in $\pi _{1L}$, and since G is strictly increasing in both its arguments, $\partial G(\beta (\pi _{1L},m-\varepsilon ),\beta (\pi _{1L},m+\varepsilon ))/\partial \pi _{1L}|_{\pi _{1L}=\tilde{\pi } _{1L}(\varepsilon ,\alpha )}<0$. The implicit function theorem then yields that $\tilde{\pi }_{1L}(\varepsilon ,\alpha )$ is differentiable in $\alpha$ with

$$\begin{aligned} \frac{\partial \tilde{\pi }_{1L}(\varepsilon ,\alpha )}{\partial \alpha } =\left. -\frac{\partial G(\beta (\pi _{1L},m-\varepsilon ),\beta (\pi _{1L},m+\varepsilon ))/\partial \alpha }{\partial G(\beta (\pi _{1L},m-\varepsilon ),\beta (\pi _{1L},m+\varepsilon ))/\partial \pi _{1L}} \right| _{\pi _{1L}=\tilde{\pi }_{1L}(\varepsilon ,\alpha )}\text {.} \end{aligned}$$

As the denominator is negative, $\partial \tilde{\pi }_{1L}(\varepsilon ,\alpha )/\partial \alpha$ shares sign with $\partial G(\beta (\pi _{1L},m-\varepsilon ),\beta (\pi _{1L},m+\varepsilon ))/\partial \alpha$.

In case (B), we have $\Delta _{0}\ge 0\ge \Delta _{1}$ and, therefore, $\Delta _{0}-\Delta _{1}>0$. In case (C), we have $\Delta _{0}\le 0\le \Delta _{1}$ and, therefore, $\Delta _{0}-\Delta _{1}<0$. The explicit formula for $\partial G(\mu _{1},\mu _{2})/\partial \alpha$ in the main text then implies that $\partial G(\beta (\pi _{1L},m-\varepsilon ),\beta (\pi _{1L},m+\varepsilon ))/\partial \alpha$ shares sign with $\Delta _{0}-\Delta _{1}$. $\square$

Proof of lemma 2

Suppose $I=\{\varepsilon \in [ 0,\bar{\varepsilon }):\tilde{\pi } _{1L}(\varepsilon ,\alpha )<1\}\not =\varnothing$. Continuity of $\tilde{\pi } _{1L}(\varepsilon ,\alpha )$ follows from the maximum theorem, as $\tilde{\pi }_{1L}(\varepsilon ,\alpha )=\max \{\pi _{1L}\in [ 0,1]:G(\beta (\pi _{1L},m-\varepsilon ),\beta (\pi _{1L},m+\varepsilon ))\ge 0\}$ and G is continuous. Differentiability of $\tilde{\pi }_{1L}(\varepsilon ,\alpha )$ in $\varepsilon$ on I follows by an argument similar to the one for the differentiability of $\tilde{\pi }_{1L}(\varepsilon ,\alpha )$ in $\alpha$ in the proof of Proposition 2. The remainder of the proof consists of three steps.$\square$

Step 1. The set I is a right-open interval.

Proof

The result is obvious if $I=[0,\bar{\varepsilon })$, so suppose $\tilde{\pi }_{1L}(\tilde{\varepsilon },\alpha )=1$ for some $\tilde{ \varepsilon }\in [ 0,\bar{\varepsilon })$. Notice that

$$\begin{aligned}&G(m-\varepsilon ,m+\varepsilon )\\&\quad =\left\{ \begin{array}{l} (m-\varepsilon )\left( \alpha \Delta _{1}+(1-\alpha )\Delta _{0}\right) +(m+\varepsilon )\left( \alpha \Delta _{0}+(1-\alpha )\Delta _{1}\right) -\Delta _{L} \\ (1-\alpha )(m-\varepsilon )(\Delta _{0}+\Delta _{1})+\alpha (m+\varepsilon )(\Delta _{0}+\Delta _{1})-\Delta _{L} \\ \alpha (m-\varepsilon )(\Delta _{0}+\Delta _{1})+(1-\alpha )(m+\varepsilon )(\Delta _{0}+\Delta _{1})-\Delta _{L} \end{array} \begin{array}{c} \text {(A)} \\ \,\text {(B)} \\ \text {(C)} \end{array} \right. \end{aligned}$$

is a linear function of $\varepsilon$. Therefore, either $G(m-\varepsilon ,m+\varepsilon )\ge 0$ (and $\tilde{\pi }_{1L}(\varepsilon ,\alpha )=1$) on some segment [0, x] and $G(m-\varepsilon ,m+\varepsilon )<0$ (and $\tilde{ \pi }_{1L}(\varepsilon ,\alpha )<1$) on $(x,\bar{\varepsilon })$, or $G(m-\varepsilon ,m+\varepsilon )<0$ on some segment [0, x) and $G(m-\varepsilon ,m+\varepsilon )\ge 0$ on $[x,\bar{\varepsilon })$. $\square$

Step 2. If $\partial \tilde{\pi }_{1L}(\tilde{\varepsilon },\alpha )/\partial \varepsilon =0$ for some $\tilde{\varepsilon }\in I$, then $\partial ^{2}\tilde{\pi }_{1L}(\tilde{\varepsilon },\alpha )/\partial \varepsilon ^{2}<0$.

Proof

Suppose $\partial \tilde{\pi }_{1L}(\tilde{\varepsilon } ,\alpha )/\partial \varepsilon =0$ and $\tilde{\pi }_{1L}(\tilde{ \varepsilon },\alpha )<1$. The implicit function theorem implies that $\tilde{ \pi }_{1L}$ is differentiable at $\tilde{\varepsilon }$ with derivative

$$\begin{aligned} \frac{\partial \tilde{\pi }_{1L}(\tilde{\varepsilon },\alpha )}{\partial \varepsilon }=\left. -\frac{\partial G(\beta (\pi _{1L},m-\tilde{ \varepsilon }),\beta (\pi _{1L},m+\tilde{\varepsilon }))/\partial \varepsilon }{\partial G(\beta (\pi _{1L},m-\tilde{\varepsilon }),\beta (\pi _{1L},m+\tilde{\varepsilon }))/\partial \pi _{1L}}\right| _{\pi _{1L}= \tilde{\pi }_{1L}(\tilde{\varepsilon },\alpha )}\text {.} \end{aligned}$$

Since the denominator is strictly negative, $\partial \tilde{\pi }_{1L}( \tilde{\varepsilon },\alpha )/\partial \varepsilon$ shares sign with the numerator. It follows that $\partial G(\beta (\pi _{1L},m-\tilde{ \varepsilon }),\beta (\pi _{1L},m+\tilde{\varepsilon }))/\partial \varepsilon =0$.

With a slight abuse of notation, abbreviate $\partial G(\beta (\pi _{1L},m-\varepsilon ),\beta (\pi _{1L},m+\varepsilon ))/\partial x$ as $\partial G/\partial x$. The second (implicit) derivative at $\varepsilon = \tilde{\varepsilon }$ is given by

$$\begin{aligned} \frac{\partial ^{2}\tilde{\pi }_{1L}(\tilde{\varepsilon },\alpha )}{\partial \varepsilon ^{2}}= & {} \left. -\frac{\frac{\partial ^{2}G}{\partial \varepsilon ^{2}}\left( \frac{\partial G}{\partial \pi _{1L}}\right) ^{2}-2 \frac{\partial ^{2}G}{\partial \varepsilon \partial \pi _{1L}}\frac{\partial G}{\partial \pi _{1L}}\frac{\partial G}{\partial \varepsilon }+\frac{ \partial ^{2}G}{\partial \pi _{1L}^{2}}\left( \frac{\partial G}{\partial \varepsilon }\right) ^{2}}{\left( \frac{\partial G}{\partial \pi _{1L}} \right) ^{3}}\right| _{\pi _{1L}=\tilde{\pi }_{1L}(\tilde{\varepsilon } ,\alpha ),\varepsilon =\tilde{\varepsilon }} \\= & {} \left. -\frac{\partial ^{2}G(\beta (\pi _{1L},m-\tilde{\varepsilon } ),\beta (\pi _{1L},m+\tilde{\varepsilon }))/\partial \varepsilon ^{2}}{ \partial G(\beta (\pi _{1L},m-\tilde{\varepsilon }),\beta (\pi _{1L},m+\tilde{ \varepsilon }))/\partial \pi _{1L}}\right| _{\pi _{1L}=\tilde{\pi }_{1L}( \tilde{\varepsilon },\alpha )}, \end{aligned}$$

which shares sign with $\partial ^{2}G(\beta (\pi _{1L},m-\tilde{\varepsilon } ),\beta (\pi _{1L},m+\tilde{\varepsilon }))/\partial \varepsilon ^{2}$ . Straightforward differentiation gives

$$\begin{aligned} \frac{\partial ^{2}G}{\partial \varepsilon ^{2}}=\left\{ \begin{array}{c} {-2\pi }_{1L}{(1-\pi }_{1L})\left( \frac{\alpha \Delta _{1}+(1-\alpha )\Delta _{0}}{\left( m-\varepsilon +\pi _{1L}(1-m+\varepsilon )\right) ^{3}}+\frac{\alpha \Delta _{0}+(1-\alpha )\Delta _{1}}{\left( m+\varepsilon +\pi _{1L}(1-m-\varepsilon )\right) ^{3}} \right) \text { (A)} \\ {-2\pi }_{1L}{(1-\pi }_{1L})\left( \frac{(1-\alpha )(\Delta _{0}+\Delta _{1})}{\left( m-\varepsilon +(1-m+\varepsilon )\pi _{1L}\right) ^{3}}+\frac{\alpha (\Delta _{0}+\Delta _{1})}{\left( m+\varepsilon +(1-m-\varepsilon )\pi _{1L}\right) ^{3}}\right) \text { (B)} \\ {-2\pi }_{1L}{(1-\pi }_{1L})\left( \frac{\alpha (\Delta _{0}+\Delta _{1})}{\left( m-\varepsilon +(1-m+\varepsilon )\pi _{1L}\right) ^{3}}+\frac{(1-\alpha )(\Delta _{0}+\Delta _{1})}{\left( m+\varepsilon +(1-m-\varepsilon )\pi _{1L}\right) ^{3}}\right) \text { (C)} \end{array} \right. \text {.} \end{aligned}$$

Since $\Delta _{0}\ge 0$ and $\Delta _{1}\ge 0$ in case (A), and since $\Delta _{0}+\Delta _{1}>0$, we have that $\partial ^{2}G(\beta (\pi _{1L},m-\varepsilon ),\beta (\pi _{1L},m+\varepsilon ))/\partial \varepsilon ^{2}<0$, which proves the claim. $\square$

Step 3. Suppose $\varepsilon ,\varepsilon ^{\prime }\in I$ with $\varepsilon <\varepsilon ^{\prime }$. If $\partial \tilde{\pi } _{1L}(\varepsilon ,\alpha )/\partial \varepsilon \le 0$, then $\partial \tilde{\pi }_{1L}(\varepsilon ^{\prime },\alpha )/\partial \varepsilon <0$. If $\partial \tilde{\pi }_{1L}(\varepsilon ^{\prime },\alpha )/\partial \varepsilon \ge 0$, then $\partial \tilde{\pi } _{1L}(\varepsilon ,\alpha )/\partial \varepsilon >0$.

Proof

Suppose $\partial \tilde{\pi }_{1L}(\varepsilon ,\alpha )/\partial \varepsilon \le 0$ and, to contradiction, that

$$\begin{aligned} \partial \tilde{ \pi }_{1L}(\varepsilon ^{\prime },\alpha )/\partial \varepsilon >0 \end{aligned}$$

for some $\varepsilon ^{\prime }>\varepsilon$. Let $\tilde{\varepsilon } =\max \{x\in [ \varepsilon ,\varepsilon ^{\prime }]:\partial \tilde{\pi }_{1L}(x,\alpha )/\partial \varepsilon =0\}$, where $\tilde{\varepsilon }$ is well defined by the hypothesis, and because $\partial \tilde{\pi }_{1L}(x,\alpha )/\partial \varepsilon$ is differentiable on I and, therefore, continuous in its first argument on I. Step 2 implies $\partial ^{2}\tilde{\pi }_{1L}(\tilde{ \varepsilon },\alpha )/\partial \varepsilon ^{2}<0$, and, combining with the definition of $\tilde{\varepsilon }$, we obtain $\partial \tilde{\pi } _{1L}(x,\alpha )/\partial \varepsilon <0$ on $(\tilde{\varepsilon },\varepsilon ^{\prime }]$, a contradiction. Therefore, $\partial \tilde{\pi }_{1L}(\varepsilon ^{\prime },\alpha )/\partial \varepsilon \le 0$.

Suppose $\partial \tilde{\pi }_{1L}(\varepsilon ^{\prime },\alpha )/\partial \varepsilon =0$. Step 2 implies $\partial ^{2}\tilde{\pi } _{1L}(\varepsilon ^{\prime },\alpha )/\partial \varepsilon ^{2}<0$, and there is then an interval $(z,\varepsilon ^{\prime })$ on which $\partial \tilde{\pi }_{1L}(x,\alpha )/\partial \varepsilon >0$, which produces the same contradiction as in the previous paragraph. Hence, $\partial \tilde{\pi } _{1L}(\varepsilon ^{\prime },\alpha )/\partial \varepsilon <0$. $\square$

The statement that we must have either case (i), (ii), or (iii), follows by combining Step 1 and Step 3. More precisely, suppose $\partial \tilde{\pi } _{1L}(\varepsilon ^{*},\alpha )/\partial \varepsilon =0$ for some $\varepsilon ^{*}\in I$. If $\varepsilon ^{*}$ is an element of the interior of I, then Step 3 implies that we are in case (iii), and otherwise $\varepsilon ^{*}=\min I$ and we are in case (ii). If there is no such $\varepsilon ^{*}$, then the continuity of $\partial \tilde{\pi }_{1L}(\varepsilon ,\alpha )/\partial \varepsilon$ in $\varepsilon$ implies that we are either in case (i) or (ii).

Proof of Proposition 3

Suppose $I=\{\varepsilon \in [ 0,\bar{\varepsilon }):\tilde{\pi } _{1L}(\varepsilon ,\alpha )<1\}\not =\varnothing$. We prove each direction of the statement separately.

Step 1. If either (a) $\Delta _{0}=\Delta _{1}$, (b) $\alpha =1/2$, (c) $\Delta _{0}>\Delta _{1}$ and $\alpha <1/2$, or (d) $\Delta _{0}<\Delta _{1}$ and $\alpha >1/2$, then $\partial \tilde{\pi }_{1L}(\varepsilon ,\alpha )/\partial \varepsilon \le 0$ for all $\varepsilon \in I$.$\square$

Proof

Suppose, to contradiction, $\partial \tilde{\pi } _{1L}(\varepsilon ,\alpha )/\partial \varepsilon >0$ for some $\varepsilon \in I$. Lemma 2 then implies $\tilde{\pi }_{1L}(0,\alpha )<1$ and $\partial \tilde{\pi }_{1L}(0,\alpha )/\partial \varepsilon >0$. The implicit function theorem implies that $\partial \tilde{\pi }_{1L}(0,\alpha )/\partial \varepsilon$ shares sign with

$$\begin{aligned}&\left. \frac{\partial G(\beta (\pi ,m-\varepsilon ),\beta (\pi ,m+\varepsilon ))}{\partial \varepsilon }\right| _{\varepsilon =0}\\&\quad =\left\{ \begin{array}{c} \frac{\tilde{\pi }_{1L}(0,\alpha )(2\alpha -1)(\Delta _{0}-\Delta _{1})}{ \left( m+\tilde{\pi }_{1L}(0,\alpha )(1-m)\right) ^{2}}\text { if}\ \Delta _{0}\ge 0\text { and }\Delta _{1}\ge 0\text { (A)} \\ \frac{\tilde{\pi }_{1L}(0,\alpha )(2\alpha -1)(\Delta _{0}+\Delta _{1})}{ \left( m+\tilde{\pi }_{1L}(0,\alpha )(1-m)\right) ^{2}}\text { if }\Delta _{0}\ge 0\text { and }\Delta _{1}\le 0\text { (B)} \\ \frac{\tilde{\pi }_{1L}(0,\alpha )(1-2\alpha )(\Delta _{0}+\Delta _{1})}{ \left( m+\tilde{\pi }_{1L}(0,\alpha )(1-m)\right) ^{2}}\text { if }\Delta _{0}\le 0\text { and }\Delta _{1}\ge 0\text { (C)} \end{array} \right. \end{aligned}$$

Therefore, we must have either $\Delta _{0}>\Delta _{1}$ and $\alpha >1/2$ or $\Delta _{0}<\Delta _{1}$ and $\alpha <1/2$, which is the negation of the hypothesis in the claim. $\square$

Step 2. If $\partial \tilde{\pi }_{1L}(\varepsilon ,\alpha )/\partial \varepsilon \le 0$ for all $\varepsilon \in I$, then either (a) $\Delta _{0}=\Delta _{1}$, (b) $\alpha =1/2$, (c) $\Delta _{0}>\Delta _{1}$ and $\alpha <1/2$, or (d) $\Delta _{0}<\Delta _{1}$ and $\alpha >1/2$.

Proof

Suppose $\partial \tilde{\pi }_{1L}(\varepsilon ,\alpha )/\partial \varepsilon \le 0$ for all $\varepsilon \in I$. If $\tilde{\pi } _{1L}(0,\alpha )<1$, then $\partial \tilde{\pi }_{1L}(0,\alpha )/\partial \varepsilon \le 0$, and the formula derived in Step 1 immediately implies the result. Suppose, therefore, that $\tilde{\pi }_{1L}(0,\alpha )=1$, which implies $G(m,m)\ge 0$. We have that

$$\begin{aligned} \frac{\partial G(m-\varepsilon ,m+\varepsilon )}{\partial \varepsilon } =\left\{ \begin{array}{l} (2\alpha -1)(\Delta _{0}-\Delta _{1}) \\ (2\alpha -1)(\Delta _{0}+\Delta _{1}) \\ (1-2\alpha )(\Delta _{0}+\Delta _{1}) \end{array} \begin{array}{c} \text {if}\ \Delta _{0}\ge 0\text { and }\Delta _{1}\ge 0 \\ \text {if }\Delta _{0}\ge 0\text { and }\Delta _{1}\le 0 \\ \text {if }\Delta _{0}\le 0\text { and }\Delta _{1}\ge 0 \end{array} \begin{array}{c} \text {(A)} \\ \text {(B)} \\ \text {(C)} \end{array} .\right. \end{aligned}$$

If $\Delta _{0}>\Delta _{1}$ and $\alpha >1/2$ we are either in case (A) or (B) and $\partial G(m-\varepsilon ,m+\varepsilon )/\partial \varepsilon >0$. If $\Delta _{0}<\Delta _{1}$ and $\alpha <1/2$, we are either in case (A) or (C) and $\partial G(m-\varepsilon ,m+\varepsilon )/\partial \varepsilon >0$. But then $G(m-\varepsilon ,m+\varepsilon )\ge 0$ for all $\varepsilon \in [ 0,\bar{\varepsilon })$, which contradicts the assumption that $\tilde{ \pi }_{1L}(\varepsilon ,\alpha )<1$ for some $\varepsilon \in [ 0,\bar{ \varepsilon })$. Therefore, one of conditions (i)–(iv) in the statement of the claim must hold. $\square$

Proof of Proposition 4

Let $G(\mu _{1},\mu _{2},\alpha )$ be the sender’s constraint function G such that $\alpha$ is explicitly included in the list of arguments. We first prove the following claim.

Claim. Suppose $\tilde{\pi }_{1L}(\varepsilon ,\alpha )<1$ and $\partial \tilde{\pi }_{1L}(\varepsilon ,\alpha )/\partial \varepsilon =0$. If $\Delta _{0}>\Delta _{1}$, then $\partial ^{2}\tilde{\pi }_{1L}(\varepsilon ,\alpha )/\partial \varepsilon \partial \alpha >0$. If $\Delta _{0}<\Delta _{1}$, then $\partial ^{2}\tilde{\pi }_{1L}(\varepsilon ,\alpha )/\partial \varepsilon \partial \alpha >0$.$\square$

Proof

Suppose $\tilde{\pi }_{1L}(\varepsilon ,\alpha )<1$, $\partial \tilde{\pi }_{1L}(\varepsilon ,\alpha )/\partial \varepsilon =0$ and $\Delta _{0}>\Delta _{1}$. With a slight abuse of notation, abbreviate $\partial G(\beta (\pi _{1L},m-\varepsilon ),\beta (\pi _{1L},m+\varepsilon ),\alpha )/\partial x$ as $\partial G/\partial x$. Notice that

$$\begin{aligned} \frac{\partial \tilde{\pi }_{1L}(\varepsilon ,\alpha )}{\partial \varepsilon }&=\left. -\frac{\frac{\partial G(\beta (\pi _{1L},m-\varepsilon ),\beta (\pi _{1L},m+\varepsilon ),\alpha )}{\partial \varepsilon }}{\frac{\partial G(\beta (\pi _{1L},m-\varepsilon ),\beta (\pi _{1L},m+\varepsilon ),\alpha ) }{\partial \pi _{1L}}}\right| _{\pi _{1L}=\tilde{\pi }_{1L}(\varepsilon ,\alpha )}\implies \\ \frac{\partial \tilde{\pi }_{1L}(\varepsilon ,\alpha )}{\partial \varepsilon \partial \alpha }&=\left. \frac{\left( \frac{\partial ^{2}G}{\partial \pi _{1L}^{2}}\frac{\partial \tilde{\pi }_{1L}}{\partial \alpha }+\frac{ \partial ^{2}G}{\partial \pi _{1L}\partial \alpha }\right) \frac{\partial G}{ \partial \varepsilon }-\left( \frac{\partial ^{2}G}{\partial \varepsilon \partial \pi _{1L}}\frac{\partial \tilde{\pi }_{1L}}{\partial \alpha }+\frac{\partial ^{2}G}{\partial \varepsilon \partial \alpha }\right) \frac{\partial G}{\partial \pi _{1L}}}{\left( \frac{\partial G}{\partial \pi _{1L}}\right) ^{2}}\right| _{\pi _{1L}=\tilde{\pi }_{1L}(\varepsilon ,\alpha )} \\&=\left. -\frac{\frac{\partial ^{2}G}{\partial \varepsilon \partial \pi _{1L}}\frac{\partial \tilde{\pi }_{1L}}{\partial \alpha }+\frac{\partial ^{2}G }{\partial \varepsilon \partial \alpha }}{\frac{\partial G}{\partial \pi _{1L}}}\right| _{\pi _{1L}=\tilde{\pi }_{1L}(\varepsilon ,\alpha )}, \end{aligned}$$

where the second line uses the fact that $\partial \tilde{\pi } _{1L}(\varepsilon ,\alpha )/\partial \varepsilon =0\Leftrightarrow \partial G/\partial \varepsilon =0$. We have that $\partial G/\partial \pi _{1L}<0$, so the sign of $\partial ^{2}\tilde{\pi }_{1L}(\varepsilon ,\alpha )/\partial \varepsilon \partial \alpha$ coincides with that of the numerator. Proposition 2 implies that $\partial \tilde{\pi }_{1L}/\partial \alpha >0$. The remaining terms need to be computed.

We compute²⁷

$$\begin{aligned}&\frac{\partial G}{\partial \varepsilon }\\&\quad ={\left\{ \begin{array}{ll} \pi _{1L}\left( -\frac{\left( \alpha \Delta _{1}+(1-\alpha )\Delta _{0}\right) }{\left( m-\varepsilon +\pi _{1L}(1-m+\varepsilon )\right) ^{2}}+ \frac{\left( \alpha \Delta _{0}+(1-\alpha )\Delta _{1}\right) }{\left( m+\varepsilon +\pi _{1L}(1-m-\varepsilon )\right) ^{2}}\right) \text { if}\ \Delta _{0}\ge 0\text { and }\Delta _{1}\ge 0\text { (A)} \\ \pi _{1L}\left( -\frac{1-\alpha }{\left( m-\varepsilon +\pi _{1L}(1-m+\varepsilon )\right) ^{2}}+\frac{\alpha }{\left( m+\varepsilon +\pi _{1L}(1-m-\varepsilon )\right) ^{2}}\right) \text { if }\Delta _{0}\ge 0 \text { and }\Delta _{1}\le 0\text { (B)}. \end{array}\right. } \end{aligned}$$

Therefore,

$$\begin{aligned}&\frac{\partial G}{\partial \varepsilon }=0\nonumber \\&\quad \Leftrightarrow \,{\left\{ \begin{array}{ll} \frac{\left( \alpha \Delta _{1}+(1-\alpha )\Delta _{0}\right) }{\left( m-\varepsilon +\pi _{1L}(1-m+\varepsilon )\right) ^{2}}=\frac{\left( \alpha \Delta _{0}+(1-\alpha )\Delta _{1}\right) }{\left( m+\varepsilon +\pi _{1L}(1-m-\varepsilon )\right) ^{2}}\text { if}\ \Delta _{0}\ge 0\text { and } \Delta _{1}\ge 0\text { (A)} \\ \frac{1-\alpha }{\left( m-\varepsilon +\pi _{1L}(1-m+\varepsilon )\right) ^{2}}=\frac{\alpha }{\left( m+\varepsilon +\pi _{1L}(1-m-\varepsilon )\right) ^{2}}\text { if }\Delta _{0}\ge 0\text { and }\Delta _{1}\le 0\text { (B)}. \end{array}\right. } \end{aligned}$$

(B.1)

We compute

$$\begin{aligned}&\frac{\partial ^{2}G}{\partial \varepsilon \partial \pi _{1L}}\nonumber \\&\quad ={\left\{ \begin{array}{ll} 2\pi _{1L}\left( \frac{(1-m+\varepsilon )\left( \alpha \Delta _{1}+(1-\alpha )\Delta _{0}\right) }{\left( m-\varepsilon +\pi _{1L}(1-m+\varepsilon )\right) ^{3}}-\frac{(1-m-\varepsilon )\left( \alpha \Delta _{0}+(1-\alpha )\Delta _{1}\right) }{\left( m+\varepsilon +\pi _{1L}(1-m-\varepsilon )\right) ^{3}}\right) \text {(A)} \\ 2\pi _{1L}\left( \frac{\alpha (1-m+\varepsilon )\left( \Delta _{1}+\Delta _{0}\right) }{\left( m-\varepsilon +\pi _{1L}(1-m+\varepsilon )\right) ^{3}}- \frac{(1-\alpha )(1-m-\varepsilon )\left( \Delta _{0}+\Delta _{1}\right) }{ \left( m+\varepsilon +\pi _{1L}(1-m-\varepsilon )\right) ^{3}}\right) \text {(B)} \end{array}\right. }\nonumber \\&={\left\{ \begin{array}{ll} 2\pi _{1L}\frac{\left( \alpha \Delta _{1}+(1-\alpha )\Delta _{0}\right) }{ \left( m-\varepsilon +\pi _{1L}(1-m+\varepsilon )\right) ^{2}}\left( \frac{ (1-m+\varepsilon )}{\left( m-\varepsilon +\pi _{1L}(1-m+\varepsilon )\right) }-\frac{(1-m-\varepsilon )}{\left( m+\varepsilon +\pi _{1L}(1-m-\varepsilon )\right) }\right) \text {(A)} \\ 2\pi _{1L}\frac{\alpha \left( \Delta _{1}+\Delta _{0}\right) }{\left( m-\varepsilon +\pi _{1L}(1-m+\varepsilon )\right) ^{2}}\left( \frac{ (1-m+\varepsilon )}{\left( m-\varepsilon +\pi _{1L}(1-m+\varepsilon )\right) }-\frac{(1-m-\varepsilon )}{\left( m+\varepsilon +\pi _{1L}(1-m-\varepsilon )\right) }\right) \text {(B)}, \end{array}\right. } \end{aligned}$$

(B.2)

where the first line follows by applying the product rule and (B.1 ), and the second line follows by again applying (B.1). Hence, $\partial ^{2}G/\partial \varepsilon \partial \pi _{1L}>0$.

Next,

$$\begin{aligned}&\frac{\partial ^{2}G}{\partial \varepsilon \partial \alpha }\\&\quad ={\left\{ \begin{array}{ll} \pi _{1L}\left( \frac{\Delta _{0}-\Delta _{1}}{\left( m-\varepsilon +\pi _{1L}(1-m+\varepsilon )\right) ^{2}}-\frac{\Delta _{1}-\Delta _{0}}{\left( m+\varepsilon +\pi _{1L}(1-m-\varepsilon )\right) ^{2}}\right) \text { if}\ \Delta _{0}\ge 0\text { and }\Delta _{1}\ge 0\text { (A)} \\ \pi _{1L}\left( \frac{\Delta _{0}+\Delta _{1}}{\left( m-\varepsilon +\pi _{1L}(1-m+\varepsilon )\right) ^{2}}+\frac{\Delta _{0}+\Delta _{1}}{\left( m+\varepsilon +\pi _{1L}(1-m-\varepsilon )\right) ^{2}}\right) \text { if } \Delta _{0}\ge 0\text { and }\Delta _{1}\le 0\text { (B)} \end{array}\right. } \end{aligned}$$

Since $\Delta _{0}>\Delta _{1}$, we have $\partial ^{2}G/\partial \varepsilon \partial \alpha >0$.

Summing up, we have $\partial ^{2}G/\partial \varepsilon \partial \pi _{1L}>0$, $\partial \tilde{\pi }_{1L}/\partial \alpha >0$ and $\partial ^{2}G/\partial \varepsilon \partial \alpha >0$ and, therefore,

$$\begin{aligned} \frac{\partial ^{2}G}{\partial \varepsilon \partial \pi _{1L}}\frac{\partial \tilde{\pi }_{1L}}{\partial \alpha }+\frac{\partial ^{2}G}{\partial \varepsilon \partial \alpha }\,>0\text {.} \end{aligned}$$

Hence, $\partial ^{2}\tilde{\pi }_{1L}(\varepsilon ,\alpha )/\partial \varepsilon \partial \alpha >0$. The proof for the case $\Delta _{1}>\Delta _{0}$ is virtually identical, and is omitted. $\square$

The remainder of the proof establishes statement 1 in Proposition 4 (the proof of statement 2 is analogous). Therefore, suppose $\Delta _{0}>\Delta _{1}$ and $\alpha >1/2$, and that $\tilde{\pi } _{1L}(\varepsilon ,\alpha )<1$ for some $(\varepsilon ,\alpha )\in [ 0, \bar{\varepsilon })\times (1/2,1]$. Notice that we have $G(m,m,1/2)<0$. To see this, suppose, to contradiction, $G(m,m,1/2)\ge 0$. But

$$\begin{aligned}&G(m-\varepsilon ,m+\varepsilon ,\alpha \\&\quad ={\left\{ \begin{array}{ll} (m-\varepsilon )\left( \alpha \Delta _{1}+(1-\alpha )\Delta _{0}\right) +(m+\varepsilon )\left( \alpha \Delta _{0}+(1-\alpha )\Delta _{1}\right) -\Delta _{L} &{}\text {(A)}\\ (1-\alpha )(m-\varepsilon )(\Delta _{0}+\Delta _{1})+\alpha (m+\varepsilon )(\Delta _{0}+\Delta _{1})-\Delta _{L} &{} \text {(B)} \end{array}\right. } \end{aligned}$$

is increasing in $\varepsilon$ and $\alpha$ if $\Delta _{0}>\Delta _{1}$ and $\alpha >1/2$. Therefore, $G(m-\varepsilon ,m+\varepsilon ,\alpha )\ge 0$ for all admissible values of $(\varepsilon ,\alpha )$, which contradicts $\tilde{\pi }_{1L}(\varepsilon ,\alpha )<1$ for some $(\varepsilon ,\alpha )$.

We now prove the first part of statement 1 in Proposition 4.

Step 1. There is some $\hat{\alpha }\in (1/2,1]$ such that there is an interior ambiguity bliss point $\varepsilon (\alpha )$ for all $\alpha < \hat{\alpha }$ and such that $\tilde{\pi }_{1L}(\varepsilon ,\alpha )$ is increasing in $\varepsilon$ for any $\alpha >\hat{\alpha }$. It holds that $\varepsilon ^{\prime }(\alpha )>0$ on $(1/2,\hat{\alpha })$ and $\lim _{\alpha \downarrow 1/2}\varepsilon (\alpha )=0$.

Proof

First, the implicit function theorem implies that if $\tilde{ \pi }_{1L}(\varepsilon ,\alpha )<1$ and $\partial \tilde{\pi } _{1L}(\varepsilon ,\alpha )/\partial \varepsilon =0$, then this equality defines an implicit function $\varepsilon (\alpha )$ on some neighborhood around $\alpha$, which satisfies

$$\begin{aligned} \varepsilon ^{\prime }(\alpha )=-\frac{\frac{\partial ^{2}\tilde{\pi } _{1L}(\varepsilon ,\alpha )}{\partial \varepsilon \partial \alpha }}{\frac{ \partial ^{2}\tilde{\pi }_{1L}(\varepsilon ,\alpha )}{\partial \varepsilon ^{2}}}\text {.} \end{aligned}$$

The Claim and Step 2 in the proof of Lemma 2 imply $\varepsilon ^{\prime }(\alpha )>0$. That is, if an interior bliss point $\varepsilon (\alpha )$ exists, then $\varepsilon ^{\prime }(\alpha )>0$.

Since $G(m,m,1/2)<0$, we can confirm that $\tilde{\pi }_{1L}(0,1/2)<1$, and consequently, by the proof of Proposition 3, $\partial \tilde{ \pi }_{1L}(0,1/2)/\partial \varepsilon =0$. Combining with the arguments in the previous paragraph, the implicit function theorem implies existence of an interior bliss point $\varepsilon (\alpha )$ on some interval $(1/2,\hat{ \alpha })$.

The proof of Step 1 concludes by demonstrating that there is some $\hat{ \alpha }\in (1/2,1]$ such that an interior bliss point exists for all $\alpha <\hat{\alpha }$ but not for any $\alpha >\hat{\alpha }$. To this end, suppose $\alpha _{2}>\alpha _{1}$ and that $\partial \tilde{\pi }_{1L}(\varepsilon (\alpha _{2}),\alpha _{2})/\partial \varepsilon =0$. Notice that

$$\begin{aligned} G(\beta (\pi _{1L},m-\varepsilon (\alpha _{2})),\beta (\pi _{1L},m+\varepsilon (\alpha _{2})),\alpha )=0 \end{aligned}$$

defines $\tilde{\pi }_{1L}(\varepsilon (\alpha _{2}),\alpha )$ as an implicit function of $\alpha$ globally over the interval $(1/2,\alpha _{2}]$. This follows because $G(m-\varepsilon (\alpha _{2}),m+\varepsilon (\alpha _{2}),\alpha _{2})<0$ implies $G(m-\varepsilon (\alpha _{2}),m+\varepsilon (\alpha _{2}),\alpha )<0$ for all $\alpha <\alpha _{2}$. Since $\partial \tilde{\pi }_{1L}(\varepsilon (\alpha _{2}),\alpha )/\partial \varepsilon =0$ implies $\partial ^{2}\tilde{\pi }_{1L}(\varepsilon (\alpha _{2}),\alpha )/\partial \varepsilon \partial \alpha >0$, and $\alpha _{2}>\alpha _{1}$, it must be that $\partial \tilde{\pi }_{1L}(\varepsilon (\alpha _{2}),\alpha _{1})/\partial \varepsilon <0$. Therefore, $\tilde{\pi }_{1L}(\varepsilon ,\alpha _{1})$ is not increasing in $\varepsilon$. Since we know that either $\tilde{\pi }_{1L}(\varepsilon ,\alpha _{1})$ is increasing in $\varepsilon$, or there is an interior ambiguity bliss point $\varepsilon (\alpha _{1})$, we conclude the latter. Since $\alpha _{2}$ and $\alpha _{1}$ were arbitrary, whenever $\varepsilon (\alpha _{2})$ exists, $\varepsilon (\alpha _{1})$ exists for any $\alpha _{1}<\alpha _{2}$. This implies that there is some $\hat{\alpha }\in (1/2,1]$ such that there is an interior bliss point for all $\alpha <\hat{\alpha }$, but not for any $\alpha >\hat{\alpha }$. $\square$

Step 2. There is some $\hat{\Delta }_{0}>\Delta _{1}$ such that the value $\hat{\alpha }$ in the statement of Step 1 satisfies $\hat{\alpha }=1$ if $\Delta _{0}<\hat{\Delta }_{0}$ and $\hat{\alpha }<1$ if $\Delta _{0}>\hat{ \Delta }_{0}$.

Proof

Let $\tilde{\varepsilon }(\Delta _{0})$ denote the interior ambiguity bliss-point as a function of $\Delta _{0}$ with $\alpha$ fixed at $\alpha =1$. The comparative statics with respect to $\alpha$ in Step 1 and 2 can be repeated (virtually identically), to conclude that (1) $\tilde{ \varepsilon }^{\prime }(\Delta _{0})>0$ whenever $\tilde{\varepsilon }(\Delta _{0})$ exists; (2) if $\Delta _{0}$ is sufficiently close to $\Delta _{1}$, then there is an interior bliss point $\tilde{\varepsilon }(\Delta _{0})$; (3) if $\tilde{\varepsilon }(\Delta _{0})$ exists, then $\tilde{\varepsilon } (\Delta _{0}^{\prime })$ exists for any $\Delta _{0}^{\prime }\in (\Delta _{1},\Delta _{0})$.

To conclude the argument, it suffices to notice that there is some $\hat{ \Delta }_{0}$ such that if $\Delta _{0}>\hat{\Delta }_{0}$, then $\tilde{\pi } _{1L}(\varepsilon ,1)=1$ for some (large) values of $\varepsilon$. In this case, $\tilde{\pi }_{1L}(\varepsilon ,1)$ is increasing, i.e., there is no $\tilde{\varepsilon }(\Delta _{0})$. Therefore, there must be a value $\hat{ \Delta }_{0}$ such that $\tilde{\varepsilon }(\Delta _{0})$ exists if $\Delta _{0}<\hat{\Delta }_{0}$, and $\tilde{\pi }_{1L}(\varepsilon ,1)$ is increasing if $\Delta _{0}>\hat{\Delta }_{0}$. Combining with Step 1, $\hat{\alpha }=1$ if $\Delta _{0}<\hat{\Delta }_{0}$, and $\hat{\alpha }<1$ if $\Delta _{0}>\hat{ \Delta }_{0}$. $\square$