(Noisy) communication

Abstract

Communication is central to many settings in marketing and economics. A focal attribute of communication is miscommunication. We model this key characteristic as a noise in the messages communicated, so that the sender of a message is uncertain about its perception by the receiver, and then identify the strategic consequences of miscommunication. We study a model where competing senders (of different types) can invest in improving the precision of the informative but noisy message they send to a receiver, and find that there exists a separating equilibrium where senders’ types are completely revealed. Thus, although communication is noisy it delivers perfect results in equilibrium. This result stems from the fact that a sender’s willingness to invest in improving the precision of their messages can itself serve as a signal. Interestingly, the content of the messages is ignored by the receiver in such a signaling equilibrium, but plays a central role by shaping her beliefs off the equilibrium path (and thus, enables separation between the types). This result also illustrates the uniqueness of the signaling model presented here. Unlike other signaling models, the suggested model does not require that the costs and benefits of the senders will be correlated with their types to achieve separation. The model’s results have implications for various marketing communication tools such as advertising and sales forces.

Appendix

1.1 Appendix A: Uniqueness of the simple model

There are two types of separating equilibria: (1) sender’s strategies depend on their type, and (2) sender’s strategies do not depend on their type.

In Appendix A.1 we show that for the first case there is a unique sequential equilibrium (the one presented in Section 2) in which c > 1 − q.

In Appendix A.2 we show that for the second case there is a unique sequential equilibrium in which c < 1 − q.

Thus, for c > 1 − q there is a unique separating (sequential) equilibrium as stated in Proposition 2.

1.1.1 A.1 Strategies depend on senders’ types

Here, we study a separating equilibrium in which sender’s strategies depend on their types.

We start by characterizing the only set of consistent beliefs in such a separating equilibrium. Then we show that for these beliefs and the given parameter values, there is a unique separating equilibrium.

In any separating equilibrium, either H sends a message and L not, or the reverse is true. We consider each case in turn.

Case 1

H sends a message, and L does not.

Let H send a message with probability 1 − ε _H , and L send a message with probability ε _L (both ε _H and ε _L are greater than 0). Recall that the prior probability that player s is H is $\mu _{s}^{0}(r_{s}).$ Then (using Bayes rule) the receiver’s beliefs at each of her four information sets are:

$$\mu ^{\varepsilon }_{s} = {\left\{ {\begin{array}{*{20}l} {{\frac{{(1 - \varepsilon _{H} )(1 - \varepsilon _{L} )\mu ^{0}_{s} (r_{s} )}}{{(1 - \varepsilon _{H} )(1 - \varepsilon _{L} )\mu ^{0}_{s} (r_{s} ) + \varepsilon _{H} \varepsilon _{L} (1 - \mu ^{0}_{s} (r_{s} ))}}} \hfill} & {{{\text{if}}} \hfill} & {{A_{s} = (1,0)} \hfill} \\ {{\frac{{\varepsilon _{H} \varepsilon _{L} \mu ^{0}_{s} (r_{s} )}}{{\varepsilon _{H} \varepsilon _{L} \mu ^{0}_{s} (r_{s} ) + (1 - \varepsilon _{H} )(1 - \varepsilon _{L} )(1 - \mu ^{0}_{s} (r_{s} ))}}} \hfill} & {{{\text{if}}} \hfill} & {{A_{s} = (0,1)} \hfill} \\ {{\frac{{(1 - \varepsilon _{H} )\varepsilon _{L} \mu ^{0}_{s} (r_{s} )}}{{(1 - \varepsilon _{H} )\varepsilon _{L} \mu ^{0}_{s} (r_{s} ) + (1 - \varepsilon _{H} )\varepsilon _{L} (1 - \mu ^{0}_{s} (r_{s} ))}}} \hfill} & {{{\text{if}}} \hfill} & {{A_{s} = (1,1)} \hfill} \\ {{\frac{{\varepsilon _{H} (1 - \varepsilon _{L} )\mu ^{0}_{s} (r_{s} )}}{{\varepsilon _{H} (1 - \varepsilon _{L} )\mu ^{0}_{s} (r_{s} ) + (1 - \varepsilon _{L} )\varepsilon _{H} (1 - \mu ^{0}_{s} (r_{s} ))}}} \hfill} & {{{\text{if}}} \hfill} & {{A_{s} = (0,0)} \hfill} \\ \end{array} } \right\}}$$

(5)

where A _s is an indicator vector in which the first variable is equal to 1 if s sends a message and zero otherwise, and the second variable is equal to 1 if s’s competitor sends a message and zero otherwise.

It is straightforward to show that the limit of μ ^ε as ε _H ,ε _L →0 is:

$$ \mu _{s}=\left\{ \begin{array}{c} {\kern20pt}1\;\text{ if }\;A_{s}=(1,0) \\ {\kern20pt}0\;\text{ if }\;A_{s}=(0,1) \\ \mu _{s}^{0}(r_{s})\;\text{ if }\;A_{s}=(1,1) \\[2pt] \mu _{s}^{0}(r_{s})\;\text{ if }\;A_{s}=(0,0) \end{array} \right\} $$

which are exactly the beliefs in (B).

Recall that under these beliefs, the following table represents the net payoff functions of both senders.

Now, one can see that when c is not in the interval [1 − q,0.5], there is no separating equilibrium in which H sends a message and L does not: (a) when c < 1 − q, sender L finds it profitable to imitate H; (b) when c > 0.5 , sender H deviates.

Case 2

L sends a message, and H does not.

In this case, it is easy to show that the only consistent beliefs are:

$$ \mu _{s}=\left\{ \begin{array}{c} {\kern20pt}0\;\text{ if }\;A_{s}=(1,0) \\ {\kern20pt}1\;\text{ if }\;A_{s}=(0,1) \\ \mu _{s}^{0}(r_{s})\;\text{ if }\;A_{s}=(1,1) \\[2pt] \mu _{s}^{0}(r_{s})\;\text{ if }\;A_{s}=(0,0) \end{array} \right\} $$

The following table represents the net payoff functions of both senders.

It is clear that in this case, it is not optimal for either sender to send a message.

1.1.2 A.2 Strategies do not depend on senders’ types

Lemma 5

A separating equilibrium in which sender’s strategies do not depend on their type, exists if and only if 1 − q > c.

Proof

Without loss of generality, consider the case where sender 1 sends a message and sender 2 does not.

First, we characterize beliefs. To obtain consistent beliefs, we describe the players strategies.

Player 1 sends a message with probability 1 − ε _1H if he is H and 1 − ε _1L if he is L. Player 2 sends a message with probability ε _2H if he is H and ε _2L if he is L. Denote the prior probability that player 1 is H by p. Then (using Bayes rule) the beliefs that player 1 is H are:

$$ \mu _{1}^{\varepsilon }=\left\{\begin{array}{*{20}r}{\frac{(1-\varepsilon _{1H})(1-\varepsilon _{2L})p}{(1-\varepsilon _{1H})(1-\varepsilon_{2L})p+(1-\varepsilon _{1L})(1-\varepsilon _{2H})(1-p)}\;\text{ if }\;A_{1}=(1,0)} \\{\frac{\varepsilon _{1H}\varepsilon _{2L}p}{\varepsilon _{1H}\varepsilon_{2L}p+\varepsilon _{1L}\varepsilon _{2H}(1-p)}\;\text{ if }\;A_{1}=(0,1)} \\{\frac{(1-\varepsilon _{1H})\varepsilon _{2L}p}{(1-\varepsilon_{1H})\varepsilon _{2L}p+(1-\varepsilon _{1L})\varepsilon _{2H}(1-p)}\;\text{ if }\;A_{1}=(1,1)} \\{\frac{\varepsilon _{1H}(1-\varepsilon _{2L})p}{\varepsilon _{1H}(1-\varepsilon _{2L})p+\varepsilon_{1L}(1-\varepsilon _{2H})(1-p)} \;\text{ if }\;A_{1}=(0,0)}\end{array}\right\} $$

(6)

This can be rewritten as:

$$ \mu _{1}^{\varepsilon }=\left\{\begin{array}{*{20}r}{\frac{p}{p+\frac{(1-\varepsilon _{1L})(1-\varepsilon _{2H})}{(1-\varepsilon_{1H})(1-\varepsilon _{2L})}(1-p)}\;\text{ if }\;A_{1}=(1,0)} \\{\frac{p}{p+\frac{\varepsilon _{1L}\varepsilon _{2H}}{\varepsilon_{1H}\varepsilon _{2L}}(1-p)}\;\text{ if }\;A_{1}=(0,1)} \\{\frac{p}{p+\frac{(1-\varepsilon _{1L})\varepsilon _{2H}}{(1-\varepsilon_{1H})\varepsilon _{2L}}(1-p)}\;\text{ if }\;A_{1}=(1,1)} \\{\frac{p}{p+\frac{\varepsilon _{1L}(1-\varepsilon _{2H})}{\varepsilon _{1H}(1-\varepsilon_{2L})}(1-p)}\;\text{ if }\;A_{1}=(0,0)}\end{array}\right\} $$

(7)

It is clear that the limit of $\mu _{1}^{\varepsilon }$ for A ₁ = (1,0) is p. The $\mu _{1}^{\varepsilon }$ of the other elements can be either 0 or 1 (depending on the the ratios $\frac{\varepsilon _{1L}}{\varepsilon _{1H}}$ and $\frac{\varepsilon _{2H}}{\varepsilon _{2L}}.)$

Using these beliefs, we can now check for optimality of sender strategies. Note that we can ignore the case where the limit of $\mu _{1}^{\varepsilon }$ for A ₁ = (0,0) is 1, since in this case, it is optimal for any type of player 1 to deviate. Thus, we only focus on cases where the limit of $\mu _{1}^{\varepsilon }$ for A ₁ = (0,0) is 0.

Thus, we are interested in two cases [notice also that the limit of $\mu _{1}^{\varepsilon }$ for A ₁ = (0,1) is irrelevant for the Nash equilibrium].

Case 1

μ is given by:

$$ \mu =\left\{ \begin{array}{r} p \quad\text{if}\quad A_{1}=(1,0) \\ \text{Not relevant} \quad\text{if}\quad A_{1}=(0,1) \\ 0 \quad\text{if}\quad A_{1}=(1,1) \\ 0 \quad\text{if}\quad A_{1}=(0,0) \end{array} \right\} $$

In this case, the expected net payoffs to senders are:

Irrespective of the choice by nature, there is no (separating) equilibrium in this case: (a) If p > c, then it is optimal for 2 to deviate. (b) But if p < c, then it is optimal for 1 to deviate.

Case 2

μ is given by:

$$ \mu =\left\{ \begin{array}{r} p\quad\text{if}\quad A_{1}=(1,0) \\ \text{Not relevant}\quad\text{if}\quad A_{1}=(0,1) \\ 1\quad\text{if}\quad A_{1}=(1,1) \\ 0\quad\text{if}\quad A_{1}=(0,0) \end{array} \right\} $$

In this case, the expected net payoffs to senders are:

In this case, 2 has no incentive to deviate. The conditions that assure that 1 will not deviate are: (a) if he is of type H, it must be that q > c and if he is of type L it must be that 1 − q > c. Thus, a necessary condition to sustain a (separating) equilibrium is that 1 − q > c.□

1.2 Appendix B: Non-informative messages

Here, we show that when the messages are non-informative (i.e., q = 0.5), there is no separating (sequential) equilibrium.

Lemma 6

When q = 0.5 a separating sequential equilibrium does not exist.

Proof

There are two types of potential separating equilibria: (1) H sends a message and L does not, and (b) the other way around.

Case 1

H send a message with probability 1 − ε _H and L sends a message with probability ε _L . It is easy to show that the consistent beliefs are:

$$ \mu _{s}=\left\{\begin{array}{*{20}l}{1\quad\text{ if }\;A_{s}=(1,0)} \\{0\quad\text{ if }\;A_{s}=(0,1)} \\{ 0.5\;\text{ otherwise}}\end{array}\right\} $$

(8)

Thus, the expected net payoffs are:

When c < 0.5, then the only equilibrium of this game is (1,1) and when c > 0.5, then the only equilibrium of this game is (0,0). Thus, there is no separating equilibrium that is consistent with these beliefs.

Case 2

L sends a message with probability 1 − ε _L and H sends a message with probability ε _H . It is easy to show that the consistent beliefs are:

$$ \mu _{s}=\left\{\begin{array}{*{20}l}{0\quad\text{ if }\;A_{s}=(1,0)} \\{1\quad\text{ if }\;A_{s}=(0,1)} \\{0.5\;\text{ otherwise}}\end{array}\right\} $$

(9)

and the only equilibrium is (0,0) irrespective of the cost.□

1.3 Appendix C: Endogenous precision: non-strategic equilibrium

Proposition 3

(Non strategic equilibrium)

Proof

We start by demonstrating that sending a non-informative message (q = 0.5) is a dominant strategy for L. Then we show that as a result, it is optimal for H to send an informative message (q > 0.5). To simplify the presentation, assume (without loss of generality) that player 1 is H.

If q ₂ ≤ q ₁ (i.e., the precision of the message sent by L is at most as precise as the message sent by H), the probability that L is selected (irrespective of q ₂) is 1 − q ₁.²⁴

And, if q ₂ > q ₁, the probability that L is selected is 1 − q ₂, which is lower than 1 − q ₁.

Thus, the probability that L is selected is a non-increasing function in q ₂, while his cost is an increasing function in q ₂. It follows that q ₂ = 0.5 is a dominant strategy for L.

Next, we show that if C ^′(0.5) < 1, it is optimal for H to send an informative message (q > 0.5). It is easy to show that E(d ₁|q ₁,q ₂ = 0.5) = q ₁ and thus:

$$ \frac{\partial E\left[ \pi _{1}(q_{1});q_{2}=0.5\right] }{\partial q_{1}} =1-C^{\prime }(q_{1}) $$

Let q ⁰ denote the optimal precision for H. Specifically, q ⁰ satisfies the following condition 1 − C ^′(q ⁰) = 0. Since C ^″ > 0 and C ^′(0.5) < 1, it immediately follows that q ⁰ > 0.5.□

1.4 Appendix D: Endogenous precision: strategic equilibrium

Lemma 7

There exists a $\underline{q}$ where $0.5<\underline{q}<1$ that satisfies the condition: $(1-\underline{q})-C(\underline{q})=0$ . Furthermore, for any $q>\underline{q}$ , (1 − q) − C(q) < 0.

Proof

The function (1 − q) − C(q) is decreasing in q, and is positive at q = 0.5 and negative at q = 1.□

Lemma 8

There exists a $\overline{q}$ where $\overline{q} >0.5$ that satisfies the condition: $1-C(\overline{q}) =\max\limits_{0.5\leq q\leq 1}q-C(q)$ . Furthermore, for any $q<\overline{q}$ , $1-C(q) >\max\limits_{0.5\leq q\leq 1}q-C(q)$ .

Proof

The function $\begin{array}{cc} 1-C(q) \end{array}$is decreasing in q and is positive at q = 0.5.□

Lemma 9

$\overline{q}>$ q.

Proof

The function $\begin{array}{cc} 1-C(q) \end{array}$is decreasing in q. Next, we show that it is positive at q.

If $\underline{q}<q^{0}$, $\begin{array}{cc} 1-C(\underline{q}) \end{array} =\left[ 1-q^{0}\right] +\left[ C(q^{0})-C(\underline{q})\right]$ where both elements are positive.

If $\underline{q}>q^{0}$, $\begin{array}{cc} 1-C(\underline{q}) \end{array} >\left[ 1-\underline{q}\right] -C(\underline{q})+C(q^{0})>\left[ 1- \underline{q}\right] -C(\underline{q})=0$.□

We are now ready to prove Proposition 4:

Proof

(Proposition 4) To simplify the presentation, assume (without loss of generality) that player 1 is H. Given the beliefs and $q_{1}=q^{\ast }$, where $\underline{ q}<q^{\ast }<\overline{q}$, L will optimally choose q ₂ = 0.5 since: (a) choosing any q such that 0.5 < q < q ^∗ or q > q ^∗ involves a cost without any revenues, and (b) choosing $q_{2}=q^{\ast }$ leads to losses from Lemma 8.

Given the beliefs and q ₂ = 0.5, H will optimally choose $q_{1}=q^{\ast }$ since the highest payoff from any q ≠ q ^∗ is q ⁰ − C(q ⁰) which is smaller than the equilibrium payoff 1 − C(q ^∗) as illustrated by Lemma 9.

It is trivial to show that beliefs agree with senders’ strategies.□