Receiver’s sensitivity and strategic information transmission in multi-sender cheap talk

Abstract

This study examines how a CEO’s sensitivity affects information transmission from multiple unit managers when they communicate strategically during decision-making for technology investment. We develop a simple cheap talk model comprising one CEO and K unit managers, wherein the CEO makes a decision to maximize the units’ profit after the managers reveal their preference for the decision through cheap talk. We demonstrate that the expected total profit improves when sensitivity to the opinion of a manager whose expected preference is moderate among all managers increases. The driving force of the results is termed as alignment effect; increasing sensitivity to a moderate opinion decreases the difference between the expected CEO’s decision and the managers’ ideals. Furthermore, some implications on the decision-making process in technology investment within firms are discussed.

Appendix

1.1 Derivation of experts’ strategy and (2)

We first show that the equilibrium communication strategy in any perfect Bayesian equilibrium is of the interval form. Let two different posterior expectation of $\theta _i$ induced in an equilibrium be $m', m''$ and suppose $m' < m''$. Given that $d = d^*$, $\frac{ \partial ^2 E [\pi _i | \theta _i, m_i ] }{ \partial \theta _i^2 } = - 1 < 0$ and $\frac{ \partial ^2 E [\pi _i | \theta _i, m_i ] }{ \partial m_i \partial \theta _i } = \frac{ 2\alpha _i}{\sum _l \alpha _l} > 0$. Then, for $m'$ and $m''$, there exists, at most, one type of expert who is indifferent between both. Suppose that $\theta _i' < \theta _i''$ and $E [ \pi _i | \theta _i', m'' ] \ge E [ \pi _i | \theta _i', m' ]$ and $E [ \pi _i | \theta _i'', m' ] > E [ \pi _i | \theta _i'', m'' ]$. Then, $E [ \pi _i | \theta _i', m'' ] - E [ \pi _i | \theta _i', m' ] > E [ \pi _i | \theta _i'', m'' ] - E [ \pi _i | \theta _i'', m' ]$, which is contradiction.

When the expert induces $m'$, the expected DM’s decision from expert i’s perspective is given by $E[ d | m_i = m' ] = \frac{ \alpha _i }{ \sum _{l} \alpha _l } (m' + c_i) + \sum _{k \ne i} \frac{ \alpha _k }{ \sum _{l} \alpha _l } (E[m_k] + c_k)$. Substituting this into the indifferent condition, we obtain equation (2) as follows;

$$\begin{aligned}&\frac{ \alpha _i }{ \sum _{k} \alpha _k } ( ( m_{i,j+1} + c_i )^2 - ( m_{i,j} + c_i )^2 ) + 2 (m_{i,j+1} - m_{i,j}) \sum _{k \ne i} ( E [ m_k ] + c_k ) \\&\qquad - 2 (m_{i,j+1} - m_{i,j}) (a_{i,j} + c_i) = 0 \\&\rightarrow m_{i,j+1} + m_{i,j} + 2 c_i + 2 \frac{ \sum _l \alpha _l }{ \alpha _i } \sum _{k \ne i} ( E [ m_k ] + c_k ) - 2 \frac{ \sum _l \alpha _l }{ \alpha _i } ( a_{i,j} + c_i )= 0 \\&\rightarrow a_{i,j+1} - a_{i,j} = a_{i,j} - a_{i,j-1} + 4 \frac{ \sum _{k \ne i} \alpha _k (a_{i,j} + c_i) - \sum _{k \ne i} \alpha _k ( E [ m_k ] + c_k ) }{ \alpha _i }. \end{aligned}$$

Together with the boundary conditions $a_{i,0} = -s_i$ and $a_{i,N_i} = s_i$, the second-order difference Eq. (2) has the following solution in explicit form;

$$\begin{aligned} a_{i,j} = \frac{ \displaystyle x (\alpha _i)^{j} - y (\alpha _i)^{j} }{ \displaystyle x (\alpha _i)^{N_i} - y (\alpha _i)^{N_i} } (s_i - q_i (\alpha ) ) + \frac{ \displaystyle x (\alpha _i)^{N_i-j} - y (\alpha _i)^{N_i-j} }{ \displaystyle x (\alpha _i)^{N_i} - y (\alpha _i)^{N_i} } (- s_i - q_i (\alpha ) ) + q_i (\alpha )\nonumber \\ \end{aligned}$$

(10)

where $x (\alpha _i) = - \left( 1 - \frac{2 \sum _l \alpha _l }{\alpha _i} \right) + \sqrt{ \left( 1 - \frac{2 \sum _l \alpha _l }{\alpha _i} \right) ^2 - 1}$ and $y (\alpha _i) = - \left( 1 - \frac{2 \sum _l \alpha _l}{\alpha _i} \right) - \sqrt{ \left( 1 - \frac{2 \sum _l \alpha _l}{\alpha _i} \right) ^2 - 1}$. Note that $x (\alpha _i) \cdot y (\alpha _i) = 1$.

1.2 The properties regarding to communication strategy and the proofs

Property 1

For $i=1,2,...,K$, the upper bound of $N_i$ does not exist if $q_i ( \alpha ) \in [ - s_i, s_i ]$.

Proof

It is straightforward to check that $\{ a_{i,j} \}_{j = 0,1,...N_i}$ satisfy boundary constraints that $a_{i,0} = - s_i$ and $a_{i,N_i} = s_i$ for any $N_i$. Then, we show that $\{ a_{i,j} \}_{j = 0,1,...N_i}$ satisfy the order constraints. Since $x (\alpha _i) > y (\alpha _i)$, $\displaystyle x (\alpha _i)^{j} - y (\alpha _i)^{j}$ is strictly increased and $\displaystyle x (\alpha _i)^{N_i-j} - y (\alpha _i)^{N_i-j}$ is strictly decreased as j increases. Then, for any $N_i$, the first term of (10) is not decreasing in j if $s_i \ge q_i (\alpha )$ and strictly increasing in j if $s_i > q_i (\alpha )$, and the second term of (10) is not decreasing in j if $- s_i \le q_i (\alpha )$ and strictly increasing in j if $- s_i < q_i (\alpha )$. Then, if $- s_i \le q_i (\alpha ) \le s_i$, $a_{i,j}$ is strictly increasing in j. $\square$

Property 2

$E[ m_i^2]$ is increasing in $N_i$.

Proof

After lengthy calculations, we obtain

$$\begin{aligned} E[m_i^2]= & {} \sum _{j=0}^{N_i-1} \int _{a_{i,j}}^{a_{i,j+1}} \left( \frac{a_{i,j+1} + a_{i,j}}{2} \right) ^2 \frac{1}{2 s_i } d \theta _i \nonumber \\= & {} \frac{1}{8 s_i} \sum _{j=0}^{N_i-1} ( a_{i,j+1}^3 + a_{i,j+1}^2 a_{i,j} - a_{i,j+1} a_{i,j}^2 - a_{i,j}^3 ) \nonumber \\= & {} \frac{1}{4} \frac{x(\alpha _i)^2 + 2 x(\alpha _i) + 1}{x(\alpha _i)^2 + x(\alpha _i) + 1} s_i^2 - \frac{1}{4} \frac{x(\alpha _i)^2 - 2 x(\alpha _i) + 1}{x(\alpha _i)^2 + x(\alpha _i) + 1} q_i(\alpha )^2 \nonumber \\&- \frac{1}{4} \frac{ (x(\alpha _i)^2-1)^2}{ x(\alpha _i) (x(\alpha _i)^2 + x(\alpha _i) + 1) } \left( \frac{ x(\alpha _i)^{N_i} }{ (x(\alpha _i)^{N_i}+1)^2 } (s_i^2 - q_i (\alpha )^2) \nonumber \right. \\&\left. + \frac{ 4 x(\alpha _i)^{2N_i} }{ (x(\alpha _i)^{2N_i}-1)^2 } s_i^2 \right) . \end{aligned}$$

(11)

The first and second term in (11) is independent of $N_i$. Since $x (\alpha _i) > 1$ and $s_i> q_i(\alpha ) > -s_i$, the third term in (11) is positive.

To check whether the third term in (11) is decreasing in $N_i$, we define

$$\begin{aligned} f(X) = \frac{1}{4} \frac{ (x(\alpha _i)^2-1)^2}{ x(\alpha _i) (x(\alpha _i)^2 + x(\alpha _i) + 1) } \left( \frac{ X }{ (X+1)^2 } (s_i^2 - q_i(\alpha )^2) + \frac{ 4 X^2 }{ (X^2-1)^2 } s_i^2 \right) , \end{aligned}$$

and note that

$$\begin{aligned} f'(X) = \dfrac{1}{4} \dfrac{ (x(\alpha _i)^2-1)^2}{ x(\alpha _i) (x(\alpha _i)^2 + x(\alpha _i) + 1) } \left( \dfrac{ -X^2 + 1 }{ (X+1)^4 } (s_i^2 - q_i(\alpha )^2) + \dfrac{ -2 X^5 - 2X }{ (X^2-1)^4 } s_i^2 \right) . \end{aligned}$$

Since $f'(X)>0$ for $X>1$, the third term in (11) is strictly decreasing in $N_i$, then $E[m_i^2]$ strictly increases in $N_i$. Equation (5) can be derived by substituting $x (\alpha _i) = - \left( 1 - \frac{2 \sum _l \alpha _l }{\alpha _i} \right) + \sqrt{ (1 - \frac{2 \sum _l \alpha _l}{\alpha _i} )^2 - 1}$ into (11), and taking the limit of $N_i$ (then, the third term in (11) become zero). $\square$

Property 3

$E[ \pi ]$ and $E[ \pi _i ]$ are increasing in $E[ m_i^2]$.

Proof

First, we show that $E[ \pi _i ]$ is increasing in $E[ m_i^2]$. Substituting $d^* = \sum _{k} \frac{ \alpha _k }{ \sum _{l} \alpha _l } (m_k + c_k)$ into $E[ \pi _i ]$, we obtain

$$\begin{aligned} E[ \pi _i ]= & {} - E \left[ \left( \sum _{k} \frac{ \alpha _k }{ \sum _{l} \alpha _l } (m_k + c_k) - \theta _i - c_i \right) ^2 \right] \nonumber \\= & {} \frac{ \alpha _i }{ \sum _{l} \alpha _l } \left( 2 - \frac{ \alpha _i }{ \sum _{l} \alpha _l } \right) E [ m_i^2 ] - E [ \theta _i^2 ] - \left( \frac{ \alpha _i }{ \sum _{l} \alpha _l } - 1 \right) ^2 c_i^2 \nonumber \\&- \sum _{k \ne i} \left( \frac{ \alpha _k }{ \sum _{l} \alpha _l } \right) ^2 ( E[m_k^2] + c_k^2 ) \nonumber \\&- 2 \left( \frac{ \alpha _i }{ \sum _{l} \alpha _l } - 1 \right) \sum _{k \ne i} \frac{ \alpha _k }{ \sum _{l} \alpha _l } c_i c_k - \sum _{j \ne i} \sum _{k \ne i, j} \frac{ \alpha _j \alpha _k }{ (\sum _{l} \alpha _l)^2 } c_j c_k. \end{aligned}$$

(12)

The second equality follows from the facts $E[ m_i ] = 0$, $E[ \theta _i m_i ] = E [ E[ \theta _i m_i | r_i ] ] = E [ m_i E[ \theta _i | r_i ] ] = E[ m_i^2 ]$, and $E[ \theta _i m_j ] = E[ \theta _i] E[ m_j ] = 0$ for $j \ne i$. Immediately, $E[ \pi _i ]$ is increasing in $E [ m_i^2 ]$.

Finally, we show that $E[ \pi ]$ is increasing in $E[ m_1^2],...,E[ m_K^2]$. Summing (12) with i, we obtain

$$\begin{aligned} E[ \pi ] = \sum _i \alpha _i \left( - E[ \theta _i^2] + \frac{ \alpha _i }{ \sum _{l} \alpha _l } E[ m_i^2] - \frac{1}{2} \sum _{k \ne i} \frac{ \alpha _k }{ \sum _{l} \alpha _l } (c_i - c_k)^2 \right) . \end{aligned}$$

(13)

Then, $E[ \pi ]$ is increasing in $E[ m_1^2],...,E[ m_K^2]$. $\square$

1.3 Proof of Proposition 2

The first term in (6) is given by

$$\begin{aligned} \left. \frac{\partial M_i}{\partial \alpha _i}(\alpha , q) \right| _{ \displaystyle q = q_i (\alpha ) }= & {} \frac{ \sum _{k \ne i} \alpha _k }{ (4 \sum _{k \ne i} \alpha _k + 3\alpha _i)^2 } s_i^2 \nonumber \\&+ \frac{ 3 \sum _{k \ne i}\alpha _k }{ (4 \sum _{k \ne i} \alpha _k + 3\alpha _i)^2 } \left( \frac{ \sum _{k \ne i} \alpha _k (c_k - c_i) }{ \sum _{k \ne i} \alpha _k } \right) ^2 \end{aligned}$$

(14)

and is always positive. Since $q_i (\alpha )$ is independent of $\alpha _i$, $\frac{ \partial q_i }{ \partial \alpha _i } (\alpha ) = 0$. Then, the second term in (6) becomes zero. Thus, the marginal effect of $\alpha _i$ on ${{\hat{M}}}_i ( \alpha )$ is always positive. $\square$

1.4 Proof of Proposition 3

The first term in (7) is given by

$$\begin{aligned} \left. \frac{\partial M_j}{\partial \alpha _i}(\alpha , q) \right| _{ \displaystyle q = q_j (\alpha ) }= & {} - \frac{ \alpha _j }{ (4 \sum _{k \ne j} \alpha _k + 3\alpha _j)^2 } s_j^2 \nonumber \\&- \frac{ 3 \alpha _j }{ (4 \sum _{k \ne j} \alpha _k + 3\alpha _j)^2 } \left( \frac{ \sum _{k \ne j} \alpha _k (c_k - c_j) }{ \sum _{k \ne j} \alpha _k } \right) ^2 \end{aligned}$$

(15)

and is always negative. The second term in (7) is given by

$$\begin{aligned} \frac{ \partial M_j }{ \partial q_j } ( \alpha , q_j (\alpha ) ) \cdot \frac{ \partial q_j }{ \partial \alpha _{i} } (\alpha ) = \frac{ 2 (\sum _{k \ne j} \alpha _k (c_k - c_j) ) ( \sum _{k \ne j} \alpha _k (c_k - c_i) ) }{ ( 4 \sum _{ k \ne j} \alpha _k + 3 \alpha _j ) (\sum _{k \ne j} \alpha _k)^2}. \end{aligned}$$

(16)

Summing (15) and (16) and rearranging it, we obtain $\frac{ \partial {{\hat{M}}}_j }{ \partial \alpha _i } ( \alpha )$ as follows;

$$\begin{aligned} \frac{ \partial {{\hat{M}}}_j }{ \partial \alpha _i } ( \alpha )= & {} \frac{ 1 }{ \left( 4 \sum _{k \ne j} \alpha _k + 3 \alpha _j \right) ^2 \left( \sum _{k \ne j} \alpha _k \right) ^2 } \nonumber \\&\left( \underbrace{ - \alpha _j \left( \sum _{k \ne j} \alpha _k \right) ^2 s_j^2 - 3 \alpha _j \left( \sum _{k \ne j} \alpha _k (c_j - c_k) \right) ^2 }_{\text {The sensitivity effect}} \right. \nonumber \\&+ \left. \underbrace{2 \left( 4\sum _{k \ne j} \alpha _k + 3 \alpha _j \right) \left( \sum _{k \ne j} \alpha _k (c_k - c_j) \right) \left( \sum _{k \ne j} \alpha _k (c_k - c_i) \right) }_{\text {The alignment effect}} \right) . \nonumber \\ \end{aligned}$$

(17)

The sign of the alignment effect equals to the sign of $\sum _{k \ne j} \alpha _k (c_k - c_j) \cdot \sum _{k \ne j} \alpha _k (c_k - c_i)$. Then, it is strictly positive if and only if $c_i > \frac{\sum _{k \ne j} \alpha _k c_k}{ \sum _{k \ne j} \alpha _k }$ and $c_j > \frac{\sum _{k \ne j} \alpha _k c_k}{ \sum _{k \ne j} \alpha _k }$ or $c_i < \frac{\sum _{k \ne j} \alpha _k c_k}{ \sum _{k \ne j} \alpha _k }$ and $c_j < \frac{\sum _{k \ne j} \alpha _k c_k}{ \sum _{k \ne j} \alpha _k }$.

Finally, we provide an example in which the cross effect is positive. Let $\alpha _i = 1$ for $i = 1,...,K$ and $c_i = c_j$. Substituting these into (17) and utilizing $q_j (\alpha ) = \frac{\sum _{k \ne j} (c_k - c_j)}{ K-1 }$, the parts bracketed in (17) can be represented as

$$\begin{aligned} (K-1)^2 ( - s_j^2 + (8K-5) ( q_j (\alpha ) )^2). \end{aligned}$$

(18)

Since $s_j^2 > ( q_j (\alpha ) )^2$ under our assumption $s_i > \max _{ k \in \{ 1,...,K \} } \left\{ 2 |c_k| \right\}$, (18) can be positive if the absolute value of $q_j (\alpha )$ is sufficiently large (say, $q_j (\alpha ) = \frac{s_j}{2}$). $\square$

1.5 Proof of Proposition 4

For a given $\beta _1,...,\beta _K$, the DM’s decision is given by $\sum _{i} \frac{ \beta _i }{ \sum _{l} \beta _l } (m_i + c_i)$. We denote $\beta = (\beta _1,...,\beta _K)$. Then, the quality of communication with expert i is given by $M_i (\beta , q_i( \beta ) )$ and the DM’s expected payoff is given as

$$\begin{aligned} E[\pi ]= & {} - \sum _{i} \left( \alpha _i E[\theta _i^2] + \left( \left( \frac{ \beta _i }{ \sum _{l} \beta _l } \right) ^2 \sum _l \alpha _l - 2 \alpha _i \frac{ \beta _i }{ \sum _{l} \beta _l } \right) M_i (\beta , q_i( \beta ) ) \right. \\&- \left( \alpha _i \left( \frac{ \beta _i }{ \sum _{l} \beta _l } - 1 \right) ^2 + \left( \frac{ \beta _i }{ \sum _{l} \beta _l } \right) ^2 \sum _{k \ne i} \alpha _k \right) c_i^2 \\&\left. - \sum _{k \ne i} \left( \frac{ \beta _i \beta _k }{ (\sum _{l} \beta _l)^2 } \sum _l \alpha _l - \alpha _i \frac{ \beta _k }{ \sum _{l} \beta _l } - \alpha _k \frac{ \beta _i }{ \sum _{l} \beta _l } \right) c_i c_k \right) . \end{aligned}$$

The first derivative of $E[\pi ]$ with $\beta _i$ at $\beta _k = \alpha$ for all k is given by

$$\begin{aligned} \left. \frac{\partial E[\pi ] }{\partial \beta _i} \right| _{ \displaystyle (\beta _1,...,\beta _K) = (\alpha ,...,\alpha ) }= & {} \left. \frac{ \alpha }{ K } \sum _k \frac{ \partial {{\hat{M}}}_k }{ \partial \beta _i }(\beta ) \right| _{ \displaystyle (\beta _1,...,\beta _K) = (\alpha ,...,\alpha ) }. \end{aligned}$$

According to (14), (15), and (16), each marginal effect is as follows;

$$\begin{aligned} \left. \frac{ \partial {{\hat{M}}}_i }{ \partial \beta _i }(\beta ) \right| _{ \displaystyle (\beta _1,...,\beta _K) = (\alpha ,...,\alpha ) } = \frac{ K-1 }{ \alpha (4K-1)^2 } s_i^2 + \frac{ 3 \sum _{k \ne i}(c_k - c_i)^2 }{ \alpha (4K-1)^2 (K-1) } \end{aligned}$$

and

$$\begin{aligned} \left. \frac{ \partial {{\hat{M}}}_j }{ \partial \beta _i }(\beta ) \right| _{ \displaystyle (\beta _1,...,\beta _K) = (\alpha ,...,\alpha ) }= & {} - \frac{ 1 }{ \alpha (4K-1)^2 } s_j^2 - \frac{ 3 \sum _{k \ne j}(c_k - c_j)^2 }{ \alpha (4K-1)^2 (K-1)^2 } \\&+ \frac{ 2 \sum _{k \ne j}(c_k - c_i) \sum _{k \ne j}(c_k - c_j) }{ \alpha (4K-1) (K-1)^2 }. \end{aligned}$$

Summing them, we derive (9) as follows;

$$\begin{aligned}&\left. \sum _{k} \frac{ \partial {{\hat{M}}}_k }{ \partial \beta _i } ( \beta ) \right| _{ \displaystyle (\beta _1,...,\beta _K) = (\alpha ,...,\alpha ) } \\&\quad = \frac{ 1 }{ \alpha (4K-1)^2 } \left( (K-1) s_i^2 - \sum _{k \ne i} s_k^2 \right) \\&\qquad + \frac{ 3(K-1) ( \sum _{k \ne i }( c_k-c_i) )^2 - \sum _{j \ne i} ( 3 ( \sum _{k \ne j} (c_k-c_j) )^2 - 2(4K-1) ( \sum _{k \ne j} (c_k-c_i) \sum _{k \ne j} ( c_k - c_j ) ) ) }{ \alpha (4K-1)^2 (K-1)^2 } \\&\quad = \frac{ 1 }{ \alpha (4K-1)^2 } \left( (K-1) s_i^2 - \sum _{k \ne i} s_k^2 \right) \\&\qquad + \frac{5K-2}{ \alpha (4K-1)^2 (K-1)^2 } \left( -(K-2)(K-1)c_i^2 + (K-2) \sum _{j \ne i} c_j^2 \right. \\&\left. \qquad + 2 (K-2) c_i \sum _{j \ne i}c_j - 2 \sum _{j \ne i}\sum _{k \ne i, j} c_j c_k \right) \\&\quad = \frac{ 1 }{ \alpha (4K-1)^2 } \left( (K-1) s_i^2 - \sum _{k \ne i} s_k^2 \right) \\&\qquad + \frac{ (5K-2)(K-2) }{ \alpha (K-1)(4K-1)^2 } \left( \frac{ 1 }{ (K-1)(K-2) } \sum _{j \ne i} \sum _{k \ne i,j} (c_j-c_k)^2 - \frac{ 1 }{ (K-1) } \sum _{k \ne i} (c_i-c_k)^2 \right) . \end{aligned}$$

$\square$

1.6 Proof of observation 1

Without loss of generality, we assume $c_i = 0$ and $c_j > 0$. Define $g_i ( c ) = \sum _{j \ne i} \sum _{k \ne i,j} (c_j-c_k)^2 - (K-1) \sum _{k \ne i} (c_i-c_k)^2$ where $c = (c_1,...,c_K)$. We have $\frac{ \partial g_i }{ \partial c_k } = -4 \sum _{l \not \in {\mathbb {I}} } c_l + 2 K c_j > -4 \sum _{l \not \in {\mathbb {I}} } c'_l + 2 K c_j$ where $c'_l$ is defined as $c'_l = c_l$ if $c_l > 0$ and $c'_l = - c_l$ otherwise. If $c_j \ge c'_l$ for all l, $\sum _{l \not \in {\mathbb {I}} } c'_l < (K-I) c_j$. Then, $\frac{ \partial g_i }{ \partial c_k } > (4 I - 2K) c_j$. Thus, $g_i ( c )$ is increasing in $c_j$ if $I > K/2$.

Let $c'' = (c''_1,...,c''_K)$ where $c''_l$ is defined as $c''_l = c_l$ for $l \in {\mathbb {I}}$ and $c''_l = \min _{l \not \in {\mathbb {I}} } c'_l$ otherwise. It is immediate that $g_i (c) > g_i (c'')$. Then, we have

$$\begin{aligned} g_i (c'') = ( 2 (I-1) (K-I) - (K-2) (K-I) ) (\min _{l \not \in {\mathbb {I}} } c'_l)^2 \end{aligned}$$

If $I > K/2$, $g_i (c'') > 0$ therefore $g_i (c) > 0$. $\square$