Strategic compliments in sales

Abstract

Salespersons often spend time and money giving prospective buyers compliments such as kind words, meals and gifts. Though prior research has shown that compliments will influence a prospective buyer’s decision, it is unknown the extent to which salespersons should make these investments. In this paper, we develop an analytical model to examine how seller and buyer characteristics affect the equilibrium provision of compliments by the seller. We establish that the optimal magnitude of compliments is non-monotonic in the buyer’s sensitivity to compliments. We identify conditions for when a seller of a high-quality product will offer greater (or lesser) compliments than a seller of a lower quality product. We show that, under certain conditions, an uninformed buyer earns greater utility than a buyer who knows the quality of the seller’s product. The findings have implications for sellers in their choice of compliments and buyers in the inferences they draw from the compliments received.

Appendix A: proofs of lemmas and propositions

1.1 Proof of observation 1

The derivative of k ^∗ with respect to α is

$$\frac{\frac{1}{\sqrt{1-\frac{4 c_{j}}{\alpha M}}}-Arcosh\left( \frac{\alpha M}{2 c_{j}}-1\right)+T_{j}}{\alpha^{2}} $$

This is a decreasing function of α. Furthermore, it is strictly positive for $\alpha = \frac {4c_{j}}{M}+\epsilon$, for sufficiently small 𝜖>0, and is strictly negative for α=1 (given that $T_{j} < Arcosh\left (\frac {M}{2 c_{j}}-1\right )-\sqrt {\frac {M}{M - 4 c_{j}}}$ ). Therefore, there exists $0 < \bar {\alpha } < 1$ for which this expression becomes zero. Though $\bar {\alpha }$ does not have a closed-form solution, it is the unique solution to

$$1+T\sqrt{1-\frac{4 c_{j}}{\bar{\alpha} M}} = Arcosh\left( \frac{\bar{\alpha} M}{2 c_{j}}-1\right)\sqrt{1-\frac{4 c_{j}}{\bar{\alpha} M}} $$

Therefore, k ^∗ is an increasing function of α for $\alpha < \bar {\alpha }$, and a decreasing function of α for $\alpha > \bar {\alpha }$.

When $T_{j}> Arcosh\left (\frac {M}{2 c_{j}}-1\right )-\sqrt {\frac {M}{M - 4 c_{j}}}$, the derivative of k ^∗ with respect to α is always positive.

1.2 Proof of proposition 1

First assume that c _L >c _H . Derivative of $k_{H}^{\ast } - k_{L}^{\ast }$ with respect to M is

$$\frac{\sqrt{1-\frac{4 c_{H}}{\alpha M}}}{\alpha M-4 c_{H}}-\frac{\sqrt{1-\frac{4 c_{L}}{\alpha M}}}{\alpha M-4 c_{L}} $$

which is always negative. If $M = \frac {4 c_{L}+\epsilon }{\alpha }$, for sufficiently small 𝜖>0, then $\lim _{\epsilon \rightarrow 0}k_{H}^{\ast } - k_{L}^{\ast } = \frac {\Delta T (\alpha -1) + Arcosh(\frac {2 c_{L}}{c_{H}}-1)}{\alpha }$. This is positive if and only if $\frac {c_{L}}{c_{H}} > \frac {1+Cosh({\Delta } T (1-\alpha ))}{2}$. We also have $\lim _{M \rightarrow \infty }k_{H}^{\ast } - k_{L}^{\ast } = \frac {\Delta T (\alpha -1) + Log(\frac {\alpha }{c_{H}})-Log(\frac {\alpha }{c_{L}})}{\alpha }$. This is negative if and only if $\frac {c_{L}}{c_{H}} < e^{\Delta T (1-\alpha )}$. In other words, $\frac {c_{L}}{c_{H}} < e^{\Delta T (1-\alpha )}$ implies that as M grows, $k_{H}^{\ast } - k_{L}^{\ast }$ becomes strictly negative. Therefore, when $\frac {1+Cosh({\Delta } T (1-\alpha ))}{2} < \frac {c_{L}}{c_{H}} < e^{\Delta T (1-\alpha )}$, there exists a unique $\bar {M}$ which is the solution to $k_{H}^{\ast } = k_{L}^{\ast }$. $\bar {M}$ does not have a closed-form solution; however, it is the unique solution to the following equation

$$\frac{\sqrt{\alpha \bar{M} (\alpha \bar{M}- 4 c_{H})}+\alpha \bar{M}- 2 c_{H}}{\sqrt{\alpha \bar{M} (a \bar{M}-4 c_{L})}+\alpha \bar{M}- 2 c_{L}} = \frac{c_{H}}{c_{L}}e^{\Delta T (1-\alpha)} $$

If $\frac {c_{L}}{c_{H}} \leq \frac {1+Cosh({\Delta } T (1-\alpha ))}{2}$, then $k_{H}^{\ast } - k_{L}^{\ast }$ is always non-positive; and if $\frac {c_{L}}{c_{H}} \geq e^{\Delta T (1-\alpha )}$, then $k_{H}^{\ast } - k_{L}^{\ast }$ is always non-negative.

For the case where c _L ≤c _H , since k ^∗ is a decreasing function of T _j and a decreasing function of c _j , we always have $k_{H}^{\ast } \leq k_{L}^{\ast }$.

1.3 Proofs of lemmas 1 and 2

First, we prove that $\bar {\kappa }_{H} \geq \bar {\kappa }_{L}$. This shows that the set of strategies in which the high-type gives compliments of magnitude $\max (\bar {\kappa }_{L}, k_{H}^{\ast H})$ and the low-type give compliments of magnitude $k_{L}^{\ast L}$ is a separating equilibrium.

To prove that $\bar {\kappa }_{H} \geq \bar {\kappa }_{L}$, it suffices to show that ${u_{L}^{L}}(k_{L}^{\ast L}) \geq {u_{L}^{H}}(\bar {\kappa }_{H})$. Since $k_{L}^{\ast L}$ is the value that maximizes ${u_{L}^{L}}(.)$, we know that

$${u_{L}^{L}}(k_{L}^{\ast L}) \geq {u_{L}^{L}}(k_{H}^{\ast L}) $$

By expanding the right-hand-side, we get

$${u_{L}^{L}}(k_{L}^{\ast L}) \geq {u_{H}^{L}}(k_{H}^{\ast L}) - k_{H}^{\ast L} (c_{L} - c_{H}) $$

Using the definition of $\bar {\kappa }_{H}$, we can write this inequality as

$${u_{L}^{L}}(k_{L}^{\ast L}) \geq {u_{H}^{H}}(\bar{\kappa}_{H}) - k_{H}^{\ast L} (c_{L} - c_{H}) $$

Using the definition of ${u_{L}^{H}}$ and ${u_{H}^{H}}$, we have ${u_{H}^{H}}(k) = {u_{L}^{H}}(k) + k (c_{L} - c_{H})$ for any k. Therefore, we can write the above inequality as

$${u_{L}^{L}}(k_{L}^{\ast L}) \geq {u_{L}^{H}}(\bar{\kappa}_{H}) + \bar{\kappa}_{H} (c_{L} - c_{H}) - k_{H}^{\ast L} (c_{L} - c_{H}) $$

Since ${u_{H}^{H}}(k) \geq {u_{H}^{L}}(k)$ for any k, we have ${u_{H}^{H}}(k_{H}^{\ast L}) \geq {u_{H}^{L}}(k_{H}^{\ast L})$. This implies that $\bar {\kappa }_{H} \geq k_{H}^{\ast L}$. Therefore, the above inequality implies

$${u_{L}^{L}}(k_{L}^{\ast L}) \geq {u_{L}^{H}}(\bar{\kappa}_{H}) $$

This implies that $\bar {\kappa }_{H} \geq \bar {\kappa }_{L}$, which completes the first part of the proof.

Next, we show that no other separating equilibrium can survive D1 criterion refinement. Given the definitions of $\bar {\kappa }_{L}$ and $\bar {\kappa }_{H}$, we know that there is no separating equilibrium in which the high-type gives compliments of magnitude lesser than $\bar {\kappa }_{L}$, otherwise, the low-type would benefit from deviating. Furthermore, according to D1 criterion refinement,¹⁵ the weight of the buyer’s belief L must be zero for any compliments of magnitude at least $\bar {\kappa }_{L}$. Therefore, the high-type always gives compliments of magnitude $\max (\bar {\kappa }_{L}, k_{H}^{\ast H})$ in any separating equilibrium that survives D1 criterion refinement. It is easy to see that the low-type gives compliments of magnitude $k_{L}^{\ast L}$ in any separating equilibrium. Therefore, the only separating equilibrium that could possibly survive D1 criterion refinement is the one in which the high-type gives a compliment of magnitude $\max (\bar {\kappa }_{L}, k_{H}^{\ast H})$ and the low-type gives a compliment of magnitude $k_{L}^{\ast L}$.

Next, we prove that this equilibrium does survive D1 criterion refinement. We have already shown that the buyer’s belief on any compliments of magnitude greater than $\bar {\kappa }_{L}$ is H. It remains to show that the buyer’s belief on any compliments of magnitude lesser than $\bar {\kappa }_{L}$ is L. Consider a compliment of magnitude less than $\kappa < \bar {\kappa }_{L}$, and suppose that the buyer’s belief is z>0, i.e., a seller who sends a compliment of size κ is high-type with probability z>0. Note that the low-type is indifferent between giving compliments of magnitude $\bar {\kappa }_{L}$ and being perceived as high-type, and giving compliments of magnitude $k_{L}^{\ast L}$ and being perceived as low-type. Therefore, a hypothetical deviation from $k_{L}^{\ast L}$ to $\bar {\kappa }_{L}$ leaves the utility of the low-type unchanged. Now, a deviation from compliments of magnitude $\bar {\kappa }_{L}$ to compliments of magnitude κ has the same effect on probability of winning the contract (r(.)) for both types. However, the low-type gains more in terms of lowering the cost of compliment (since c _H <c _L ). Therefore, whenever such deviation is weakly profitable for the high-type, it is strictly profitable for the low-type. Therefore, according to D1 criterion, the buyer’s belief on out-of-equilibrium compliment κ must be L, i.e., z=0.

Finally, we show that no pooling equilibrium survives D1 criterion refinement. Assume for sake of contradiction that a pooling equilibrium exists. Let κ be the magnitude of compliments in this pooling equilibrium. For any k>κ, the high-type gains more (or loses less) by deviating to k, for any buyer’s belief weights on k, because c _H <c _L . Therefore, using D1 equilibrium refinement, the buyer’s belief on k must be H (i.e., someone using k must be high-type). In other words, since the high-type always gains more (or loses less) by increasing the magnitude of compliments, the buyer’s out-of-equilibrium belief about any compliments greater than κ must be high. Therefore, if k = κ + 𝜖 for sufficiently small 𝜖>0, both types benefit from deviating from κ to k because they get a discrete increase in their utility (by moving from belief X to belief H) but only lose c 𝜖 in terms of costs. This contradicts the equilibrium assumption.

The proof of Lemma 2 is very similar to that of Lemma 1. We can show that $\underline {\kappa }_{H} < \underline {\kappa }_{L}$. Then, as long as $\underline {\kappa }_{L} > 0$, the specified separating equilibrium exists. The proof of uniqueness subject to D1 criterion refinement is also very similar: the high-type always gains more (or loses less) by deviating to lesser compliments. Therefore, out-of-equilibrium beliefs on such deviations is always high. A pooling equilibrium could only exist if the magnitude of the compliment cannot be lowered, i.e., when both types give zero compliments.

1.4 Proof of proposition 2

Note that when c _H <c _L , the only equilibrium that survives D1 criterion refinement is a separating equilibrium. Therefore, in equilibrium, the buyer will rationally infer the sender’s type. The magnitude of compliments that sellers offer in the case of partial information is greater than or equal to that of full information case (low-type’s compliments are of the same magnitude, and high-type offers greater (or equal) compliments). Therefore, when c _H <c _L , the buyer’s utility in partial information setting is always greater than or equal to that of full information setting.

On the other hand, when c _H >c _L , the equilibrium is not always separating. Furthermore, the magnitude of compliments in partial information case is always less than or equal to that of full information case. Therefore, when c _H >c _L , the buyer is always (weakly) worse off when he does not know the seller’s type.

1.5 Proof of lemma 3

Note that when c _H >c _L we have $\hat {k}_{H} = \min (\underline {\kappa }_{L}, k_{H}^{\ast H})$, and when c _H <c _L we have $\hat {k}_{H} = \max (\bar {\kappa }_{L}, k_{H}^{\ast H})$. We already know that $k_{H}^{\ast H}$ is an increasing function of M. It remains to be proven that $\bar {\kappa }_{L}$ and $\underline {\kappa }_{L}$ are increasing functions of M.

$\bar {\kappa }_{L}$ is defined as the solution $\bar {\kappa }_{L}$ to

$${u_{L}^{L}}(k_{L}^{\ast L}) = {u_{L}^{H}}(\bar{\kappa}_{L}). $$

This equation expands to

$$\frac{M e^{(1-\alpha) T_{L}}}{e^{(1-\alpha) T_{L}}+1}=\frac{M e^{\alpha \bar{\kappa}_{L}+(1-\alpha) T_{H}}}{e^{\alpha \bar{\kappa}_{L}+(1-\alpha) T_{H}}+1}-c_{L} \bar{\kappa}_{L} $$

for the case where $K_{L}^{\ast L}=0$, and to

$$\begin{array}{@{}rcl@{}} &&-\frac{M}{e^{Arcosh\left( \frac{\alpha M}{2 c_{L}}-1\right)}+1}-\frac{c_{L} Arcosh\left( \frac{\alpha M}{2 c_{L}}-1\right)}{\alpha}\\ &&+\left( \frac{1}{\alpha}-1\right) c_{L} T_{L}+M=\frac{M e^{\alpha \bar{\kappa}_{L}+(1-\alpha) T_{H}}}{e^{\alpha \bar{\kappa}_{L}+(1-\alpha) T_{H}}+1}-c_{L} \bar{\kappa}_{L} \end{array} $$

for the case where $K_{L}^{\ast L} > 0$. If we take derivative of the first equation with respect to M we get

$$\frac{ e^{(1-\alpha) T_{L}}}{e^{(1-\alpha) T_{L}}+1}=\frac{ e^{\alpha \bar{\kappa}_{L}+(1-\alpha) T_{H}}}{e^{\alpha \bar{\kappa}_{L}+(1-\alpha) T_{H}}+1}+ \frac{\alpha M}{2 \cosh (\alpha \bar{\kappa}_{L}-\alpha T_{H}+T_{H})+2}\frac{\partial \bar{\kappa}_{L}}{\partial M}-c_{L} \frac{\partial \bar{\kappa}_{L}}{\partial M} $$

This implies that

$$\frac{\partial \bar{\kappa}_{L}}{\partial M} = \frac{\frac{ e^{(1-\alpha) T_{L}}}{e^{(1-\alpha) T_{L}}+1}-\frac{ e^{\alpha \bar{\kappa}_{L}+(1-\alpha) T_{H}}}{e^{\alpha \bar{\kappa}_{L}+(1-\alpha) T_{H}}+1}}{\frac{\alpha M}{2 \cosh (\alpha \bar{\kappa}_{L}-\alpha T_{H}+T_{H})+2}-c_{L}} $$

It is easy to see that the numerator of the right hand side is negative; furthermore, given that we are considering the case where $\bar {\kappa }_{L} \geq k_{H}^{\ast H}$ and c _H <c _L , using basic calculus we can show that the denominator of the right hand side is negative as well; therefore, $\frac {\partial \bar {\kappa }_{L}}{\partial M}$ is positive. If we take the derivative of the second equation (the case where $K_{L}^{\ast L} > 0$) with respect to M, and use the same technique, we get

$$\frac{\partial \bar{\kappa}_{L}}{\partial M} = \frac{\frac{1}{2} \sqrt{1-\frac{4 c_{L}}{\alpha M}}-\frac{1}{2} \tanh \left( \frac{1}{2} (\alpha (\bar{\kappa}_{L}-T_{H})+T_{H})\right)}{\frac{\alpha M}{2 \cosh (\alpha \bar{\kappa}_{L}-\alpha T_{H}+T_{H})+2}-c_{L}} $$

The denominator is the same as before, which we already know is negative. Given that we are considering the case where $\bar {\kappa }_{L} \geq k_{H}^{\ast H}$, we can use basic calculus to show that the numerator is negative as well; therefore, $\frac {\partial \bar {\kappa }_{L}}{\partial M}$ is positive. Finally, note that using the closed-form expression for $k_{H}^{\ast H}$, it is easy to see that $\lim _{M \rightarrow \infty } k_{H}^{\ast H} = \infty$; therefore, we have $\lim _{M \rightarrow \infty } \hat {k}_{H} = \infty$ as well. This completes the proof for the case of c _H <c _L .

The analysis for the case of c _L <c _H is very similar. We have

$${u_{L}^{L}}(k_{L}^{\ast L}) = {u_{L}^{H}}(\underline{\kappa}_{L}). $$

By taking derivative of both sides with respect to M and rearranging the terms (same technique as before) we get

$$\frac{\partial \underline{\kappa}_{L}}{\partial M} = \frac{\frac{1}{2} \sqrt{1-\frac{4 c_{L}}{\alpha M}}-\frac{1}{2} \tanh \left( \frac{1}{2} (\alpha (\underline{\kappa}_{L}-T_{H})+T_{H})\right)}{\frac{\alpha M}{2 \cosh (\alpha \underline{\kappa}_{L}-\alpha T_{H}+T_{H})+2}-c_{L}} $$

Using c _H >c _L and $\underline {\kappa }_{L} \leq k_{H}^{\ast H}$ (because it is where $\underline {\kappa }_{L}$ is relevant), it is easy to show that both the numerator and the denominator of the right hand side are positive. Therefore, $\underline {\kappa }_{L}$ is an increasing function of M.

Next, we will prove that $\lim _{M \rightarrow \infty } \hat {k}_{H} = \infty$. Using basic calculus, it is easy to see that $\lim _{M \rightarrow \infty } \frac {{u_{L}^{L}}(k_{L}^{\ast L})} {M}= 1$. Therefore, using the definition of $\underline {\kappa }_{L}$, we must have $\lim _{M \rightarrow \infty } \frac {{u_{L}^{H}}(\underline {\kappa }_{L})}{M}= 1$. But, using the definition of ${u_{L}^{H}}(.)$, this cannot happen if $\lim _{M \rightarrow \infty } \underline {\kappa }_{L}$ is bounded; therefore, we must have $\lim _{M \rightarrow \infty } \underline {\kappa }_{L} =\infty$. Also, using the closed-form expression for $k_{H}^{\ast H}$ we already know that $\lim _{M \rightarrow \infty } k_{H}^{\ast H} = \infty$ as well; therefore, since $\hat {k}_{H} = \min (\underline {\kappa }_{L}, k_{H}^{\ast H})$ we have $\lim _{M \rightarrow \infty } \hat {k}_{H} = \infty$. This completes the proof for the case of c _H >c _L as well.

1.6 Proof of proposition 3

We first identify the values of M such that we can rule out the possibilities that the cap has a positive effect on the organization’s actual utility. If $T_{L}<\alpha \hat {k}_{L}+(1-\alpha )T_{L}<p T_{H}+(1-p)T_{L}$, then the uncapped buying agent never uniquely makes Type I or Type II errors for low-type sellers. Given T _L ≤0, the first inequality holds trivially for all $\hat {k}_{L}$. From Lemmas 1 and 2, $\hat {k}_{L}$ is a weakly increasing function of M, $\hat {k}_{L}=0$ at M=0, and $\hat {k}_{L} \rightarrow \infty$ as $M \rightarrow \infty$. Thus, p>α is sufficient to ensure that $\alpha \hat {k}_{L}+(1-\alpha )T_{L}<p T_{H}+(1-p)T_{L}$ at M=0 and there exists $\hat {M}>0$ such that $\alpha \hat {k}_{L}+(1-\alpha )T_{L}<p T_{H}+(1-p)T_{L}$ if and only if $M<\hat {M}$.

If $p T_{H}+(1-p)T_{L}<\alpha \hat {k}_{H}+(1-\alpha )T_{H}<T_{H}$, then the uncapped buying agent never uniquely makes Type I or Type II errors for high-type sellers. From Lemma 3, $\hat {k}_{H}=0$ at M=0 and $\hat {k}_{H}$ is weakly increasing in M. Therefore, there exists $M^{\prime }$ such that $\alpha \hat {k}_{H}+(1-\alpha )T_{H}<T_{H}$ is satisfied if and only if $M<M^{\prime }$. The condition p<1−α is sufficient to ensure that $p T_{H}+(1-p)T_{L}<\alpha \hat {k}_{H}+(1-\alpha )T_{H}$ for all M>0.

Therefore, if α<p<(1−α), then we can rule out the possibilities that allowing the buying agent to accept compliments has a negative effect on the organization’s actual utility for all $M<\bar {M}\:= min(\hat {M},M^{\prime })$. Moreover, these conditions imply the ranges exist such allowing the buying agent to accept compliments has a positive effect on the objective buyer’s actual utility for 𝜖 satisfying $\max \{-T_{H},-(\alpha \hat {k}_{H}+(1-\alpha )T_{H})\} < \epsilon < -(p T_{H}+(1-p )T_{L})$ or $-(p T_{H}+(1-p )T_{L})<\epsilon <\min \{-T_{L}, -(\alpha \hat {k}_{L}+(1-\alpha )T_{L})\}$.