Why are Some Salespeople More Aggressive than Others?

Abstract

Why are salespeople in certain industries (such as cars and mattresses) aggressive, while their counterparts in other industries (such as luxury boutiques) relatively customer-oriented? Using a principal-agent-customer model, this paper demonstrates that the level of salesperson aggressiveness depends on: (1) the proportion of customers with a high willingness to pay; (2) whether or not customers are well-informed and (3) repeat customers. If the proportion of customers with a high willingness to pay is relatively large, then salespeople tend to be customer-oriented. By contrast, if the proportion of customers with a high willingness to pay is relatively small, then salespeople tend to be aggressive toward uninformed customers while well-informed customers shun the store. Finally, if the proportion of customers with a high willingness to pay is relatively small, then in an infinitely repeated game, the agent can close sales with well-informed customers without being aggressive, provided that the principal is patient enough about future profit.

Appendices

Appendices

Appendix A Empirical Information

Sales Commission

This section shows that the reward to salespeople is based on commission in most luxury boutiques. The data are collected from The Galleria in Houston, Texas in spring 2009. A mystery shopper visited 19 boutiques (Table 1) to inquire whether or not their salespeople “work on commission.” Every agent in every store gave an answer “yes” or “no.” Note that the reward to salespeople in more than 84 % of the retailers depends on sales commission.

Table 1 Companies with Commission-Based Rewards

Sales Person Attitude

When gathering data for Table 1), the mystery shopper pretended to be a customer who intended to buy one item. None of the salespeople in any of the boutiques in Table 1 urged the mystery shopper buy an item “today.” All salespeople gave a business card upon request, and the salespeople at Salvatore Ferragamo, Stuart Weitzman, and Yves Saint Laurent also wrote down product information on the card.

Appendix B Finding Equilibrium for Section 2 --- Backward Induction

Appendix B.1 The Agent’s Optimal Effort Level

Given s _P  = (p,w,k) and $\widehat{s}$ _i (making a purchase if WTP _i (e) ≥ p), the agent will only choose among e = 0, e = e ₁ (p), or e = e ₂ (p), whichever gives the agent the highest payoff.

Sketch of a Proof

Given P, w, and k, the agent can choose from three different ranges of effort level; that is, 0 ≤ e < e ₁ (p), e ₁ (p) ≤ e < e ₂ (p), and e ₂ (p) ≤ e < ∞ (Fig. 1). The effort level must be greater than or equal to e _i (p), i = 1 or 2, for the agent to close the sale with a type i customer. If 0 ≤ e < e _i (p), then s _i  = N∀ _i . Therefore, the agent’s payoff is

$$ {U}_A\left( sp=\left( p, w, k\right),{s}_A= e\in \left[0,{e}_1(p)\right),{s}_i= N\right)= w- e. $$

The agent’s payoff is decreasing in e; therefore, if e ∈ [0, e ₁ (p)), she would choose e = 0 to maximize her payoff. In this case,

$$ {U}_A\left({s}_P=\left( p, w, k\right),{s}_A=0,{s}_i= N\right)= w. $$

(14)

Alternatively, the agent can choose an effort level such that e ₁ (p) ≤ e < e ₂ (p). In this effort range, the agent can close the sale if and only if the customer belongs to type 1. In particular, the agent’s expected payoff will be

$$ {U}_A\left({s}_P=\left( p, w, k\right),{s}_A= e\in \left[{e}_1,{e}_2\right),{s}_1= Y,{s}_2= N\right)= w+\pi kp- e, $$

where π is the probability that the customer belongs to type 1. The agent’s payoff is decreasing in e, so choosing e ₁ (p) will maximize her payoff. Therefore, the agent’s payoff will be

$$ {U}_A\left({s}_P=\left( p, w, k\right),{s}_A={e}_1(p),{s}_1= Y,{s}_2= N\right)= w+\pi kp-{e}_1(p). $$

(15)

Finally, the agent can choose an effort level such that e ₂ (p) ≤ e < ∞. If so, the agent can close the sale with the customer, regardless of her type. Hence, the payoff to the agent will be

$$ {U}_A\left({s}_P=\left( p, w, k\right),{s}_A= e\in \left[{e}_2(p),\infty \right),{s}_i= Y\right)= w+ kp- e. $$

The agent will choose e ₂ (p) to maximize her payoff. Specifically, her payoff will be

$$ {U}_A\left({s}_P=\left( p, w, k\right),{s}_A={e}_2,{s}_i= Y\right)= w+ kp-{e}_2(p). $$

(16)

Therefore, given s _P  = (p, w, k) and $\widehat{s}$ _i , the agent will choose among three different effort levels: e =0, e ₁ (p), or e ₂ (p). The corresponding payoffs to the agent are w, w + πkp –e ₁ (p), and w + kp - e ₂ (p), respectively. In sum, the agent will choose

• 1.
  $$ e={e}_2(p)\mathrm{if}\ \max\ \left( w+ kp-{e}_2(p), w+\pi kp-{e}_1(p), w\right)= w+ kp-{e}_2(p); $$
• 2.
  $$ \mathrm{e} = {\mathrm{e}}_1(p)\ \mathrm{if}\ \max\ \left( w+ kp-{e}_2(p), w+\pi kp-{e}_1(p), w\right)= w+\pi kp-{e}_1(p); $$
• 3.
  $$ \mathrm{e} = 0\ \mathrm{if}\ \max\ \left( w+ kp-{e}_2(p), w+\pi kp-{e}_1(p), w\right)= w. $$

Note that if w > 0 and k = 0, then the dominant strategy is e = 0. If the agent chooses e ₂ over e ₁ , that means (1-π) kp > e ₂(p) - e ₁ (p). That is,

$$ \pi <1-\frac{e_2(p)-{e}_1(p)}{kp}. $$

(17)

Appendix B.2 The Principal’s Optimal Strategy

Given the choice of the agent, the principal will determine the optimal P, w and k. Suppose the agent chooses e ₂ (p) (that is, π < 1 - $\frac{e2(p)- e1(p)}{\kappa p}$). The principal will maximize

$$ {U}_P\left({s}_A={e}_2(p),{s}_i={\widehat{s}}_i\right)=\left(1- k\right) p- w. $$

(18)

The principal ensures that the labor participation constraint holds; that is, w + kp -e ₂ (p) = W. Rearranging the labor participation constraint, κ = $\frac{\left( W- w\right)+ e2(p)}{p}$ . Substituting κ into Eq. (18), then U _P (s _A  = e ₂ (p), s _i = $\widehat{s}$ _i  = p - e ₂ (p)- W. Take the first order condition with respect to p to find the optimal price, one can verify that the optimal price p ^** is such that e’ ₂ (p ^**) = 1

Hence, if π < 1- $\frac{e2(p) - e1(p)}{\kappa p}$ and s _i = $\widehat{s}$ _i , then s _A (e) = e ₂(p), s _P  = (p ^**, w ^**, k ^**) such that p = p ^**, w ^** = [0, W), and k ^** = $\frac{\left( W- w**\right)+ e2\left( p**\right)\ }{p**}$, where e’ ₂ (p ^**) = 1. Using the same method, one can easily verify that if π > 1 - $\frac{e2(p) - e1(p)}{\kappa p}$ and s _i $\widehat{s}$= _i , then s _A  = e ₁ (p), s _P  = (p ^*, w ^*, k ^*) such that p = p ^*, w^* ∈ [0,W), and k ^* = $\frac{\left( W- w*\right)+ e1\left( p*\right)\ }{\pi\ p*}$ , where e’ ₁ (p ^*) = π.

Appendix B.3 Equilibrium

Recall that π ₀ = $\frac{\left( W- w\right)+ e1(p)}{\left( W- w\right)+ e2(p)}$. Plugging the principal’s optimal strategy into κ ^** (or κ ^*) into Eq. (17), one can easily verify that, if 0 < π ≤ π ₀ , then there exists an equilibrium where s _p  = (p ^**, k ^**, w ^**), s _A  = e ₂(p), and s _i = $\widehat{s}$ _i . However, if π ₀   ≤ π < 1, then there exists an equilibrium where s _p  = (p ^* , k ^* , w ^*), s _A  = e ₁ (p), and s _i = $\widehat{s}$ _i .

Appendix C Equilibrium: Stage Game with an Exit

This section shows equilibria of the stage game with an exit, given different signs of Eqs. (7), (8), and (11). Although these equilibria do not depict typical stylized facts, they are included here for completeness. The reader will be able to see the details of these equilibria at the author’s website.

Appendix D Example 2

This section proves that, in Example 2, the agent knows that $\frac{\left(100- w*\right)+ e1(p)}{\left(100- w*\right)+ e2(p)}$ < $\frac{3}{4}$. The agent knows that the principal will choose two possible prices: either p = p ^** such that eʹ ₂ (p ^** ) = 1, or p = p ^* such that eʹ ₁ (p ^* ) = π. That is, p ^** = 2,000 or p ^* = 2,250. If p = p** = 2,000, then $\frac{\left(100- w*\right)+ e1(2000)}{\left(100- w*\right)+ e2(2000)}$ ∈ [ $\frac{2}{3}$, 0.70]. However, if p = p = 2,250, then $\frac{\left(100- w*\right)+ e1(2250)}{\left(100- w*\right)+ e2(2250)}$ ∈ $\frac{2}{3}$[, 0.69]. In either case, π = $\frac{3}{4}$ is greater than $\frac{\left(100- w*\right)+ e1(2000)}{\left(100- w*\right)+ e2(2000)}$ .