An economic rationale for firing whistleblowers

Abstract

Organization members disclosing their superiors’ violations of duty are characterized by high ethical standards and, surprisingly often, fired for blowing the whistle. This paper provides an economic rationale for firing whistleblowers in a model where only the “ethical” type of agent can internally report the manager’s violation of duty. Revelation of an ethical type in the organization increases the perceived future probability of detecting and punishing the manager. Replacing the ethical type by an agent of unknown type restores this probability to its initial level, and this is optimal in organizations in which the standard of proof in establishing the manager’s violation of duty is low.

Appendix

1.1 Proof of Lemma 1

Consider part (ii) first. If the statement is wrong, one of the following three cases must arise when r = 0 is recommended:

• (a) The principal decides d = A for both σ = 0 and σ = 1 submissions, that is, always reverses the recommendation to r = 1. Then the ethical agent will set q(r = 0, σ) = σ and always blow the whistle (for both σ = 0 and σ = 1) which means that the hierarchy always implements a = 1. The principal can achieve this outcome without the manager, instructing the agent to implement a = 1, always.

• (b) The principal decides d = M for both σ = 0 and σ = 1 submissions. Then the manager’s optimal strategy is r(s) = 0 for all s and the hierarchy implements action a = 0 with probability one. The principal can achieve this outcome without the manager and instructing the agent to implement a = 0, always.

• (c) The principal decides d = M if σ = 1 is submitted, d = A if σ = 0 is submitted, given recommendation r = 0. Such a policy cannot be optimal because $prob(s=1 | r=0, \sigma = 0) < prob(s=1 | r=0, \sigma = 1).$

The only possibility left is that in equilibrium the principal decides d = M (d = A) only if σ = 0 (σ = 1) is submitted. Part (i) is straightforward: If there is an equilibrium in which all manager types make the correct recommendation, the ethical agent will set $q(r, {\sigma}) =\emptyset$ for all r and σ and blow no whistle because the principal will decide d = M. Then, however, the manager will deviate to recommending the wrong action, which upsets the proposed equilibrium. Nor can there be an equilibrium in which all manager types make the wrong recommendation because the principal can implement such an outcome by firing the manager. Part (iii) is now obvious, because the strategy q(r = 0, σ = 0) = 0 of submitting the signal σ = 0 when r = 0 will produce the decision d = M, the consequence of which is a fine f _A imposed on the whistleblower. Q.E.D.

1.2 Proof of Proposition 1

Several features of the optimal contract are transparent. First, the optimal contract must be binding the participation constraint of the type-0 manager. If this constraint is not binding the principal can increase his payoff by slightly decreasing the wage w _M , which still induces acceptance with probability one without affecting the strategies. Thus, from (5),

$$w^*_M = p^1_0 \lambda f^* - prob(a=0 | r_t (s)) \cdot \underline{t} z,$$

(13)

where

$$prob(a=0 | r_t (s)) \cdot \underline{t}z = [\lambda (p_0^0 +p_1^0 (1-\tau)) + (1- \lambda )(1- (p_1^0 +p_1^1 )\tau )]\cdot \underline{t}z. $$

Second, the optimal contract always binds the ethical agent’s participation constraint, thus, expected reward payments $\lambda (p^1_0 +(1-\tau)p^1_1)R_A$ plus the agent’s base wage w _A equals λ(1 − τ)p ⁰₁ L:

$$\lambda (p^1_0 +(1-\tau)p^1_1)R_A + w_A = \lambda (1- \tau)p^0_1 L.$$

(14)

The principal can control τ (see (6)) through the penalty f _M and, to keep the manager’s participation constraint satisfied, adjust the wage w _M according to (13). As τ changes, the interval in (3), to which the IDC probability β must belong, also changes. Since f _M affects τ directly, the optimal τ can be determined from the principal’s expected payoff stated solely in terms of τ through substitutions in (8). To this end, using (13) one obtains

$$\lambda (p^1_0 +(1-\tau)p^1_1 )f_M -\lambda p^1_0 f_M = \lambda (1-\tau)p^1_1 f_M .$$

Since f _M and τ are related through (6), using this condition and simplifying, the expression above becomes

$$\lambda (p^1_0 +(1-\tau)p^1_1 )f_M -\lambda p^1_0 f_M = \tau (1-\tau)(p_1 - \lambda p^1_1 )z.$$

(15)

The expression in (15) enters the principal’s objective in (8) with a positive sign, whereas the expression in (14), with a negative sign. It will be useful to arrange the coefficient of B in (8) as a sum of two terms, one that depends on τ plus one that does not:

$$ \tau [ (1-\lambda)(1 -p_0)+\lambda (p_1 - p_1^1)] + [(1- \lambda )p_0 +\lambda (p^1_1 +p^0_0)].$$

Using the fact that p ₁ + p ₀ = 1, the coefficient of τ above simplifies to p ₁ − λp ¹₁ . Now, combining these terms and substitutions from (14) and (15) into (8) yields

$$\begin{array} {l}U_P = \tau(p_1 - \lambda p^1_1)B + [(1- \lambda )p_0 +\lambda (p^1_1 +p^0_0)]B\\ + \tau (1-\tau)(p_1 -\lambda p^1_1 )z + prob(a=0 | r_t (s)) \cdot \underline{t} z - \lambda (1-\tau)p^0_1 L. \end{array}$$

(16)

This expression is concave and increasing in $\tau{\text:}\;\partial U/\partial \tau = (p_1 - \lambda p^1_1 )(B+(1- 2\tau)z - \underline tz) + \lambda p^0_1 L > 0$ for all τ ∈ [0, 1] because p ₁ >  λp ¹ ₁ and $B > (\underline t + 1) z.$

It follows that the principal prefers τ as large as possible, hence, that he should increase f _M as much as possible. This, however, will shift the interval in (3) to the left and eventually violate the constraint $\beta \geq \underline {\beta}.$ Recall that β must belong to the interval in (3) because, by part (ii) of Lemma 1, in any equilibrium in which the probability of an ethical type is positive the decision must be d = A if σ = 1, d = M if σ = 0, when r = 0. Given this, the principal should increase τ and decrease β accordingly until β is equal to both $\underline{\beta}$ and the upper bound of the interval in (3). The maximal (and optimal) τ that can be induced is³³

$$ \tau^* = 1 - \frac{\underline{\beta} p^1_0} {(1- \underline {\beta})p^1_1},$$

which is positive if $\underline {\beta} < p^1_1/(p^1_1 + p^1_0),$ as assumed in (4). The optimal value of β is then $\beta^* = \underline {\beta}.$

Consider now the optimal incentive schemes. The case where the principal screens out standard agents is straightforward: set w _A <  0 and pick a corresponding reward R _A > 0 that satisfies (7) with equality. Assume that the principal attracts both types of agents. It is easy to see that the agent’s optimal base wage must be w ^* _A = 0. Though the principal can satisfy the participation constraints of both agent types with a wage offer w _A  > 0, a reduction in w _A coupled with an increase in R _A that keeps both participation constraints satisfied will increase the principal’s expected payoff. A strictly positive wage gives a positive rent to the standard agent and thus cannot be optimal. The optimal reward R ^* _A is found through the participation constraint of the ethical agent, holding with equality for w _A = 0. Q.E.D.

1.3 Proof of Proposition 2

To show existence of a critical $\underline {\beta}^C > 1/2$ such that the inequality in (10) is reversed for $\underline {\beta} < \underline {\beta}^C,$ note first that B − z(p ₁ τ^*/p ¹₁ ) is positive for all possible values of τ^* because by assumption, B > z(p ₁/p ¹₁ ). Consider now the sign of (1 − τ^*)p ¹₁  − p ¹₀ . Substituting for τ^* yields

$$(1-\tau^*) p^1_1 - p^1_0 = p^1_0 \left[\frac{\underline {\beta}} {(1- \underline {\beta})}- 1\right],$$

which is positive if and only if $\underline {\beta} > 1/2.$ That is, the first term in (10) is positive if and only if $\underline {\beta} > 1/2.$

Note that (1 − τ^*)p ⁰₁ L and $(1-p^0 -p_1^1 \tau^* )\underline{t}z$ are decreasing in τ^*, hence, increasing in $\underline{\beta}.$ In particular, for $\underline{\beta} = 1/2,$

$$(1- \tau^*) p^0_1 L = \frac{p^1_0 p^0_1}{p^1_1}L,\quad {and} \quad (1-p^0 -p_1^1 \tau^* )\underline{t}z = (1-p^0 - (p_1^1 -p_0^1 )) \underline{t}z,$$

which implies that the expression in (10) is negative for $\underline {\beta} =1/2,$ and by continuity, at the right neighborhood of $\underline {\beta} = 1/2$ as well. As $\underline {\beta}$ is further increased above 1/2, the first term in (10) is positive and increasing, while the second and third terms are negative and decreasing. Then, either a critical level of $\underline {\beta}$ exists such that the expression in (10) becomes zero or the upper admissible bound for $\underline {\beta},$ namely, p ¹ ₁/(p ¹ ₁ + p ¹ ₀) < 1, is reached. The value of $\underline {\beta}^C$ is defined as the maximum of the two values of $\underline {\beta}.$ Q.E.D.