An information-based model for the differential treatment of gains and losses

Abstract

This study defines reporting conservatism as a higher verification standard for probable gains compared to losses and builds a model that endogenously generates optimal behavior resembling an asymmetric preference for gains versus losses. Our model considers the setting where one party produces a resource and another tries to expropriate it. The key factor determining the extent of the gain-loss asymmetry is the level of information asymmetry or trust between the two parties. The information asymmetry-based results of our model provide a simpler explanation for the vast empirical literature on conservatism, where the bulk of the economic relationships among the parties appear to be information-based with little direct relation to explicit debt contracts, a factor that has been the focus of theoretical arguments. We also suggest new empirical analyzes.

Appendix

Proof

Proposition 1

Let a _s (σ _s ) be the stealer’s strategy as a function of his signalσ _s ∈{T, W}for T = Tough and W = Weak. Let a _p (𝜃) be the producer’s strategy as a function of her type 𝜃 ∈{T, W}. Let b be the vector of beliefs at each information set in the extended game. We seek to find the perfect Bayesian equilibrium$(a_{s}^{*}(\sigma _{s}),a_{p}^{*}(\theta ))$.

To solve, work backwards. Figure 2 presents the extensive form of the game. Observe that the producer has a nontrivial choice only when the stealer chooses to fight. For a producer of type 𝜃= T, she earns y − μ _{p T} by fighting and 0 by leaving, regardless of the stealer’s signal. Sincey > μ _{p T} by assumption, ap∗(T) = F. Similarly, a Weak producer makes a choice only when the stealer chooses to fight. This producer earns 0 by leaving and− μ _{p W} by fighting, regardless of the stealer’s signal. Thus ap∗(W) = L. This proves parts 3 and 4.

To solve for the stealer’s strategy, it is necessary to compute the stealer’s beliefs, according to Bayes’s rule. Let b _{i j} denote the stealer’s belief that 𝜃 = i when the stealer’s signal σ _s = j. Calculated by Bayes’s rule, this yields the following.

$$b_{TT}\ = \frac{\phi q_{s}}{\phi q_{s} + (1 - \phi)(1 - q_{s})} $$

$$b_{TW}\ = \frac{\phi(1 - q_{s})}{\phi(1-q_{s}) + (1 - \phi)q_{s}} $$

$$b_{WT}\ = \frac{(1 - \phi)(1 - q_{s})}{\phi q_{s} + (1 - \phi) (1 - q_{s})} $$

$$b_{WW}\ = \frac{(1 - \phi)q_{s}}{\phi(1 - q_{s}) + (1 - \phi)q_{s}}. $$

Suppose σ _s = W. Given $a_{p}^{*}(\theta )$, if the stealer fights, he earns: Δ _W = b _{W W} y − b _{T W} μ _{s T} . He earns nothing if he leaves. Thus $a_{s}^{*}(W)=F$ if and only if Δ _W > 0, and $a_{s}^{*}(W)=L$ otherwise.

Now suppose σ _s = T. Given $a_{p}^{*}(\theta )$, the stealer earns 0 if he leaves and if he fights he earns, in expectation,${\Delta }_{T}=b_{WT}y-b_{TT}\mu _{s_{T}}$. Thus $a_{s}^{*}(T)=F$ if and only if Δ _T > 0, and $a_{s}^{*}(T)=L$ otherwise. This proves parts 1 and 2.

To see uniqueness, first note that the producer’s equilibrium strategy in the actionspace of Fig. 2 is strictly dominant. Given the producer’s strategy, the conditions for Δ _W > 0 and Δ _T > 0 are necessary and sufficient to determine the stealer’s best response. The equilibrium is thus unique.□

Proof

Proposition 2 Note from Eq. 1 that f(y; e) is linear in e; its second derivative with respect to e is therefore zero. Because the integrand and its e derivatives are piecewise integrably bounded, Leibniz’s rule applies to this integral.

$$ \frac{\partial^{2}}{\partial e^{2}} \left[ {\int}_{\underline{y}}^{\bar{y}} B(y) f(y;e)\,dy - C(e) \right] \,=\, {\int}_{\underline{y}}^{\bar{y}} B(y) \frac{\partial^{2}}{\partial e^{2}}f(y;e)\,dy \,-\, C^{\prime\prime}(e) \,=\, - C^{\prime\prime}(e) \!<\! 0. $$

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The optimization problem of the proposition with respect to e is therefore strictly concave and yields a unique solution for the optimal e. This unique solution is less than $\bar {e}$ because $C^{\prime }(\bar {e}) = \infty$. The presence of discontinuous drops in B(y) implies that we cannot assert that the optimal e is always greater than$\underline {e}$. However, the possibility that the optimal e can equal$\underline {e}$ has no impact on our analysis. Moreover, there are a wide variety of settings in which the optimal$e > \underline {e}$. To see this, first note that $\underline {y}, \bar {y}$ are assumed to be greater than μ _{p T} , whereas y _W , y _T do not depend on μ _{p T} but on the other free exogenous parameter μ _{s T} . As a result, any of the following three possibilities, among others, are settings where we can show that the optimal $e > \underline {e}$: 1)$\underline {y} < \bar {y} < y_{W}$ (the stealer never attacks); 2) $y_{W} < \underline {y} < \bar {y} < y_{T}$ (the stealer attacks only if he receives the Weak signal); and 3)$\underline {y} > y_{T}$ (the stealer always attacks). In each case, B(y) is a linear function with positive slope. Calculation shows that$\left .\frac {\partial }{\partial e}{\int }_{\underline {y}}^{\bar {y}} B(y) f(y;e)\,dy\right |_{e=\underline {e}}>0$, and $C^{\prime }(\underline {e}) = 0$. So, $e=\underline {e}$ cannot be the solution in any of these cases. □

Proof

Proposition 3

Suppose the producer sees the stealer and learns her own type (Tough or Weak). The expected payoff to the Tough producer after meeting the stealer, but before the fight, is as follows.

$$ B_{T}(y) = \left\{ \begin{array}{lr} y & : y \leq y_{W} \\ y - (1 - q_{s})\mu_{pT} & : y_{W} < y \leq y_{T} \\ y - \mu_{pT} & : y > y_{T}. \end{array} \right. $$

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The expected payoff to the Weak producer after meeting the stealer but before the fight is as follows.

$$ B_{W}(y) = \left\{ \begin{array}{lr} y & : y \leq y_{W} \\ (1 - q_{s})y & : y_{W} < y \leq y_{T} \\ 0 & : y > y_{T}. \end{array} \right. $$

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□

Note that because $q_{s} > \frac {1}{2}$ and y > μ _{p T} ,∀y : B _T (y) ≥ B _W (y).

Suppose the producer can offer to share a part of the output in the hope that the stealer will leave. The stealer receives the transfer, observes the size of the rest of the output, and then gets his signal about the producer’s type. He then decides whether to continue the stealing game for the rest of the output. We solve for an equilibrium in this game.

Separating equilibrium

First consider separating equilibria. Let (t _W , t _T ) be the equilibrium transfers offered in a separating equilibrium by the Weak and the Tough producer respectively. Let P(t) be the stealer’s probability assessment that the producer is the Tough type. By consistency of beliefs, P(t _W ) = 0 and P(t _T ) = 1. Thus, by submitting t _W , the Weak producer gets 0 (the stealer will attack for the rest of the output for sure), whereas, by deviating to t _T and pooling with the Tough producer, she will be weakly better off than zero. We will show that she earns B _W (⋅), which from Eq. 9 is weakly greater than zero). This deviation is profitable, and therefore no separating equilibria exist, i.e., t _W = t _T .

We next determine the optimal t _W = t _T and show that it is unique in y. We first specify the off-equilibrium beliefs of the stealer.

Pooling equilibrium’s off-equilibrium beliefs

Given that the equilibrium t _W = t _T is pooling, we assume that P(t) = ϕ for any feasible t. That is, the transfer itself does not change the priors of the stealer. We will prove later that our equilibrium based on this belief structure satisfies the intuitive criterion.

We consider the Tough producer’s case first. This producer offers t _T such that the following holds.

$$ t_{T} = \operatorname{arg\,max}_{0 \leq t \leq y} B_{T}(y - t). $$

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This recursive formulation occurs because, by assumption, an offering of t does not change the stealer’s priors from ϕ. Therefore, for the remaining output y − t, for all feasible t, the stealer acts sequentially rationally, and the Tough producer expects to get, by definition of B _T (.) above, B _T (y − t).

If B _T (y) were monotonically increasing, t _T = 0. However, there are regions where it is not. Given the structure of the exogenous variables, there are two cases to consider: B _T (y _W ) ≤ B _T (y _T ) and B _T (y _W ) > B _T (y _T ). We solve for each case separately.

Case I:

B _T (y _W ) ≤ B _T (y _T )

This condition implies the following.

$$\begin{array}{@{}rcl@{}} y_W & \leq& y_T - (1-q_s) \mu_{pT}\\ \text{ or } \mu_{pT} & \leq& \frac{y_T - y_W}{1 - q_s} = \frac{\phi(2q_s-1)}{(1-\phi)(1-q_s)^2q_s}. \end{array} $$

Figure 4 plots an example B _T (y). Note in this figure that $\mu _{pT} = 4 < \frac {18-8}{1-0.6} = 25$.

Fig. 4

The benefit function B _T (y) for m = 0.75,ϕ = 0.5,q _s = 0.6,μ _{p T} = 4, and μ _{s T} = 12,y ∈ [5,25]. The horizontal lines are the regions where the Tough producer shares to increase her payoff

The smaller local maximum of B _T (y) is attained at B _T (y _W ), and the larger local maximum is attained at B _T (y _T ). We also note that:

$$B_{T}(y_{W}) = y_{W} > B_{T}(y), y \in (y_{W}, y_{W} + (1- q_{s}) \mu_{pT}). $$

So, for y _W < y ≤ y _W + (1 − q _s )μ _{p T} , the Tough producer will make a transfer t _T = y − y _W and get the payoff y _W . The Weak producer will mimic that transfer and also earn y _W , because the stealer will not attack when remaining output is less than or equal to y _W . Further note that, in this case, y _W + (1 − q _s )μ _{p T} ∈ (y _W , y _T ], the middle continuous range of B _T (y).

We also see from Eq. 8 the following.

$$B_{T}(y_{T}) = y_{T} - (1 - q_{s}) \mu_{pT} > B_{T}(y), y \in (y_{T}, y_{T} + q_{s} \mu_{pT}). $$

So, for y _T < y ≤ y _T + q _s μ _{p T} , the Tough producer will transfer t _T = y − y _T and get the payoff B _T (y _T ). The Weak producer will mimic that transfer and earn (1 − q _s )y _T . Note that, because B _T (y _T ) ≥ B _T (y _W ) by assumption, the Tough producer will not transfer the larger amount y − y _W .

The complete specification of the equilibrium for Case I is as follows.

• 1. Set the default t _T = t _W = 0. Then, apply the following changes in order.

• 2. If y ≤ y _W , t _T = t _W = 0.

• 3. If y _W < y ≤ y _W + (1 − q _s )μ _{p T} and y _W is feasible (i.e., $y_{W} > \underline {y}$), then t _T = t _W = y − y _W . If y _W is not feasible, then t _T = t _W = 0.

• 4. If y _T < y ≤ y _T + q _s μ _{p T} and y _T is feasible (i.e., $y_{T} > \underline {y}$), then t _T = t _W = y − y _T . If y _T is not feasible, then t _T = t _W = 0.

• 5. The Tough producer earns B _T (y − t _T ), and the Weak producer earns B _W (y − t _W ).

Case II

B _T (y _W ) > B _T (y _T )

This condition implies the following.

$$\begin{array}{@{}rcl@{}} y_W & >& y_T - (1-q_s) \mu_{pT}\\ \textit{ or } \mu_{pT} & >& \frac{y_T - y_W}{1 - q_s} = \frac{\phi(2q_s-1)}{(1-\phi)(1-q_s)^2q_s} > 0. \end{array} $$

Figure 5 plots an example B _T (y) for the Case II scenario. The only change in the parameters from Fig. 4 is the reduction in the accuracy of the stealer’s signal q _s from 0.6 to 0.505. The smaller local maximum of B _T (y) is attained at B _T (y _T ), and the larger local maximum is attained at B _T (y _W ). If the Tough producer’s current payoff is less than these maxima, the Tough producer will move to the relevant maximum.

Fig. 5

The benefit function B _T (y) for m = 0.75,ϕ = 0.5,q _s = 0.505,μ _{p T} = 4, and μ _{s T} = 12,y ∈ [5,25]. The horizontal line is the region where the Tough producer shares to increase her payoff

The complete specification of the equilibrium for Case II is as follows.

• 1. Set the default t _T = t _W = 0. Then apply the following changes in order.

• 2. If y ≤ y _W , t _T = t _W = 0.

• 3. If y _W < y ≤ y _T and y _W is feasible (i.e., $y_{W} > \underline {y}$), then t _T = t _W = y − y _W . If y _W is not feasible, then t _T = t _W = 0.

• 4. If y _T < y ≤ y _W + μ _{p T} and y _W is feasible (i.e., $y_{W} > \underline {y}$), then t _T = t _W = y − y _W . (Note that y _W + μ _{p T} > y _T − (1 − q _s )μ _{p T} + μ _{p T} > y _T .)

• 5. If y _W is not feasible but y _T is, then, as in Case I, if y _T < y ≤ y _T + q _s μ _{p T} , then t _T = t _W = y − y _T , and if y > y _T + q _s μ _{p T} , then t _T = t _W = 0.

• 6. If y _T is not feasible, then t _T = t _W = 0.

• 7. The Tough producer earns B _T (y − t _T ), and the Weak producer earns B _W (y − t _W ).

Note that, in Case II, the equilibrium t _T = t _W = 0 when y > y _W + μ _{p T} : the Tough producer gains more at y than at y _W .

Finally, we observe several aspects of the equilibrium common to both cases.

• 1. The equilibrium t _T = t _W is unique in y in both Case I and Case II.

• 2. The equilibrium t _T = t _W is 0 if the range $[\underline {y}, \bar {y}]$ is such that B _T (y) is continuous in the entire range.

• 3. As $q_{s} \rightarrow 1, y_{W} \equiv \frac {(1-q_{s})\phi }{q_{s}(1-\phi )} \mu _{sT}\rightarrow 0$ monotonically, and $y_{T} \equiv \frac {q_{s}\phi }{(1-q_{s})(1-\phi )} \mu _{sT} \rightarrow +\infty$ monotonically. The limiting values of y _W , y _T also imply that, for any given admissible values of $\underline {y}, \bar {y}$, there exists a threshold $\bar {q_{s}}, \frac {1}{2} < \bar {q_{s}} < 1$, above which $y_{W} < \underline {y} < \bar {y} < y_{T}$, thus resulting in the equilibrium t _T = t _W = 0.

• 4. In both Case I and Case II, the regions where t _T = t _W > 0 are controlled by the size of μ _{p T} . As μ _{p T} shrinks, the chances of non-zero t _T = t _W shrinks. (See the horizontal regions and their boundaries in Figs. 4 and 5.)

• 5. With probability 1, a random choice of exogenous parameters will admit a range of y where t _T = t _W = 0. To see this, the probability that $\underline {y} = y_{W}$ is 0. If $\underline {y} < y_{W}$, then there exists an 𝜖 > 0 such that the optimal t _T for y ∈ (y _W − 𝜖, y _W ) is 0. A similar $(\underline {y}, \underline {y} + \epsilon )$ range can be found if $\underline {y} > y_{W}$.

Intuitive criterion

We argue that our equilibrium satisfies the intuitive criterion. Recall that a (sequential) equilibrium fails to satisfy the intuitive criterion if, for some non-equilibrium transfer t, the following condition is met: for all possible beliefs of the stealer upon receiving the non-equilibrium t, the Weak (Tough) producer is worse off than in equilibrium, and therefore the stealer infers upon receiving t that the producer is Tough (Weak). This inference is reinforcing in that, because of this inference, the Tough (Weak) producer is better off than in equilibrium, and so the Tough (Weak) producer will deviate to the non-equilibrium t (Kreps 1990, p.436, p.818).

In our model, consider a t that makes the Tough producer worse off (compared to the equilibrium), even when the stealer does not attack. The Tough producer will not issue this t because it is strictly equilibrium dominated. A Weak producer will not issue this t either, for the stealer will put probability one that she is Weak and attack for sure. This gives her 0, and we have shown that the she is weakly better off than 0 in equilibrium.

On the other hand, suppose there is an off-equilibrium t that makes the Weak producer worse off (compared to equilibrium), even in the best-case possibility that the stealer receiving this t does not attack. That is, y − t is less than what she earns in equilibrium. This implies that the stealer, upon receiving t, infers that the producer is definitely Tough. We have shown using Eqs. 8 and 9 that ∀y : B _T (y) ≥ B _W (y), so the Tough producer earns weakly more than the Weak producer in the pooling equilibrium. Therefore the best the Tough producer can do by deviating to t is y − t (that is, when the stealer does not attack), which is less than what the Weak producer earns in equilibrium, which is less than what the Tough producer earns in equilibrium. So the Tough producer will not deviate to the off-equilibrium t.

A final point is to show that our equilibrium is sequential. Consider the sequence 0 < {𝜖 _n } < 1, where both the Tough and Weak producers issue the equilibrium t^∗ with probability 1 − 𝜖 _n , and issue all other t ∈ [0,y] −{t^∗} using a uniform distribution of total probability measure 𝜖 _n . This is a strictly mixed sequence of actions with strictly positive Bayes-consistent beliefs, i.e., the stealer’s posterior about the Tough type after receiving any t is still ϕ, which is strictly positive because 0 < ϕ < 1. This sequence’s limit 𝜖 _n → 0 yields our equilibrium actions and beliefs, and this equilibrium, by construction, is sequentially rational. Our equilibrium satisfies the intuitive criterion.

Proof

Proposition 4 Suppose the producer is Tough, which occurs with probabilityϕ.In this state of the world, work backward from the last node in the gametree, which is the producer’s action. If the stealer fights, the producer earnsy − μ _{p T} > 0 by fighting and nothing by leaving, so the producer prefers to fight. The stealer earns− μ _{s T} . The stealer therefore prefers to leave. Thus the producer earns all the surplus y and the stealer earns nothing.

Suppose the producer is Weak. If the stealer fights, a Weak producer will lose against the stealer and earn− μ _{p W} by fighting and 0 by leaving. Therefore the producer will leave. Knowing this, the stealer will fight, earning the full surplus y, while the producer gets nothing. This establishes the Nash equilibrium of the extensive form game. The same logic shows it is unique and sub-game perfect.□

Proof

Proposition 5 Consider the certification game.In this game, immediately after Nature reveals 𝜃 and before any other event in Fig. 2, a third party can reveal 𝜃 to the stealer for a certification fee of k < μ _{p T} , which the producer pays.

The stealer will not attack a known Tough producer (the stealer will lose) but will attack a known Weak producer (the producer will leave). Suppose,in addition, the stealer’s belief is that a noncertifying producer is Weak with probability 1. With this belief, the stealer will always attack a noncertifying producer. Consequently, upon not certifying, the Tough producer earns y − μ _{p T} , which is less than y − k, the payoff from certifying. So the Tough producer will always certify. The Weak producerearns 0 from not certifying (the stealer will attack based on his prior beliefs), and −k upon certifying (the producer pays k and then gets attacked upon being certified as Weak). The Weak producer will therefore never certify. The stealer’s beliefs are thus consistent with the producers’ actions, and everyone’s actionis the best response given the beliefs. There are no fights in this equilibrium.□