Two-sided unobservable investment, bargaining, and efficiency

Abstract

Asymmetric information can lead to inefficient outcomes in many bargaining contexts. It is sometimes natural to think of asymmetric information as emerging from imperfect observation of previously taken actions (e.g., obtaining compliments or substitutes for the item being bargained over). How do such strategic investment choices prior to bargaining interact with the strategic problem of bargaining under private information? We focus on bilateral bargaining when players can make unobserved investments in the value of the item prior to their interaction. With two-sided hidden investment, strategic uncertainty induces a post-investment problem analogous to that in Myerson and Satterthwaite (J Econ Theory 29(2):265–281, 1983), and inefficiencies might be expected to arise. But, there are strong incentives to avoid investment levels that do not lead to trade and this must be anticipated by the other trader. This effect is shown to drive a form of unraveling; as a result in every equilibrium to the larger game the good ends up in the hands of the agent with the higher valuation.

Appendix

1.1 Myerson–Sattertwaite second-best mechanism with $U_s(b_s)=U_b(a_b)=0$ maximizes gains from trade

First, define constraint on the optimal mechanism $G(\alpha )$ in the same way as Myerson and Satterthwaite as:

$$\begin{aligned} G(\alpha )=&\int _{a_b}^{b_b}\int _{a_s}^{b_s}(c_b(v_b,1)-c_s(v_s,1))p^\alpha (v_s,v_b)f_s(v_s)f_b(v_b)dv_sdv_b \nonumber \\ =&U_b(a_b)+U_s(b_s), \end{aligned}$$

(17)

where

$$\begin{aligned} c_s(v_s, \alpha )=v_s+\alpha \frac{F_s(v_s)}{f_s(v_s)} \quad c_b(v_b, \alpha )=v_b-\alpha \frac{1-F_b(v_b)}{f_b(v_b)}. \end{aligned}$$

Furthermore, define $p^\alpha (v_s,v_b)$ is 1 if $c_b(v_b,\alpha )\ge c_s(v_s,\alpha )$ and zero otherwise, M-S show that $G(\alpha )$ is increasing in $\alpha$, and continuous, with $G(0)<0$ and $G(1)\ge 0$, thus ensuring there is a positive $\alpha ^*$ that satisfied the constraint from their Theorem 1 with equality, i.e., $G(\alpha ^*)=0$.

The second-best mechanism maximizes the expected gain from trade, given by:

$$\begin{aligned} \int _{a_b}^{b_b}\int _{a_s}^{b_s}(v_b-v_s)p(v_s,v_b)f_s(v_s)f_b(v_b)dv_sdv_b. \end{aligned}$$

(18)

M-S show that $\alpha ^*$ is a second-best mechanism. We can show that it is the only one by noting any function $p'(v_s, v_b)$ that differs from $p^{\alpha ^*}(v_s, v_b)$ can only do so by being less than 1 outside the “wedge” of types that donot efficiently trade as determined by $\alpha ^*$, where $p^{\alpha ^*}=1$. This is because within the wedge, we have $p^{\alpha ^*}=0$ and $c_b(v_b,\alpha ^*)-c_s(v_s,\alpha ^*)<0$. Thus, having non-zero probability of trade inside the wedge would violate the constraint in Theorem 1 of M-S. But outside the wedge, $v_b>v_s$, hence, any such function $p'(\cdot , \cdot )$ that doesn’t prescribe trade with probability one where $p^{\alpha ^*}$ does will lead to a strictly lower expected gains from trade than $p^{\alpha ^*}$.

Proof of Lemma 1

Consider an equilibrium involving the direct mechanism (x, p). To begin, consider the case of the seller. Take any two investments $v_s, v_s'$ in the support of $F_s$. Then, because the seller is mixing over these values

$$\begin{aligned}&\int _{V_b} (1-p(v_s, v_b))v_s+x(v_s,v_b) dF_b(v_b)-c_s(v_s) \\&\quad =\int _{V_b} (1-p(v_s', v_b))v_s'+x(v_s',v_b)dF_b(v_b)-c_s(v_s'). \end{aligned}$$

The left-hand side equals

$$\begin{aligned} U_s(v_s)-c_s(v_s) + v_s, \end{aligned}$$

and the right-hand side equals

$$\begin{aligned} U_s(v_s')-c_s(v_s')+v_s', \end{aligned}$$

and we can rewrite the equation above as

$$\begin{aligned}&U_s(v_s)-c_s(v_s)+v_s=U_s(v_s')-c_s(v_s')+v_s' \end{aligned}$$

(19)

$$\begin{aligned}&1+\frac{U_s(v_s)-U_s(v_s')}{v_s-v_s'}=\frac{c_s(v_s)-c_s(v_s')}{v_s-v_s'}. \end{aligned}$$

(20)

At an accumulation point of the support of $F_s$, we can take the limits as $v_i' \rightarrow v_i$ and

$$\begin{aligned} 1+U_s'(v_s)=c_s'(v_s). \end{aligned}$$

(21)

This is the first equation in the theorem. Similar calculations give the identity for the seller. $\square$

Proof of Lemma 3

The proof follows the familiar argument of Myerson and Satterthwaite. Incentive compatibility means that for all $v_s$, $v_s'$ in the support of the seller’s distribution:

$$\begin{aligned} U_s(v_s,v_s)&\ge U_s(v_s',v_s) \end{aligned}$$

(22)

$$\begin{aligned} U_s(v_s',v_s')&\ge U_s(v_s,v_s'). \end{aligned}$$

(23)

By subtracting the RHS of the second inequality from the LHS of the first and the RHS of the first from the second and canceling the payment terms, we get:

$$\begin{aligned} -\overline{p}_s(v_s)[v_s-v_s'] \ge U_s(v_s,v_s)-U_s(v_s',v_s') \ge -\overline{p}_s(v_s')[v_s-v_s']. \end{aligned}$$

For either $v_s$ or $v_s'$ in $\Theta _s$, we can stop here. For $v_s\in \mathcal {K}_s$ and $v_s\notin \underline{V}_s$, we assume $v_s>v_s'$, divide by $v_s-v_s'$ and take the limit as $v_s'\rightarrow v_s$ to obtain:

$$\begin{aligned} U_s'(v_s)=-\overline{p}_s(v_s). \end{aligned}$$

(24)

For $v_s\in \underline{V}_s$ we simply take the limit from the right, with $v_s'>v_s$ to obtain the same result. Integrating Eq. (24) within an interval $\mathcal {I}_s^j$, we obtain (10) The same method applies to the buyers. $\square$

Proof of Lemma 4

The proof proceeds analogously to the canonical case (Theorem 1 of Myerson and Satterthwaite), except that we need to make use of Lebesgue–Stieltjes integrals to account for the fact that we integrate over distributions that have gaps and atoms. First, observe that Lemma 3 implies that $U_s(b_s)\le U_s(v_s)$ for all $v_s$ in the seller’s support, and $U_b(a_b)\le U_b(v_b)$ for all $v_b$ in the buyer’s support. Next, consider the expected gains from trade under a direct mechanism (x, p).

$$\begin{aligned}&\int _{a_s}^{b_s}\int _{a_b}^{b_b}(v_b-v_s)p(v_s,v_b)dF_b dF_s=\int _{a_s}^{b_s}\int _{a_b}^{b_b}v_bp(v_s,v_b)dF_b dF_s\nonumber \\&\qquad -\int _{a_s}^{b_s}\int _{a_b}^{b_b}-v_sp(v_s,v_b)dF_b dF_s\nonumber \\&\quad =\int _{a_b}^{b_b}v_b\overline{p}(v_b)dF_b-\int _{a_s}^{b_s}v_s\overline{p}(v_s)dF_s, \end{aligned}$$

(25)

where the last line follows from integrating the two integrals in different orders, permissible by Tolleni’s theorem.

At the same time, since the payments are zero sum, the expected gains from trade is equal to the sum of the average gains of the buyers and sellers:

$$\begin{aligned} \int _{a_s}^{b_s}\int _{a_b}^{b_b}(v_b-v_s)p(v_s,v_b)dF_b dF_s=\int _{a_s}^{b_s}U_s(v_s)dF_s+\int _{a_b}^{b_b}U_b(v_b)dF_b \end{aligned}$$

(26)

Take the seller’s term, the first integral. Using the envelope theorem (Lemma 3) and using the definition of the function $\overline{\pi }_s(v_s)$ above, we can write

$$\begin{aligned} U_s(v_s)\ge U_s(b_s)+\int _{v_s}^{b_s}\overline{\pi }_s(t_s)dt_s \end{aligned}$$

(27)

So, we have:

$$\begin{aligned} \int _{a_s}^{b_s}U_s(v_s)dF_s&\ge \int _{a_s}^{b_s}\left[ U_s(b_s)+\int _{v_s}^{b_s}\overline{\pi }_s(t_s)dt_s\right] dF_s\\&= U_s(b_s)+\int _{a_s}^{b_s}\int _{v_s}^{b_s}\overline{\pi }_s(t_s)dt_sdF_s= U_s(b_s)+\int _{a_s}^{b_s}F_s(t_s)\overline{\pi }_s(t_s)dt_s, \end{aligned}$$

where the change in the order of integration again is permissible by Tolleni’s theorem. Similarly for the buyer, we have:

$$\begin{aligned} \int _{a_b}^{b_b}U_b(v_b)dF_b\ge&\int _{a_b}^{b_b}\left[ U_b(a_b)+\int _{a_b}^{t_b}\overline{\pi }_b(t_b)dt_b\right] dF_b\\ =&U_b(a_b)+\int _{a_b}^{b_b}(1-F_b(t_b))\overline{\pi }_b(t_b)dt_b. \end{aligned}$$

Putting these together, we have:

$$\begin{aligned}&\int _{a_b}^{b_b}v_b\overline{p}(v_b)dF_b-\int _{a_s}^{b_s}v_s\overline{p}(v_s)dF_s\ge U_s(b_s)+U_b(a_b)\nonumber \\&\quad +\int _{a_s}^{b_s}F_s(t_s)\overline{\pi }_s(t_s)dt_s + \int _{a_b}^{b_b}(1-F_b(t_b))\overline{\pi }_b(t_b)dt_b, \end{aligned}$$

(28)

or,

$$\begin{aligned}&\int _{a_b}^{b_b}v_b\overline{p}(v_b)dF_b-\int _{a_s}^{b_s}v_s\overline{p}(v_s)dF_s\nonumber \\&\qquad -\int _{a_s}^{b_s}F_s(t_s)\overline{\pi }_s(t_s)dt_s - \int _{a_b}^{b_b}(1-F_b(t_b))\overline{\pi }_b(t_b)dt_b \nonumber \\&\quad \ge U_s(b_s)+U_b(a_b)\ge 0. \end{aligned}$$

(29)

This proves the “only if” part of Lemma 4. To prove the “if” part, we need to show that for a function $p(\cdot , \cdot )$ satisfying (16), and when $\overline{p}_s(\cdot )$ and $\overline{p}_b(\cdot )$ are weakly decreasing and increasing, respectively, a payment function exists that makes the mechanism satisfy IC and IP. First, we observe that for $\overline{p}_s(\cdot )$ and $\overline{p}_b(\cdot )$ are weakly decreasing and increasing, respectively, $\overline{\pi }_s(\cdot )$ and $\overline{\pi }_b(\cdot )$, defined in (14) and (15) are also weakly decreasing and increasing, respectively.

Next, consider the following payment function:

$$\begin{aligned} x(v_s,v_b)=\chi _b(v_b)-\chi _s(v_s)+K, \end{aligned}$$

(30)

where $\chi _s(\cdot )$ and $\chi _b(\cdot )$ are given by the Lebesgue–Stieltjes integrals:

$$\begin{aligned} \chi _b(v_b)&=\int _{t_b=a_b}^{v_b}t_bd[\overline{\pi }_b(t_b)] \end{aligned}$$

(31)

$$\begin{aligned} \chi _s(v_s)&=\int _{t_s=a_s}^{v_s}t_sd[-\overline{\pi }_s(t_s)] \end{aligned}$$

(32)

and K is a constant. To see that this payment function satisfies incentive compatibility, consider for any pair $v_s$, $v_s'$ in the seller’s support:

$$\begin{aligned} U_s(v_s,v_s)-U_s(v_s',v_s)=-v_s(\overline{p}_s(v_s)-\overline{p}_s(v_s'))-\chi _s(v_s)+\chi _s(v_s') \end{aligned}$$

Since $\overline{\pi }_s(v_s)=\overline{p}_s(v_s)$ whenever $v_s$ is in the support of the seller, we have $\overline{p}_s(v_s)-\overline{p}_s(v_s')=-\int _{t_s=v_s'}^{v_s}d[-\overline{\pi }_s(t_s)]$, and $-\chi _s(v_s)+\chi _s(v_s')=-\int _{t_s=v_s'}^{v_s}t_sd[-\overline{\pi }_s(t_s)]$ thus we have:

$$\begin{aligned} U_s(v_s,v_s)-U_s(v_s',v_s)=&v_s\int _{t_s=v_s'}^{v_s}d[-\overline{\pi }_s(t_s)]-\int _{t_s=v_s'}^{v_s}t_sd[-\overline{\pi }_s(t_s)]\nonumber \\ =&\int _{t_s=v_s'}^{v_s}(v_s-t_s)d[-\overline{\pi }_s(t_s)]\ge 0, \end{aligned}$$

(33)

since $\overline{\pi }_s(\cdot )$ is a weakly decreasing function. The proof for the buyer proceeds analogously.

Now, consider the difference $U_s(v_s')-U_s(v_s)$ for some $v_s'\le v_s$ in the seller’s support:

$$\begin{aligned} U_s(v_s')-U_s(v_s)=&-v_s' \overline{p}_s(v_s')+v_s\overline{p}_s(v_s)-\chi _s(v_s')+\chi _s(v_s)\nonumber \\ =&-v_s' \overline{p}_s(v_s')+v_s\overline{p}_s(v_s)+\int _{t_s=v_s'}^{v_s}t_sd[-\overline{\pi }_s(t_s)]\nonumber \\ =&-v_s' \overline{p}_s(v_s')+v_s\overline{p}_s(v_s)+\int _{t_s=v_s'}^{v_s}\overline{\pi }_s(t_s)dt_s-\bigg [t_s\overline{\pi }_s(t_s)\bigg ]_{t_s=v_s'}^{v_s}\nonumber \\ =&\int _{t_s=v_s'}^{v_s}\overline{\pi }_s(t_s)dt_s, \end{aligned}$$

(34)

where the second to last step follows from integration by parts (we note that $\overline{p}_s$ is left-continuous and non-increasing under our assumptions), and the last step is due to the fact that $\overline{\pi }_s(v_s)=\overline{p}_s(v_s)$ by definition whenever $v_s$ is in the support of the seller. Thus, the payment function (30) yields for any $v_s$ in the seller’s support:

$$\begin{aligned} U_s(v_s)=U_s(b_s)+\int _{t_s=v_s}^{b_s}\overline{\pi }_s(t_s)dt_s \end{aligned}$$

(35)

A similar calculation shows that for any $v_b$ in the buyer’s support, we have:

$$\begin{aligned} U_b(v_b)=U_b(a_b)+\int _{t_b=a_b}^{v_b}\overline{\pi }_b(t_b)dt_b \end{aligned}$$

(36)

These two relations imply that under this payment function, the inequality in (27) (and the corresponding one for the buyer) is satisfied with equality, and through the steps that follow, the first inequality in (16) must also be satisfied with equality, and that if the LHS of it is non-negative, $U_s(b_s)+U_b(a_b)$ must also be non-negative.

Now consider $U_s(b_s)$. We have

$$\begin{aligned} U_s(b_s)&=\int _{a_b}^{b_b}(x(b_s,v_b)-b_sp(b_s,v_b))dF_b\nonumber \\&= \int _{a_b}^{b_b}\int _{t_b=a_b}^{v_b} t_bd[\overline{\pi }_b(t_b)]dF_b-\int _{t_s=a_s}^{b_s} t_sd[-\overline{\pi }_s(t_s)]+K-b_s\overline{p}_s(b_s) \nonumber \\&=\int _{t_b=a_b}^{b_b}(1-F_b(t_b)) t_bd[\overline{\pi }_b(t_b)]-\int _{t_s=a_s}^{b_s} t_sd[-\overline{\pi }_s(t_s)]-b_s\overline{p}_s(b_s)+K \end{aligned}$$

(37)

Setting

$$\begin{aligned} K=-\int _{t_b=a_b}^{b_b}(1-F_b(t_b)) t_bd[\overline{\pi }_b(t_b)]+\int _{t_s=a_s}^{b_s} t_sd[-\overline{\pi }_s(t_s)]+b_s\overline{p}_s(b_s) \end{aligned}$$

(38)

ensures that $U_s(b_s)=0$. Since, in addition, we assume that the LHS of (16) is non-negative, and have shown that it is equal to $U_s(b_s)+U_b(a_b)$, it must follow that $U_b(a_b)$ is also non-negative. This implies, by the envelope theorem, that the mechanism is IP for all buyer and seller types. $\square$

Proof of lemma 5

We will prove the lemma for the case of the seller; the proof works exactly the same way for the buyer.

For a certain trading probability function $p(v_s,v_b)$, that yields non-increasing and non-decreasing $\overline{p}_s(\cdot )$ and $\overline{p}_b(\cdot )$, respectively, consider a mechanism resulting in the following relationship at the focal gap in the seller’s distribution:

$$\begin{aligned} U_s(\underline{v}_s)= U_s(\overline{v}_s)+(\overline{v}_s-\underline{v}_s)(\overline{p}_s(\overline{v}_s)+\gamma ) \end{aligned}$$

where $0\le \gamma \le (\overline{p}_s(\overline{v}_s)-\overline{p}_s(\underline{v}_s))$, such that incentive compatibility is satisfied for the sellers of type $\overline{v}_s$ and $\underline{v}_s$. Under this mechanism, for all $v_s\le \underline{v_s}$, the envelope theorem will have an additional payoff increment $\gamma (\overline{v}_s-\underline{v}_s)$ for all $v_s\le \underline{v}_s$, so we need to modify inequality (27) to read:

$$\begin{aligned} U_s(v_s)\ge {\left\{ \begin{array}{ll} U_s(b_s)+\int _{v_s}^{b_s}\overline{\pi }_s(t_s)dt_s + \gamma (\overline{v}_s-\underline{v}_s)&{} \text{ for } v_s\le \underline{v}_s\\ U_s(b_s)+\int _{v_s}^{b_s}\overline{\pi }_s(t_s)dt_s &{}\text{ for } v_s>\underline{v}_s \end{array}\right. } \end{aligned}$$

(39)

Furthermore, by choosing the following payment function, we can make sure that (39) is satisfied with equality

$$\begin{aligned} x(v_s,v_b)= {\left\{ \begin{array}{ll} \chi _b(v_b)-\chi _s(v_s)+K +\gamma (\overline{v}_s-\underline{v}_s)&{} \text{ for } v_s\le \underline{v}_s\\ \chi _b(v_b)-\chi _s(v_s)+K &{} \text{ for } v_s> \underline{v}_s \end{array}\right. }, \end{aligned}$$

(40)

This statement follows straightforwardly from the same calculations as in the proof of the if part of lemma 4. To see that the payment function remains IC, note that for the buyer and $v_s, v_s'\le \underline{v}_s$ or $v_s, v_s'> \underline{v}_s$ the addition of a constant on to payment function makes no difference for incentive compatibility. For $v_s'\le \underline{v}_s<\overline{v}_s\le v_s$, we have

$$\begin{aligned}&U_s(v_s,v_s)-U_s(v_s',v_s)\nonumber \\&\quad =-v_s(\overline{p}(v_s)-\overline{p}_s(v_s'))-\chi _s(v_s)+\chi _s(v_s')-\gamma (\overline{v}_s-\underline{v}_s)\nonumber \\&\quad \ge \int _{t_s=v_s'}^{v_s}(v_s-t_s)d[-\overline{\pi }_s(t_s)]-(\overline{p}_s(\overline{v}_s)-\overline{p}_s(\underline{v}_s))(\overline{v}_s-\underline{v}_s)\ge 0, \end{aligned}$$

(41)

where the first inequality follows from the upper limit we imposed on $\gamma$, and the last one from the fact that $\overline{p}_s(\cdot )$ is a non-increasing function. It is easy to verify this payment function results in (39) being satisfied with equality.

Retracing the steps that lead up to (29) in the proof of lemma 4, we can then arrive at a modified condition:

$$\begin{aligned} G(\gamma )\equiv & {} \int _{a_b}^{b_b}v_b\overline{p}(v_b)dF_b-\int _{a_s}^{b_s}v_s\overline{p}(v_s)dF_s-\gamma (\overline{v}_s-\underline{v}_s)F_s(\underline{v}_s)\nonumber \\&-\int _{a_s}^{b_s}F_s(t_s)\overline{\pi }_s(t_s)dt_s - \int _{a_b}^{b_b}(1-F_b(t_b))\overline{\pi }_b(t_b)dt_b\nonumber \\= & {} U_s(b_s)+U_b(a_b)\ge 0, \end{aligned}$$

(42)

where we have defined the left hand side as $G(\gamma )$.

Now, the second-best mechanism is given by maximizing the aggregate welfare subject to (42), i.e., maximizing the Lagrangian through the choice of $p(\cdot , \cdot )$ and $\gamma$:

$$\begin{aligned} L=\int _{a_b}^{b_b}\int _{a_s}^{b_s}(v_b-v_s)p(v_s,v_b)dF_sdF_b+\lambda G(\gamma ), \end{aligned}$$

(43)

where $\lambda \ge 0$ is a Lagrange multiplier.¹⁵ But since $G(\gamma )$ is decreasing in $\gamma$ and $\gamma$ is bounded by zero from below (by the envelope theorem), the maximum of the Lagrangian requires $\gamma$ to be zero, which finishes the proof. $\square$

Proof of Lemma 6

We first show that in an equilibrium investment strategy with atoms there must also be a gap; this allows the remainder of the proof to focus on unraveling caused by a gap. Suppose the seller’s mixture has an atom of probability mass $q_s$ at some value $v_s^*>0$, and that there is a $v_b^*$ where the optimal mechanism prescribes that $p(v_s^*,v_b)=1$ for $v_b\ge v_b^*$ and $p(v_s^*,v_b)=0$ for $v_b<v_b^*$.¹⁶ Since there is a jump discontinuity in $\overline{p}$ (of magnitude $q_s$) at $v_b^*$ and we assume $c_b(\cdot )$ is differentiable, the mixing condition cannot be satisfied at $v_b$ just below $v_b^*$, and therefore some interval below $v_b^*$ cannot be in the support of the buyer’s mixed strategy. Furthermore, this means that for some $v_s^*>v_s>v_s^*-\epsilon$, with $\epsilon >0$, the probability of trade $\overline{p}(v_s)$ will be constant, since the gap in the buyers’ distribution means sellers with valuations slightly less than $v_s^*$ cannot trade with additional buyers relative to a seller at $v_s^*$. This again is inconsistent with the mixing condition for the sellers that states $\overline{p}(v_s)=1-c_s'(v_s)$, where $c_s'(v_s)$ is strictly increasing. Similar arguments can be made with regard to an atom in the buyer’s distribution, implying that atoms cannot be in the interior of an interval part of either player’s support. Given this, to prove Lemma 6, we will focus on a gap of the seller’s candidate mixed investment strategy in an equilibrium; similar arguments apply for the buyer. First, note that the mixing condition for the seller is given by:

$$\begin{aligned} v_s+U_s(v_s)-c_s(v_s)=v_s'+U_s(v_s')-c_s(v_s'), \end{aligned}$$

(44)

for any $v_s$, $v_s'$ in the support of the seller’s mixed strategy. Hence, for a pair of values $\underline{v}_s$, $\overline{v}_s$ bordering a gap, we must have $U_s(\underline{v}_s)-U_s(\overline{v}_s)=\overline{v}_s-\underline{v}_s+ c_s(\underline{v}_s)-c_s(\overline{v}_s)$. Dividing by $\overline{v}_s-\underline{v}_s$, we get:

$$\begin{aligned} 1+\frac{c_s(\underline{v}_s)-c_s(\overline{v}_s)}{\overline{v}_s-\underline{v}_s}=\frac{U_s(\underline{v}_s)-U_s(\overline{v}_s)}{\overline{v}_s-\underline{v}_s}=\overline{p}_s(\overline{v}_s), \end{aligned}$$

(45)

where the last equality follows from Lemma 5 for a second-best mechanism. From the convexity of the cost function, we have:

$$\begin{aligned} 1+\frac{c_s(\underline{v}_s)-c_s(\overline{v}_s)}{\overline{v}_s-\underline{v}_s}=\overline{p}_s(\overline{v}_s)<1+c_s'(\overline{v}_s), \end{aligned}$$

(46)

Now, consider a deviation from a candidate mixed-strategy equilibrium where a seller invests at $\overline{v}_s - \epsilon$, but reports $\overline{v}_s$. For small $\epsilon$, the expected change in payoff from this deviation is given by:

$$\begin{aligned} (1-\overline{p}_s(\overline{v}_s)+c'_s(\overline{v}_s)) \epsilon >0, \end{aligned}$$

(47)

meaning that such a deviation will be profitable. Hence, the candidate equilibrium is not an equilibrium strategy. This implies there cannot be any gaps in the seller’s equilibrium investment strategy.

The above argument applies when $v_s^*>0$, since it shows unraveling below $v_s^*$. The unraveling in the case with an atom at $v_s^*=0$ can be seen by showing the buyer’s support cannot have a gap. As noted on page 45, the jump in the trading probability $\overline{p}$ at some $v_b^*$ means that some interval below $v_b^*$ cannot be in the buyer’s support. This fact is true for $v_s^*=0$ as well. Since this $v_b^*$ is the lowest valuation buyer that trades with $v_s=0$, it must be that $v_b^*=a_b$, unless $a_b=0$ (since buyers that do not trade by incur a non-zero cost of investment would be better off not investing at all). If $a_b>0$, this mixing strategy is not stable (provided that first-best is not attainable), since the second-best mechanism gives the lowest valuation buyer zero expected payoff from trade,¹⁷ and therefore, this buyer is better off investing nothing. If $a_b=0$, there must be a gap between $a_b$ and $v_b^*$: since $v_b^*$ is the lowest buyer that can trade with the lowest seller, no positive investment lower than $v_b^*$ can be part of the equilibrium support. Having established a gap in the buyer’s distribution, we can use Lemma 5 and the same argument as above for the seller to show that the following deviation from the equilibrium is profitable: invest $\underline{v}_b^*+\epsilon$ with $\epsilon >0$ small and report $\underline{v}_b^*$. $\square$