The “desire to conform” and dynamic search by a committee

Abstract

We consider dynamic search by a committee where members exhibit an innate desire to conform to the committee’s decision, in addition to their economic incentives. This gives rise to multiple equilibria. “Conformal” equilibria now exist, at least one in which every member votes to continue (stop) the search except on receiving private signals from a set of small but positive measure. A “regular” equilibrium also exists, close to the unique equilibrium that occurs when there is no desire to conform. These equilibria can be Pareto ranked, however, under a certain restriction on the prior distribution of candidates, the regular equilibrium is the Pareto superior equilibrium, provided the voting occurs under a simple or a super-majority rule. Nonetheless, in this equilibrium, the desire to conform causes the committee to increase the minimum acceptable standard of a candidate for the search to stop, and increases the expected search duration.

Appendix

In the following, IVT stands for the “Intermediate Value Theorem”.

Proof

(Lemma 3) We know from AAV that $x_0$ is continuous. It is easy to see that $\psi$ is non-negative. It is also bounded, since its denominator takes values in $(1 - \delta , 1)$ and the numerator takes values in [0, 1]. It is clearly continuous. Next, observe that the numerator of $\varphi (z)$ converges to $- 1$ and 1 as z converges to 0 and 1 respectively. Using the fact that $\lim _{z \rightarrow 0} p (z, k, n) = \lim _{z \rightarrow 1} p (z, k, n) = 0$, provided $k \ne n$, we readily conclude that $\lim _{z \rightarrow 0} \varphi (z) = - \infty$ and $\lim _{z \rightarrow 1} \varphi (z) = \infty$. Since $x (z, \alpha )$ is linear in $\alpha$, and is the sum of continuous functions, the various assertions in the proposition for the case where $M\ne N$ now follow readily. $\square$

Proof

(Proposition 1.) We first prove Part 2. Choose $\varepsilon > 0$ sufficiently small so that $2 \varepsilon< y^{*} < 1 - 2 \varepsilon$. Define,

$$\begin{aligned} \delta= & {} \inf \{ | x_0 (z) - z |: z \in [\varepsilon , y^{*} - \varepsilon ] \cup [y^{*} + \varepsilon , 1 - \varepsilon ] \}. \end{aligned}$$

$\delta > 0$, since we know from Proposition 0 that $y^{*}$ is the unique fixed point of $x_0 (\cdot )$. Next, observe that $x (z, \alpha )$ converges pointwise to $x_0 (z)$ for all z in the compact interval $[\varepsilon , 1 - \varepsilon ]$. From Lemma 1, we know that $x (z, \alpha )$ is jointly continuous in both its arguments for $(z, \alpha ) \in (0, 1) \times [0, \infty )$. Hence, we can find $\alpha _{\varepsilon } > 0$ such that,

$$\begin{aligned} | x (z, \alpha ) - x_0 (z) | < \delta / 2 \forall z \in [\varepsilon , 1 - \varepsilon ] {\text { and }} \forall \alpha \in (0, \alpha _{\varepsilon }). \end{aligned}$$

(16)

From Proposition 0, $x_0 (z) > z$ for $z \in [\varepsilon , y^{*} - \varepsilon ]$ which gives,

$$\begin{aligned} x (z, \alpha ) - z > x_0 (z) - z - \delta / 2 \ge \delta / 2 \forall z \in [\varepsilon , y^{*} - \varepsilon ] \end{aligned}$$

and that $x_0 (z) < z$ for $z \in [y^{*} + \varepsilon , \varepsilon ]$, which gives,

$$\begin{aligned} x (z, \alpha ) - z < x_0 (z) - z + \delta / 2 \, \le - \delta / 2 \forall z \in [y^{*} + \varepsilon , 1 - \varepsilon ] \end{aligned}$$

Therefore, fixed points of $x (\cdot , \alpha )$, if they exist, can occur only in the regions $[0, \varepsilon )$, $(y^{*} - \varepsilon , y^{*} + \varepsilon )$ or $(1 - \varepsilon , 1]$. This proves Part 2.

Next, we turn to Part 1. From Lemma 3, we know that $\lim _{z \rightarrow 0} \, x (z, \alpha ) = - \infty$ and $\lim _{z \rightarrow 1} \, x (z, \alpha ) = \infty$. This, together with the foregoing conclusions concerning $x (z, \alpha )$ allow us to conclude the following regarding the graph of $x (z, \alpha )$: it begins from below the 45 degree line when z is close to zero, is above it in the region $[\varepsilon , y^{*} - \varepsilon ]$, is below it in the region $[y^{*} + \varepsilon , 1 - \varepsilon ]$ and again above it for z close enough to 1. Appealing to the continuity of $x (\cdot , \alpha )$ and applying the IVT, it follows that $x (\cdot , \alpha )$ intersects the 45 degree line at least once in each of the regions $(0, \varepsilon )$, $(y^{*} - \varepsilon , y^{*} + \varepsilon )$ and $(\varepsilon , 1 - \varepsilon )$, and only on those regions. Each of these points of intersection in the respective regions constitute the acceptance thresholds for a left conformal equilibrium, a regular equilibrium or a right conformal equilibrium. This completes the proof of Part 1. $\square$

Proof

(Lemma 2) Part 1. $\delta \mu > y^{*}$ is given. Set $\beta = (\delta \mu - y^{*}) / 2 \delta$. Note that $V_0 (0) - V_0 (y^{*}) = (\mu - y^{*} / \delta ) = 2 \beta > 0$ and $V_0 (0) - V_0 (1) = \mu> 2 \beta > 0$. Since $V_0 (z)$ is uniformly continuous in z in the compact interval [0, 1], there exists ${\hat{\varepsilon }} > 0$ such that

$$\begin{aligned} V_0 (z) - V_0 (z')> & {} \beta \qquad \forall z \in (0, {\hat{\varepsilon }}) \text {and } z' \in (y^{*} - {\hat{\varepsilon }}, y^{*} + {\hat{\varepsilon }}). \\ V_0 (z) - V_0 (z')> & {} \beta \qquad \forall z \in (0, {\hat{\varepsilon }}) \text {and } z' \in (1 - {\hat{\varepsilon }}, 1). \end{aligned}$$

Set ${\hat{\alpha }} = \beta (1 - \delta ) / 2$. Then, $\forall z \in (0, {\hat{\varepsilon }}) \text {and } z' \in (y^{*} - {\hat{\varepsilon }}, y^{*} + {\hat{\varepsilon }}) \cup (1 - {\hat{\varepsilon }}, 1)$

$$\begin{aligned} V (z, \alpha ) - V (z', \alpha )= & {} V_0 (z) - V_0 (z') + \alpha (\psi (z) - \psi (z')) \\{} & {} > \beta + \alpha (\psi (z) - \psi (z')) > 0 \end{aligned}$$

where the last inequality is due to the fact that $\psi (\cdot )$ is non-negative and bounded above by $1 / (1 - \delta )$ on [0, 1]. Therefore, for all $\alpha \in (0, {\hat{\alpha }})$,

$$\begin{aligned} V (z, \alpha ) - V (z', \alpha )= & {} V_0 (z) - V_0 (z') + \alpha (\psi (z) - \psi (z'))\\{} & {} \quad > 0 \qquad \forall z \in (0, {\hat{\varepsilon }}) \text {and } z' \in (y^{*} - {\hat{\varepsilon }}, y^{*} + {\hat{\varepsilon }}) \cup (1 - {\hat{\varepsilon }}, 1). \end{aligned}$$

Now, select $\varepsilon < {\hat{\varepsilon }}$, let $\alpha _{\varepsilon }$ be as in Proposition 1 and set $\alpha ^{*} = \min \{ {\hat{\alpha }}, \alpha _{\varepsilon } \}$. By Proposition 1 then, for all $\alpha \in (0, \alpha ^{*})$, only $\varepsilon$-left, $\varepsilon$-right conformal equilibria and $\varepsilon$-regular equilibria exist, and from the above, the Pareto Superior equilibrium must be the $\varepsilon$-left conformal equilibrium.

Part 2. $y^{*} > \delta \mu$ is given. Set $\beta = (y^{*} - \delta \mu ) / 2 \delta$. Observe that $V_0 (y^{*}) - V_0 (0) = (y^{*} / \delta - \mu ) = 2 \beta > 0$ and $V_0 (y^{*}) - V_0 (1) = y^{*} / \delta> 2 \beta > 0$. An anlogous argument to the one provided above for Part 1 can be used to complete the proof. $\square$

Proof

(Lemma 3) Recall that $\lim _{z\rightarrow 0}\varphi (z)=-\infty$ and $\lim _{z\rightarrow 1}\varphi (z)=\infty$. $\varphi (z)$ is also continuous in z. The proof of the lemma, including the existence of a unique ${\bar{y}}$ defined in (15) is complete upon verifying that $\varphi (z)$ is increasing.

$$\begin{aligned} \frac{p (z, k, N - 1)}{p (z, M - 1, N - 1)}= & {} \frac{\left( {\begin{array}{c}N - 1\\ k\end{array}}\right) }{\left( {\begin{array}{c}N - 1\\ M - 1\end{array}}\right) } \frac{F (z)^{N - 1 - k} (1 - F (z))^k}{F (z)^{N - M} (1 - F (z))^{M - 1}} \\= & {} \frac{\left( {\begin{array}{c}N - 1\\ k\end{array}}\right) }{\left( {\begin{array}{c}N - 1\\ M - 1\end{array}}\right) } \left( \frac{F (z)}{1 - F (z)} \right) ^{(M - 1) - k} \\= & {} \lambda (k)\theta (z)^{M-1-k}, \end{aligned}$$

where we have set $\lambda (k) = {\left( {\begin{array}{c}N - 1\\ k\end{array}}\right) }/{\left( {\begin{array}{c}N - 1\\ M - 1\end{array}}\right) }$ and $\theta (z) = {F (z)}/{1 - F (z)}$. Next, note that $\varphi (z)$ is a polynomial in $\theta (z)$ with $\lambda (k)$’s as coefficients:

$$\begin{aligned} \varphi (z)= & {} \frac{P (z, M, N - 1) - Q (z, M - 1, N - 1)}{p (z, M - 1, N - 1)} \\= & {} \sum _{j = 0}^{M - 1} \frac{p (z, j, N - 1)}{p (z, M - 1, N - 1)} - \sum _{j = M - 1}^{N - 1} \frac{p (z, j, N - 1)}{p (z, M - 1, N - 1)} \\= & {} \sum _{j = 0}^{M - 1} \lambda (j) \theta (z)^{M - 1 - j} + \sum _{j = M - 1}^{N - 1} (-\lambda (j)) \theta (z)^{M - 1 - j} \\= & {} \sum _{j = 0}^{M - 2} \lambda (j) \theta (z)^{M - 1 - j} + \sum _{j = M }^{N - 1} (-\lambda (j)) \theta (z)^{M - 1 - j} \end{aligned}$$

Clearly $\theta (z)^{M-1-j}$ is increasing in z when $j< M-1$ but is decreasing when $j> M-1$, since $\theta (z)$ is itself increasing in z. Given $\lambda (j)>0$ for all $j=0,\ldots ,N-1$, it is now clear that $\varphi (z)$ is also increasing in z. $\square$

Proof

(Proposition 2) Part 1. Let $\alpha ^{*}$ be as in Lemma 2. The $\varepsilon$-left equilibrium is Pareto superior with a MAT $y_{\alpha } \in (0, \varepsilon )$ and hence $y_{\alpha } < y^{*}$ for all $\alpha \in (0, \alpha ^{*})$. So the expected search duration decreases.

We recall that establish (17) and the Claim prior to (15). Recall $\psi (y) \ge 0$ for all $y\in [0,1]$, and from Lemma 3 that $\varphi (y)$ is increasing in y. Hence,

$$\begin{aligned} y> {\bar{y}} \Rightarrow&\varphi (y) + \psi (y) > 0 \end{aligned}$$

(17)

Again let $\alpha ^{*}$ be as in Lemma 2 so that MAT of Pareto superior equilibrium $y_{\alpha } \in (y^{*} - \varepsilon , y^{*} + \varepsilon )$. As ${\bar{y}} < y^{*}$, we may assume that $\varepsilon$ (or equivalently $\alpha ^{*}$) was chosen to be small enough so that ${\bar{y}} < y_\alpha$. Therefore, from (17) above, $\varphi (y_{\alpha }) + \psi (y_{\alpha }) > 0$. Moreover, since $y_{\alpha }$ is an equilibrium MAT of ${\mathcal {G}} (\alpha , M, N)$, we have $0 = x (y_{\alpha }, \alpha ) - y_{\alpha } = (x_0 (y_{\alpha }) - y_{\alpha }) + \alpha (\varphi (y_{\alpha }) + \psi (y_{\alpha }))$, we must have $x_0 (y_{\alpha }) < y_{\alpha }$, which in turn, by Part 2, Lemma 1, implies $y_{\alpha } > y^{*}$. Hence the expected search duration increases. $\square$

Proof

(Corollary 1) By assumption the plurality rule is a super-majority rule, i.e. $M-2 \ge N-1-M$. Noting that $\theta (\nu )=1$ and $\lambda (j) = \lambda (N-1-j)$, we have

$$\begin{aligned} \varphi (\nu )= & {} \sum _{j = 0}^{M - 2} \lambda (j) - \sum _{j = M }^{N - 1} \lambda (j) \nonumber \\= & {} \sum _{j = 0}^{M - 2} \lambda (j) - \sum _{j = 0 }^{N - 1-M} \lambda (N-1-j) \; \; = \; \sum _{j = N-M}^{M - 2} \lambda (j) \; \ge 0 \end{aligned}$$

(18)

This gives $\nu \ge {\bar{y}}$ the Claim preceding (15). Now apply Part 2, Proposition 2$\square$

Proof

Lemma 4. Let $y_1 < y^{*}$. By Proposition 0, part 2, $x_0 (y_1) > y_1$. From eq (14),

$$\begin{aligned} \alpha _1 (\varphi (y_1) + \psi (y_1))= & {} x (y_1, \alpha _1) - x_0 (y_1) \\= & {} y_1 - x_0 (y_1), \qquad \text { since }y_1\text { is a MAT of }{\mathcal {G}} (\alpha _1, M, N) \\< & {} 0. \end{aligned}$$

Hence $x (y_1, \alpha _2) > x (y_1, \alpha _1) = y_1$. Since $\lim _{y \rightarrow 0} x (y, \alpha _2) = - \infty$, there exists $y' < y_1$ such that $x (y', \alpha _2) < y'$. By IVT, we can find a $y_2 < y_1$ such that $x (y_2, \alpha _2) = y_2$. A similar argument proves the case where $y_1 > y^{*}$. $\square$ $\square$

Proof

(Proposition 3) Since Q(z, M, N) is continuously decreasing in z with $\lim _{z \rightarrow 0} Q (z, M, N) = 1$ and $\lim _{z \rightarrow 0} Q (1 - z, M, N) = 0$, there exists ${\hat{\varepsilon }} > 0$ and ${\hat{q}} > 0$ such that

$$\begin{aligned} Q ({\hat{\varepsilon }}, M, N)> \, {\hat{q}} \, > Q (1 - {\hat{\varepsilon }}, M', N) \quad \forall M, M'\ne 0. \end{aligned}$$

By Proposition 1, there exists $\alpha _{{\hat{\varepsilon }}} (M)$ sufficiently small such that both ${\hat{\varepsilon }}$-left and ${\hat{\varepsilon }}$-right exist in ${\mathcal {G}} (\alpha , M, N)$ for all $\alpha \in (0, \alpha _{{\hat{\varepsilon }}} (M))$. Set $\alpha ^{*} = \min _M \alpha _{{\hat{\varepsilon }}} (M)$ and appeal to Lemma 4 above to conclude that for all $\alpha \in (0, \alpha ^{*})$, an ${\hat{\varepsilon }}$-left conformal equilibrium and an ${\hat{\varepsilon }}$-right conformal equilibruim exist for all $M\ne 0$.

Choose any $\alpha \in (0, \alpha ^{*})$. Pick an ${\hat{\varepsilon }}$-left conformal equilibrium MAT, say y, of ${\mathcal {G}} (\alpha , M, N)$. Then, $Q (y, M, N) \ge Q ({\hat{\varepsilon }}, M, N)$ and hence $\sigma (y, M, N) < 1 / {\hat{q}}$. Likewise, for an ${\hat{\varepsilon }}$-right conformal equilibrium MAT, say $y'$, of ${\mathcal {G}} (\alpha , M', N)$, $Q (y', M', N) \ge Q (1 - {\hat{\varepsilon }}, M', N)$ and hence $\sigma (y', M', N) > 1 / {\hat{q}}$, which means M does not search-dominate $M'$. As $M, M'$ are chosen arbitrarily, this completes the proof.$\square$