Two-sided search with experts

Abstract

In this paper we study distributed agent matching in environments characterized by uncertain signals, costly exploration, and the presence of an information broker. Each agent receives information about the potential value of matching with others. This information signal may, however, be noisy, and the agent incurs some cost in receiving it. If all candidate agents agree to the matching the team is formed and each agent receives the true unknown utility of the matching, and leaves the market. We consider the effect of the presence of information brokers, or experts, on the outcomes of such matching processes. Experts can, upon payment of either a fee or a commission, perform the service of disambiguating noisy signals and revealing the true value of a match to any agent. We analyze equilibrium behavior given the fee set by a monopolist expert and use this analysis to derive the revenue maximizing strategy for the expert as the first mover in a Stackelberg game. Interestingly, we find that better information can hurt: the presence of the expert, even if the use of her services is optional, can degrade both individual agents’ utilities and overall social welfare. While in one-sided search the presence of the expert can only help, in two-sided (and general $k$-sided) search the externality imposed by the fact that others are consulting the expert can lead to a situation where the equilibrium outcome is that everyone consults the expert, even though all agents would be better off if the expert were not present. As an antidote, we show how market designers can enhance welfare by compensating the expert to change the price at which she offers her services.

Appendices

Appendix 1: Proofs

1.1 Proof of Theorem 1

The proof is based on showing that, if according to the optimal search strategy the searcher should resume her search given a signal $s$, then she must necessarily also do so given any other signal $s'<s$. Let $V$ denote the expected benefit to the searcher if resuming the search if signal $s$ is obtained. Since the optimal strategy given signal $s$ is to resume search, we know $V>E[v|s]$. Given the HSGN assumption, $E[v|s] \ge E[v | s']$ holds for $s'<s$. Therefore, $V>E[v|s']$, proving that the optimal strategy is reservation-value. Then, the expected value of the searcher when using reservation signal $t$ is given by:

$$\begin{aligned} V(t)&= -c_s + V(t) \Big (1-B\int \limits _{s=t}^\infty f_s(s) \,ds\Big )\nonumber \\&+\, B\int \limits _{s=t}^{\infty }E[v|s]f_s(s) \,ds =\frac{-c_s + B\int _{s=t}^{\infty } E[v|s]f_s(s) \,ds}{B\big (1-F_s(t)\big )} \end{aligned}$$

(35)

where B is the probability of being accepted by the other searcher, and $F_s(s)$ is the cumulative distribution function of the signal $s$. Setting the first derivative according to $t$ of Eq. 35 to zero we obtain: $V(t^*) = E[v|t^*]$ as follows:

$$\begin{aligned} \frac{d V(t)}{dt}&= \frac{d V(t)}{dt} \Big (1-B\int \limits _{s=t}^\infty f_s(s) \,ds\Big ) \\&+\, V(t) Bf_s(t)- B \cdot E[v|t] f_s(t) \\ V(t^*)&= E[v|t^*] \quad \quad f_s(t^*) \ne 0 \quad B \ne 0 \end{aligned}$$

To verify that $t^*$ is global maximum, we calculate the second derivative.

$$\begin{aligned} \frac{d^2V}{dt^2}&= \Big (1-B\int \limits _{s=t}^\infty f_s(s) \,ds\Big ) \frac{d^2V}{dt^2} + B \cdot f_s(t) \frac{dV}{dt} \\&+\, B\frac{d(V(t)f_s(t))}{dt} - B\frac{d(E[v|t] f_s(t))}{dt} \\ \\ \frac{d^2V}{dt^2}|t^*&= \frac{1}{B\big (1 - F_s(t^*)\big )} B\left( \frac{ d\big ((V(t) - E[v|t])f_s(t) \big ) }{dt}|t^* \right) \\&= -\frac{f_s(t^*)}{1-F_s(t^*)} \left( \frac{dE[v|t]}{dt}|t^* \right) \end{aligned}$$

Given the HSGN assumption, $E[v|t]$ is an increasing function in $t$. Therefore $\frac{dE[v|t]}{dt} > 0$, this implies that the value of second derivative evaluated at $t^*$ is less than zero, which confirms that $t^*$ is indeed a global maximum.

Finally substituting $V(t^*)=E[V|t^*]$ in Eq. 35 we obtain Eq. 4 as follows:

$$\begin{aligned} V(t^*)&= -c_s + V(t^*) \left( 1-B\int \limits _{s=t^*}^\infty f_s(s) \,ds \right) + B\int \limits _{s=t^*}^{\infty } E[v|s] f_s(s)\,ds \\ E[v|t^*]&= -c_s + E[v|t^*] \left( 1-B\int \limits _{s=t^*}^\infty f_s(s) \,ds \right) + B\int \limits _{s=t^*}^{\infty } E[v|s] f_s(s)\,ds \\ c_s&= B\int \limits _{s=t^*}^{\infty } \Big (E[v|s]-E[v|t^*]\Big )f_s(s) \,ds \end{aligned}$$

Since all the searchers end up using the same threshold, we get that:

$$\begin{aligned} B=\int \limits _{s=t^*}^\infty f_s(s) \,ds \end{aligned}$$

Therefore, the value $t^*$ can be calculated using the above equation.$\square$

1.2 Proof of Theorem 2

The proof extends the methodology used for proving Theorem 1. We first show that if, according to the optimal search strategy the searcher should resume her search given a signal $s$, then she must also do so given any other signal $s' < s$. Then, we show that if, according to the optimal search strategy the searcher should terminate her search given a signal $s$, then she must also necessarily do so given any other signal $s'' > s$. Again, we use $V$ to denote the expected benefit to the searcher if resuming search.

If the optimal strategy given signal $s$ is to resume search then the following two inequalities should hold, describing the superiority of resuming search over terminating search (Eq. 36) and querying the expert (Eq. 37):

$$\begin{aligned} V&> E[v|s] \end{aligned}$$

(36)

$$\begin{aligned} V&> VF_v(V|s) + \int \limits _{y=V}^{\infty } y f_v(y|s) \,dy - c_e \end{aligned}$$

(37)

Given the HSGN assumption and since $s' < s$, Eq. 36 holds also for $s'$. Similarly, notice that:

$$\begin{aligned} V&> VF_v(V|s) + \int \limits _{y=V}^{\infty } y f_v(y|s) \,dy - c_e \\&= V+\int \limits _{y=V}^{\infty } (1- F_v(y|s)) \,dy - c_e\\&> V+\int \limits _{y=V}^{\infty } (1- F_v(y|s')) \,dy - c_e\\&= VF_v(V|s') + \int \limits _{y=V}^{\infty } y f_v(y|s') \,dy - c_e \end{aligned}$$

and therefore Eq. 37 also holds for $s' < s$.

The proof for $s''>s$ is exactly equivalent: the expected cost of accepting the current opportunity can be shown to dominate both resuming the search and querying the expert. The optimal strategy can thus be described by the tuple ($t_l,t_u,V$) as stated above. Equations 6, 9, 11 are obtained after replacing the intervals $S,S'$ with the thresholds $t_l,t_u$:

$$\begin{aligned} V&= -c_s -c_e\int \limits _{s=t_l}^{s=t_u} f_s(s) \,ds+(1-AB)V+BC \\ V&= \frac{-c_s-c_e\big (F_s(t_u)-F_s(t_l)\big )+BC}{AB}\\ A&= \int \limits _{s=t_u}^{\infty } f_s(s)\,ds+\int \limits _{s=t_l}^{t_u}f_s(s)\Big (1-F_v(V|s)\Big ) \,ds \\ A&= 1-F_s(t_l)-\int \limits _{s=t_l}^{t_u}f_s(s)F_v(V|s)ds\\ C&= \int \limits _{s=t_u}^{\infty }f_s(s)E[v|s]\,ds+\int \limits _{s=t_l}^{t_u}\biggl (f_s(s) \int \limits _{y=V}^{\infty } y f_v(y|s) \,dy\biggl ) \,ds \end{aligned}$$

Equation 10 represents the fact that the system is symmetric in the way that ultimately all agents choose the same tuple $(t_l,t_u,V)$, and so the probability of being accepted is equal to the probability of accepting the match. Now we will show the correctness of Eqs. 7 and 8:

Taking the derivative of Eq. 6 w.r.t. $t_l$ and equating to zero, we obtain a $t_l$ which maximizes the expected benefit.

$$\begin{aligned} \frac{ \partial V}{ \partial t_l}&= \frac{(c_ef_s(t_l)+B\frac{ \partial C}{ \partial t_l})AB+(-c_s-c_e\big (F_s(t_u)-F_s(t_l)\big )+BC)\frac{ \partial A}{ \partial t_l}B}{A^2B^2}\\&= \frac{\Big (c_ef_s(t_l)+B\frac{ \partial C}{ \partial t_l}\Big )AB-VAB\frac{ \partial A}{ \partial t_l}B}{A^2B^2}\\&= \frac{c_ef_s(t_l)+B\frac{ \partial C}{ \partial t_l}-V\frac{ \partial A}{ \partial t_l}B}{AB}\\&= \frac{c_ef_s(t_l)+B\Big (\frac{ \partial C}{ \partial t_l}-V\frac{ \partial A}{ \partial t_l}\Big )}{AB} \end{aligned}$$

Calculating $\frac{ \partial C}{ \partial t_l}-V\frac{ \partial A}{ \partial t_l}$ separately, we get:

$$\begin{aligned} \frac{ \partial C}{ \partial t_l}&= -f_s(t_l)\int \limits _{y=V}^{\infty }yf_v(y|t_l) \,dy+\int \limits _{s=t_l}^{t_u}f_s(s)\big (-Vf_v(V|s)\big )\frac{\partial V}{\partial t_l}\,ds\\ \frac{ \partial A}{ \partial t_l}&= -f_s(t_l)+f_s(t_l)F_v(V|t_l)-\int \limits _{s=t_l}^{t_u}f_s(s)f_v(V|s)\frac{\partial V}{\partial t_l} \,ds \\&= -f_s(t_l)\int \limits _{y=V}^{\infty }f_v(y|t_l) \,dy-\int \limits _{s=t_l}^{t_u}f_s(s)f_v(V|s)\frac{\partial V}{\partial t_l} \,ds\\ \end{aligned}$$

and so

$$\begin{aligned} \frac{ \partial C}{ \partial t_l}&-V\dfrac{ \partial A}{ \partial t_l}=-f_s(t_l) \displaystyle \int \limits _{y=V}^{\infty }(y-V)f_v(y|t_l) \,dy \end{aligned}$$

and so

$$\begin{aligned} \frac{ \partial V}{ \partial t_l}&= \frac{c_ef_s(t_l)+B\big (-f_s(t_l) \int _{y=V}^{\infty }(y-V)f_v(y|t_l) \,dy\big )}{AB}\\&= \frac{f_s(t_l)\big (c_e-B\int _{y=V}^{\infty }(y-V)f_v(y|t_l) \,dy\big )}{AB} \end{aligned}$$

Substituting $\frac{\partial V}{\partial t_l}=0$ we get Eq. 7.

To confirm that this is a maximum, we compute the second derivative:

$$\begin{aligned} AB\frac{ \partial ^2 V}{\partial {t_l}^2} +\frac{ \partial A}{ \partial t_l}B\frac{ \partial V}{ \partial t_l}=&\frac{ \partial \big ( f_s(t_l)\big )}{ \partial t_l}\big (c_e-B\int \limits _{y=V}^{\infty }(y-V)f_v(y|t_l) \,dy\big )\\&-\, f_s(t_l)B\frac{ \partial \bigg (\int \limits _{y=V}^{\infty }(y-V)f_v(y|t_l) \,dy\bigg )}{ \partial t_l}\\ \end{aligned}$$

By substituting $\frac{\partial V}{\partial t_l}=0$ and $c_e = B\int \limits _{V}^{\infty } (y-V) f_v(y|t_l) \,dy$ we get:

$$\begin{aligned}&AB\frac{ \partial ^2 V}{\partial {t_l}^2}=-f_s(t_l)B\frac{ \partial \bigg (\int \limits _{y=V}^{\infty }(y-V)f_v(y|t_l) \,dy\bigg )}{ \partial t_l} \end{aligned}$$

Or alternatively:

$$\begin{aligned} \frac{ \partial ^2 V}{\partial {t_l}^2}=\frac{-f_s(t_l)\frac{ \partial \bigg (\int \limits _{y=V}^{\infty }(y-V)f_v(y|t_l) \,dy\bigg )}{ \partial t_l}}{A} \end{aligned}$$

All that is left to show is that $\int \limits _{y=V}^{\infty }(y-V)f_v(y|t_l) \,dy$ increases with $t_l$ to complete the proof of maximum.

$$\begin{aligned} \int \limits _{y=V}^{\infty }(y-V)f_v(y|t_l) \,dy&= \lim _{b\rightarrow \infty }\int \limits _{y=V}^{b}(y-V)f_v(y|t_l) \,dy\\&= \lim _{b\rightarrow \infty }[(y-V)F_v(y|t_l)]_{y=v}^{b}+\lim _{b\rightarrow \infty }\int \limits _{y=V}^{b}F_v(y|t_l) \,dy \\&= \lim _{b\rightarrow \infty }(b-V)F_v(b|t_l)+\lim _{b\rightarrow \infty }\int \limits _{y=V}^{b}F_v(y|t_l) \,dy \\&= \lim _{b\rightarrow \infty }(b-V)+\lim _{b\rightarrow \infty }\int \limits _{y=V}^{b}F_v(y|t_l) \,dy\\&= \lim _{b\rightarrow \infty }\int \limits _{y=V}^{b}(1-F_v(y|t_l)) \,dy=\int \limits _{y=V}^{\infty }\big (1-F_v(y|t_l)\big ) \,dy \end{aligned}$$

and so

$$\begin{aligned}&\frac{ \partial \bigg (\int \limits _{y=V}^{\infty }(y-V)f_v(y|t_l) \,dy\bigg )}{ \partial t_l}=\frac{ \partial \bigg (\int \limits _{y=V}^{\infty }\big (1-F_v(y|t_l)\big ) \,dy\bigg )}{ \partial t_l} \\&\quad =-\Big (\big (1-F_v(V|t_l)\big )\Big )\frac{ \partial V}{ \partial t_l}+\int \limits _{y=V}^{\infty }\frac{ \partial \big (-F_v(y|t_l)\big )}{ \partial t_l} \,dy >0 \end{aligned}$$

The inequality is given by substituting $\frac{\partial V}{\partial t_l}=0$ and due to HSGN assumption, which determines that $\forall y : \frac{ \partial \big (F_v(y|t_l)\big )}{ \partial t_l}<0$. The proof for Eq. 8 is exactly similar to the proof for Eq. 7. We have 6 Eqs. (6–11) in 6 variables. We can solve these simultaneously to calculate the value of $V$, $t_l$ and $t_u$. $\square$

1.3 Proof of Theorem 3

The proof steps resemble those used for Theorem 2. We first show that if, according to the optimal search strategy the searcher should terminate her search (and accept the partnership) given a signal $s$, then she must also necessarily do so given any other signal $s' > s$. The partnership is accepted based on signal $s$ iff the following two conditions hold:

• 1. $E[v|s]>V$

• 2. $E[v|s]>VF_v(V/(1-{\gamma }_e)|s)+\int \limits _{y=V/(1-{\gamma }_e)}^{\infty }(1-{\gamma }_e)yf_v(y|s)dy$

Or equivalently:

• 1. $E[v|s]>V$

• 2. $\int \limits _{y=-\infty }^{V/(1-{\gamma }_e)}(y-V)f_v(y|s)dy + \int \limits _{y=V/(1-{\gamma }_e)}^{\infty }{\gamma }_eyf_v(y|s)dy>0$

For every $s' > s$, we get that $E[v|s'] \ge E[v|s]>V$.

We show that for any $s' > s$:

$$\begin{aligned}&\int \limits _{y=-\infty }^{V/(1-{\gamma }_e)}(y-V)f_v(y|s')dy + \int \limits _{y=V/(1-{\gamma }_e)}^{\infty }{\gamma }_eyf_v(y|s')dy\\&\qquad \ge \int \limits _{y=-\infty }^{V/(1-{\gamma }_e)}(y-V)f_v(y|s)dy + \int \limits _{y=V/(1-{\gamma }_e)}^{\infty }{\gamma }_eyf_v(y|s)dy >0 \end{aligned}$$

Rearranging the above equation we obtain:

$$\begin{aligned}&\int \limits _{y=-\infty }^{V/(1-{\gamma }_e)}(y-V)[f_v(y|s')-f_v(y|s)]dy + \int \limits _{y=V/(1-{\gamma }_e)}^{\infty }{\gamma }_ey[f_v(y|s')-f_v(y|s)]dy \\&\quad =\big [(y-V)[F_v(y|s')-F_v(y|s)] \big ]_{y=-\infty }^{V/(1-{\gamma }_e)}- \int \limits _{y=-\infty }^{V/(1-{\gamma }_e)}[F_v(y|s')-F_v(y|s)]dy\\&\qquad +\, \big [{\gamma }_ey[F_v(y|s')-F_v(y|s)] \big ]_{y=V/(1-{\gamma }_e)}^\infty - \int \limits _{y=V/(1-{\gamma }_e)}^{\infty }{\gamma }_e[F_v(y|s')-F_v(y|s)]dy\\&\quad =[(V/(1-{\gamma }_e)-V)(F_v(V/(1-{\gamma }_e)|s')\\&\qquad -\, F_v(V/(1-{\gamma }_e)|s))-0]-\int \limits _{y=-\infty }^{V/(1-{\gamma }_e)}[F_v(y|s')-F_v(y|s)]dy\\&\qquad +\, [0-{\gamma }_eV/(1-{\gamma }_e)[F_v(V/(1-{\gamma }_e)|s')\\&\qquad \, -F_v(V/(1-{\gamma }_e)|s)]-\int \limits _{y=V/(1-{\gamma }_e)}^{\infty }{\gamma }_e[F_v(y|s')-F_v(y|s)]dy\\&\quad =-\int \limits _{y=-\infty }^{V/(1-{\gamma }_e)}[F_v(y|s')-F_v(y|s)]dy -\int \limits _{y=V/(1-{\gamma }_e)}^{\infty }{\gamma }_e[F_v(y|s')-F_v(y|s)]dy \\&\quad \ge 0+0=0 \end{aligned}$$

where the last inequality is obtained since $F_v(y|s')<F_v(y|s)$ according to the HSGN assumption. Therefore, there is a single threshold for acceptance without querying the expert. Once this is established, Eqs. 28, 30, 32 are obtained by replacing the set of intervals $S$ with the threshold $t$. Equation 31 represents the fact that the system is symmetric in the way that ultimately all agents choose the same tuple $(t,V)$, and so the probability of being accepted is equal to the probability of accepting the match. Finally, the correctness of Eq. 29 is proved by taking the derivative of Eq. 28 w.r.t. $t$, equating to zero, obtaining $t$ which maximize the expected benefit:

$$\begin{aligned} \frac{dV}{dt}&= \frac{B\frac{dC}{dt}AB-\Big (-c_s+BC\Big )\frac{dA}{dt}B}{(AB)^2}\\&= \frac{B\frac{dC}{dt}AB-\big (VAB\big )\frac{dA}{dt}B}{(AB)^2}\\&= \frac{\frac{dC}{dt}-V\frac{dA}{dt}}{A} \end{aligned}$$

however,

$$\begin{aligned} \frac{dA}{dt}&= -f_s(t)F_v(V/(1-{\gamma }_e)|t)-\int \limits _{s=t}^{\infty }f_s(s)f_v(V/(1-{\gamma }_e)|s)ds\frac{dV}{dt}\\ \\ \frac{dC}{dt}&= -f_s(t)E(v|t)+f_s(t)\int \limits _{y=V/(1-{\gamma }_e)}^{\infty }(1-{\gamma }_e)yf_v(y|t)dy\\&+\int \limits _{s=t}^{\infty } f_s(s)(-1)(1-{\gamma }_e)\frac{V}{(1-{\gamma }_e)}f_v(V/(1-{\gamma }_e)|s)ds\frac{dV}{dt}\\&= -f_s(t)E(v|t)+f_s(t)\int \limits _{y=V/(1-{\gamma }_e)}^{\infty }(1-{\gamma }_e)yf_v(y|t)dy\\&-\int \limits _{s=t}^{\infty }f_s(s)Vf_v(V/(1-{\gamma }_e)|s)ds\frac{dV}{dt} \end{aligned}$$

and so:

$$\begin{aligned} \frac{dV}{dt}&= \frac{\frac{dC}{dt}\!-\!V\frac{dA}{dt}}{A}\\&= \frac{\!-f_s(t)E(v|t)+f_s(t)\int \limits _{y=V/(1-{\gamma }_e)}^{\infty }(1\!-\!{\gamma }_e)yf_v(y|t)dy+f_s(t)VF_v(V/(1\!-\!{\gamma }_e)|t)}{A}\\&= \frac{f_s(t)\bigg (-E(v|t)+\int \limits _{y=V/(1-{\gamma }_e)}^{\infty }(1-{\gamma }_e)yf_v(y|t)dy+VF_v(V/(1-{\gamma }_e)|t)\bigg )}{A}\\&= \frac{f_s(t)\bigg (-\int \limits _{y=-\infty }^{\infty }yf_v(y|t)dy+\int \limits _{y=V/(1-{\gamma }_e)}^{\infty }(1-{\gamma }_e)yf_v(y|t)dy+V\int \limits _{y=-\infty }^{V/(1-{\gamma }_e)}f_v(y|t)dy\bigg )}{A}\\&= \frac{f_s(t)\bigg (-\int \limits _{y=-\infty }^{V/(1-{\gamma }_e)} yf_v(y|t)dy- \int \limits _{y=V/(1-{\gamma }_e)}^{\infty }yf_v(y|t)dy+\int \limits _{y=V/(1-{\gamma }_e)}^{\infty }(1-{\gamma }_e)yf_v(y|t)dy+V\int \limits _{y=-\infty }^{V/(1-{\gamma }_e)}f_v(y|t)dy\bigg )}{A}\\&= \frac{f_s(t)\bigg (\int \limits _{y=-\infty }^{V/(1-{\gamma }_e)}(V-y)f_v(y|t)dy-{\gamma }_e\int \limits _{y=V/(1-{\gamma }_e)}^{\infty }yf_v(y|t)dy\bigg )}{A} \end{aligned}$$

Substituting $\frac{dV}{dt}=0$ and assuming that $f_s(t) \ne 0$, we obtain Eq. 29

To verify maximum we compute the second derivative

$$\begin{aligned} A\frac{dV}{dt}&=f_s(t)\bigg (\int \limits _{y=-\infty }^{V/(1-{\gamma }_e)}(V-y)f_v(y|t)dy-{\gamma }_e\int \limits _{y=V/(1-{\gamma }_e)}^{\infty }yf_v(y|t)dy\bigg )\\ A\frac{d^2V}{dt^2}+\frac{dA}{dt}\frac{dV}{dt}&=\frac{df_s(t)}{dt}\bigg (\int \limits _{y=-\infty }^{V/(1-{\gamma }_e)}(V-y)f_v(y|t)dy-{\gamma }_e\int \limits _{y=V/(1-{\gamma }_e)}^{\infty }yf_v(y|t)dy\bigg )\\&+\,f_s(t)\frac{d\bigg (\int \limits _{y=-\infty }^{V/(1-{\gamma }_e)}(V-y)f_v(y|t)dy-{\gamma }_e\int \limits _{y=V/(1-{\gamma }_e)}^{\infty }yf_v(y|t)dy\bigg )}{dt}\\ \end{aligned}$$

Substituting $\frac{dV}{dt}=0$, hence $\int \limits _{y=-\infty }^{V/(1-{\gamma }_e)}(V-y)f_v(y|t)dy={\gamma }_e\int \limits _{y=V/(1-{\gamma }_e)}^{\infty }yf_v(y|t)dy$, we obtain

$$\begin{aligned} \frac{d^2V}{dt^2}&=\frac{f_s(t)\frac{d\bigg (\int \limits _{y=-\infty }^ {V/(1-{\gamma }_e)}(V-y)f_v(y|t)dy-{\gamma }_e\int \limits _{y=V/ (1-{\gamma }_e)}^{\infty }yf_v(y|t)dy\bigg )}{dt}}{A}\\ \end{aligned}$$

But since $A,f_s(t)>0$, it is enough to show that $\bigg (\int \limits _{y=-\infty }^{V/(1-{\gamma }_e)}(V-y)f_v(y|t)dy-{\gamma }_e\int \limits _{y=V/(1-{\gamma }_e)}^{\infty }yf_v(y|t)dy\bigg )$ decreases as $t$ increases. As part of the proof that the optimal strategy is based on reservation value, it was shown that:

$$\begin{aligned} \int \limits _{y=-\infty }^{V/(1-{\gamma }_e)}(y-V)f_v(y|t)dy+ {\gamma }_e\int \limits _{y=V/(1-{\gamma }_e)}^{\infty }yf_v(y|t)dy \end{aligned}$$

increases as $t$ increases.

Hence $\int _{y=-\infty }^{V/(1-{\gamma }_e)}(V-y)f_v(y|t)dy-{\gamma }_e\int _{y=V/(1-{\gamma }_e)}^{\infty }yf_v(y|t)dy$ decreases as $t$ increases.$\square$

Appendix 2: Nomenclature

Notation	Meaning
$s$	The drawn signal
$v$	The real value of the match
$f_v(y)$, $F_v(y)$	The probability density function and cumulative distribution function from which the real values of the match are drawn
$f_s(s)$, $F_s(s)$	The probability density function and cumulative distribution function from which the signals are drawn
$f_s(s|v)$, $F_s(s|v)$	The probability density function and cumulative distribution function from which the signals are drawn, given that the true value of the match is $v$
$f_v(y|s)$, $F_v(y|s)$	The probability density function and cumulative distribution function from which the real values of the matches are drawn, given that the drawn signal is $s$
$E{[}v|s{]}$	The expected value of the match, given that the drawn signal is $s$
$c_s$	The search cost. i.e., the cost of drawing another signal
$c_e$	The cost of querying the expert
$\gamma _e$	The commission (in percentages) paid to the expert
$V$	The expected utility of an agent, using the optimal strategy. By Bellman optimality principle, this value is also the threshold for the real value revealed by the expert
$t_l$	The lower threshold. Drawing a signal which is less than $t_l$, the agent rejects the match without querying the expert
$t_u$	The upper threshold. Drawing a signal which is greater than $t_u$, the agent accepts the match without querying the expert
$V(S,S^*)$	The expected utility of an agent in a noisy two-sided search with no expert present, where the agent accepts the match if and only if the signal she sees is in $S$, and the other agents accept the match if and only if the signal they see is in $S^*$
$V(S',S'')$	The expected utility of an agent in a noisy two sided search with an expert present, where the agent accepts the match if the signal she sees, denoted $s$, is in $S'$, rejects it if $s\in S''$, queries the expert if $s\notin S'\cup S''$, and whenever the expert is queried, the agent accepts if and only if the revealed value is greater or equal to $V(S,S'')$. This is defined for $S'\cap S''=\emptyset$
$d_e$	The cost to the expert of producing the exact value of a match
$\eta _{c_e}$	The expected number of expert queries a searcher performs, whenever the expert charges a fixed price $c_e$
$\eta _{\gamma _e}$	The expected number of expert queries a searcher performs, whenever the expert charges a fixed commission percentage $\gamma _e$
$\pi _e$	The expected utility of the expert
$W$	The expected social welfare, defined as $\pi _e+V$
$k$	The size of the match formed ($k=1$ implies a one-sided search, $k=2$ implies a two-sided-search etc.)