Reserve price signaling in first-price auctions with an uncertain number of bidders

Abstract

We study first-price auctions in which the number of bidders is the seller’s private information, and investigate the use of a reserve price to signal this private information. We use the D1 criterion to refine the set of equilibria and characterize a symmetric separating equilibrium outcome, where the reserve price increases with the number of bidders. The key driving force is a certain form of single-crossing property. With more bidders, the seller can afford a high reserve price that discourages competition in the auction, for the winning bid is more likely to be higher.

Appendices

Appendix 1

In this appendix section, first, we show four lemmas in order to prove Proposition 1.

Lemma 2

Let $r({\hat{n}},n)$ be a reserve price that satisfies $J(r({\hat{n}},n),{\hat{n}},n)=0$ given ${\hat{n}}$ and n. If Assumption 1holds, then we obtain

$$\begin{aligned} \frac{\partial U(r,{\hat{n}},n)}{\partial r} \gtreqless 0 \iff r \lesseqgtr r({\hat{n}},n). \end{aligned}$$

Proof

Since $\partial U(r,{\hat{n}},n)/\partial r = -f_n(r)J(r,{\hat{n}},n)$ and $J(r({\hat{n}},n),{\hat{n}},n)=0$, $\partial U(r,{\hat{n}},n)/\partial r=0$ holds at $r=r({\hat{n}},n)$. By Assumption 1, $J(r,{\hat{n}},n) \gtreqless 0$ holds if and only if the reserve price satisfies $r \gtreqless r({\hat{n}},n)$. $\square$

Lemma 3

The optimal reserve price given ${\hat{n}}$ and n decreases in ${\hat{n}}$ and increases in n:

$$\begin{aligned} \frac{\partial r({\hat{n}},n)}{\partial {\hat{n}}} = -\frac{\partial r({\hat{n}},n)}{\partial n} \le 0. \end{aligned}$$

The inequality is strict for ${\hat{n}} \not = n+1$.

Proof

The implicit function theorem yields

$$\begin{aligned} \frac{\partial r({\hat{n}},n)}{\partial {\hat{n}}} = -\frac{\partial J(r,{\hat{n}},n)/\partial {\hat{n}}}{\partial J(r,{\hat{n}},n)/\partial r}. \end{aligned}$$

Since $\partial J(r,{\hat{n}},n)/\partial r>0$ holds by Assumption 1, it suffices to prove that $\partial J(r,{\hat{n}},n)/\partial {\hat{n}} \ge 0$. First, we suppose that ${\hat{n}} \not = n+1$. By differentiating $J(r,{\hat{n}},n)$ with respect to ${\hat{n}}$, we obtain

$$\begin{aligned} \frac{\partial J(r,{\hat{n}},n)}{\partial {\hat{n}}}&= -\left( \frac{1}{n-{\hat{n}}+1} \right) ^2 \frac{F(r)}{f(r)} \frac{(n-{\hat{n}}+1)\log {F(r)}+1-F(r)^{n-{\hat{n}}+1}}{F(r)^{n-{\hat{n}}+1}} \\&= -\left( \frac{1}{n-{\hat{n}}+1} \right) ^2 \frac{F(r)}{f(r)} \frac{\log {F(r)^{ n-{\hat{n}}+1}}+1-F(r)^{n-{\hat{n}}+1}}{F(r)^{n-{\hat{n}}+1}}. \end{aligned}$$

For any ${\hat{n}} \not = n+1$, $\log {F(r)^{ n-{\hat{n}}+1}}+1-F(r)^{n-{\hat{n}}+1}<0$ holds whenever the reserve price satisfies $r<\bar{x}$. Hence, $\partial J(r,{\hat{n}},n)/\partial {\hat{n}}>0$ holds. Moreover, since $\partial J(r,{\hat{n}},n)/\partial n=- \partial J(r,{\hat{n}},n)/\partial {\hat{n}}$ holds, we obtain

$$\begin{aligned} \frac{\partial r({\hat{n}},n)}{\partial {\hat{n}}}=-\frac{\partial r({\hat{n}},n)}{\partial n}. \end{aligned}$$

Second, we suppose that ${\hat{n}}=n+1$. Since $\partial J(r,{\hat{n}},n)/\partial r=0$ holds, it follows that $\partial r({\hat{n}},n)/\partial {\hat{n}}=\partial r({\hat{n}},n)/\partial n=0$.$\square$

Lemma 4

The slope of the indifference curve given by Eq. (2) decreases in n. That is,

$$\begin{aligned} \frac{d}{d n} \left( \frac{d {\hat{n}}}{d r} \Big |_{U=const.} \right) <0. \end{aligned}$$

Proof

The numerator of the last line of Eq. (2) decreases in n, as shown in the proof of Lemma 3, and the denominator increases with n because $F_{n-1}(x)/ F_{n-1}(r)=(F(x)/F(r))^{n-1}$ and $F(x)/F(r)>1$ hold whenever $x>r$.$\square$

Lemma 5

For any symmetric separating PBE, a type 1 seller chooses $r^*$ to satisfy $J(r^*,n,n)=0$.

Proof

Suppose that $r_1 \not = r^*$ in a separating PBE. If ${\hat{n}}(r^*)=1$, then a type 1 seller benefits from choosing $r^*$ since $U(r_1,1,1)<U(r^*,1,1)=\max _{r}U(r,1,1)$ holds. If ${\hat{n}}(r^*)>1$, then a type 1 seller again benefits from choosing $r^*$ since $U(r_1,1,1)<U(r^*,1,1)<U(r^*,{\hat{n}}(r^*),1)$ holds. The last inequality holds because, for any r and n,

$$\begin{aligned} \frac{\partial U(r,{\hat{n}},n)}{\partial {\hat{n}}}=\int ^{\bar{x}}_{r}\frac{\partial b(x,r,{\hat{n}})}{\partial {\hat{n}}}f_n(x)dx>0. \end{aligned}$$

The inequality follows

$$\begin{aligned} \frac{\partial b(x,r,{\hat{n}})}{\partial {\hat{n}}}&= -\left( \frac{1}{F_{{\hat{n}}-1}(x)} \right) ^2 \\&\quad \times \left[ F_{{\hat{n}}-1}(x) \int ^{x}_{r}\frac{\partial F_{{\hat{n}}-1}(y)}{\partial {\hat{n}}}dy-\frac{\partial F_{{\hat{n}}-1}(x)}{\partial {\hat{n}}}\int ^{x}_{r}F_{{\hat{n}}-1}(y)dy \right] \\&= -\left( \frac{1}{F_{{\hat{n}}-1}(x)} \right) ^2 \\&\quad \times \left[ F_{{\hat{n}}-1}(x) \int ^{x}_{r} F_{{\hat{n}}-1}(y) \log {F(y)} dy - F_{{\hat{n}}-1}(x) \log {F(x)} \int ^{x}_{r}F_{{\hat{n}}-1}(y)dy \right] \\&= -\frac{1}{ F_{{\hat{n}}-1}(x)} \int ^{x}_{r} F_{{\hat{n}}-1}(y) [\log {F(y)}-\log {F(x)}]dy>0. \end{aligned}$$

$\square$

1.1 Proof of Proposition 1

Proof

Following the arguments of Riley (1979) and Cai et al. (2007), we see that, if there exists a unique separating equilibrium, then a schedule of reserve prices $(r_n)^N_{n=1}$ should be given by (i) $r_1=r^*$ that satisfies $J(r^*,n,n)=0$ and (ii) $r_n$ that satisfies $U(r_n,n,n)=U(r_{n+1},n+1,n)$ for $1\le n<N$. Note that $r_{n+1}>r_n$ for $1 \le n <N$.

In what follows, given the belief described in Proposition 1, we show that a type n seller has no profitable deviation. We separately consider three forms of deviations.

First, suppose that a type n seller chooses $r \in (r_n,r_{n+1})$. Bidders form a belief n by observing this reserve price. Since $r_n \ge r^*$ and U(r, n, n) decreases in r for $r>r^*$ by Lemma 2, $U(r_n,n,n)>U(r,n,n)$ holds for any $r \in (r_n,r_{n+1})$. Therefore, a type n seller cannot benefit from choosing $r \in (r_n,r_{n+1})$.

Second, suppose that a type n seller chooses $r \in [r_{n'},r_{n'+1})$ with $r_{n'+1}<r_n$. Bidders form a belief $n' \le n-1$ by observing this reserve price. Since $\partial U(r,{\hat{n}},n)/\partial {\hat{n}} >0$ holds, it may be that $U(r_n,n,n) > \max _{r} U(r,n',n)$ holds. In this case, a type n seller cannot profitably deviate by choosing $r \in [r_{n'},r_{n'+1})$.

Alternatively, we consider the case in which there exists an $r'$ that satisfies $U(r',n',n)=U(r_n,n,n)$. If two reserve prices satisfy this condition, we let $r'$ be the higher of the two reserve prices. Thus, $r' \ge r(n',n)$. The single-crossing condition (Lemma 4) implies that $r_{n'}>r'$. Thus, for any $r \in [r_{n'},r_{n'+1})$, we obtain

$$\begin{aligned} U(r_n,n,n)=U(r',n',n)>U(r,n',n). \end{aligned}$$

The inequality holds by Lemma 2 since $U(r,n',n)$ decreases in r for $r>r(n',n)$, and the above discussion yields $r_{n'}>r(n',n)$.

Moreover, if there existed an $r'<r^*$ that satisfied $U(r',1,n)=U(r_n,n,n)$, then a type n seller could benefit from choosing $r<r'$. However, $r' \ge r(1,n)>r_1=r^*$ holds by Lemma 2. Thus, we can exclude a deviation in which a type n seller chooses $r \in [\underline{x},r_1]$. Therefore, a type n seller cannot benefit from choosing $r<r_n$.

Finally, suppose that a type n seller chooses $r \in [r_{n'},r_{n'+1})$ with $r_{n'} \ge r_{n+1}$. The proof proceeds in two steps. First, we suppose that $n'=n+1$. That is, bidders form the belief $n+1$. A type n seller cannot benefit from choosing $r_{n+1}$ by construction. Moreover, for any $r \in (r_{n+1},r_{n+2})$, we obtain

$$\begin{aligned} U(r_n,n,n)=U(r_{n+1},n+1,n)>U(r,n+1,n). \end{aligned}$$

The inequality holds by Lemma 2 since $U(r,n+1,n)$ decreases in r for $r>r(n+1,n)$, and $r_{n+1}>r(n+1,n)$ holds.

Next, we suppose that $n'>n+1$. Let $r'$ be a reserve price that satisfies $U(r',n',n)=U(r_n,n,n)$. If two reserve prices satisfy this condition, we let $r'$ be the higher of the two reserve prices. Thus, $r'>r_n$. Since $\partial r({\hat{n}},n)/\partial {\hat{n}} <0$ by Lemma 3, we obtain $r_n \ge r_1=r(n,n)>r(n',n)$. The first inequality holds by construction and is strict for $n>1$. Moreover, $r_{n'}>r'$ by Lemma 4. Thus, for any $r \in [r_{n'},r_{n'+1})$, we obtain

$$\begin{aligned} U(r_n,n,n)=U(r',n',n)>U(r,n',n). \end{aligned}$$

The inequality holds by Lemma 2 since $U(r,n',n)$ decreases in r for $r>r(n',n)$, and the above discussion yields $r_{n'}>r(n',n)$. Therefore, a type n seller cannot benefit from choosing $r \ge r_{n+1}$.

The belief described in the proposition is consistent with the seller’s strategy. This completes the proof.$\square$

Appendix 2

Lemma 6

Let $\sigma (x)=f(x)/F(x)$ denote the reverse hazard rate. Eq. (1) increases with x (i.e., Assumption 1holds) if, for any x, ${\hat{n}}$, and n, $\sigma (x)$ satisfies

$$\begin{aligned} -\sigma (x)^2 (1+F(x)^{-(n-{\hat{n}}+1)}) \frac{n-{\hat{n}}+1}{F(x)^{-(n-{\hat{n}}+1)}}< \sigma '(x) < 0. \end{aligned}$$

Proof

For convenience, we let $z=n-{\hat{n}}+1$. First, suppose $z=0$. Then, we obtain

$$\begin{aligned} \frac{\partial J(x,{\hat{n}},n)}{\partial x}&= 1- \left( \frac{1}{f(x)} \right) ^2 [f(x)^2 (1+\log {F(x)})-f'(x) F(x) \log {F(x)}] \\&= 1-\frac{ f(x)^2 (1+\log {F(x)})-f'(x) F(x) \log {F(x)} }{f(x)^2} \\&= -\frac{ \log {F(x)} (f(x)^2-f'(x)F(x)) }{f(x)^2} \\&= \log {F(x)} \left( \frac{F(x)}{f(x)} \right) ^2 \frac{f'(x)F(x)-f(x)^2}{F(x)^2} \\&= \frac{ \sigma '(x) \log {F(x)} }{ \sigma (x)^2 }. \end{aligned}$$

Therefore, $\partial J/\partial x>0$ holds if $\sigma '(x)<0$.

Second, suppose $z \not = 0$. We obtain

$$\begin{aligned} \frac{\partial J(x,{\hat{n}},n)}{\partial x}&= 1-\frac{1}{z} \frac{ [(1-z)F(x)^{-z}-1]f(x)^2-f'(x)(F(x)^{1-z}-F(x)) }{ f(x)^2 } \\&= 1-\frac{1}{z} \frac{ -z f(x)^2 F(x)^{-z} + (F(x)^{-z}-1)f(x)^2-f'(x)F(x)(F(x)^{-z}-1) }{ f(x)^2 } \\&= 1+F(x)^{-z} - \frac{F(x)^{-z}-1}{z} \frac{f(x)^2-f'(x)F(x)}{f(x)^2} \\&= 1+F(x)^{-z} + \frac{F(x)^{-z}-1}{z} \frac{F(x)^2}{f(x)^2} \frac{f(x)^2-f'(x)F(x)}{F(x)^2} \\&= 1+F(x)^{-z} + \frac{F(x)^{-z}-1}{z} \frac{\sigma '(x)}{\sigma (x)^2}. \end{aligned}$$

Since $(F(x)^{-z}-1)/z>0$, $\partial J/\partial x>0$ holds if

$$\begin{aligned} \sigma '(x) > -\sigma (x)^2 (1+F(x)^{-z}) \frac{z}{F(x)^{-z}-1}. \end{aligned}$$

Note that the right-hand side is negative.$\square$

1.1 Proof of Lemma 1

Proof

Suppose that all bidders have the interim belief of $\rho =(\rho _1,\ldots , \rho _N)$ by observing r. We fix a bidder with valuation x. Suppose that all bidders other than this bidder follow $\bar{b}(x) \equiv b(x,r,\rho )$. This bidder chooses a bid p to maximize the expected payoff of

$$\begin{aligned} \sum ^{N}_{n=1}\rho _n (x-p)F_{n-1}(\bar{b}^{-1}(p)). \end{aligned}$$

Note that he takes his interim belief $\rho _n$ into consideration when computing the expected payoff. The first-order condition is given by

$$\begin{aligned} \sum ^{N}_{n=1}\rho _n \left[ -F_{n-1}(\bar{b}^{-1}(p))+(x-p)f_{n-1}(\bar{b}^{-1}(p))\frac{\partial \bar{b}^{-1}(p)}{\partial p} \right] =0. \end{aligned}$$

Since $p=\bar{b}(x)$ must hold in a symmetric equilibrium, we obtain

$$\begin{aligned}&\sum ^{N}_{n=1}\rho _n \left[ -F_{n-1}(x)+(x-\bar{b}(x))f_{n-1}(x)\frac{1}{\bar{b}'(x)} \right] =0 \\&\quad \Leftrightarrow \sum ^{N}_{n=1}\rho _n x f_{n-1}(x)=\sum ^{N}_{n=1}\rho _n[\bar{b}'(x)F_{n-1}(x)+\bar{b}(x)f_{n-1}(x)]=\sum ^{N}_{n=1}\rho _n(\bar{b}(x)F_{n-1}(x))'. \end{aligned}$$

By using $\bar{b}(r)=r$, we obtain

$$\begin{aligned} \bar{b}(x)&= \frac{1}{\sum ^{N}_{n=1}\rho _n F_{n-1}(x)} \left[ \sum ^{N}_{n=1}\rho _n r F_{n-1}(x)+\sum ^{N}_{n=1}\rho _n \int ^{x}_{r}y f_{n-1}(y)dy \right] \\&= x-\frac{1}{\sum ^{N}_{n=1}\rho _n F_{n-1}(x)}\int ^{x}_{r}\sum ^{N}_{n=1}\rho _n F_{n-1}(y)dy \\&= \sum ^{N}_{n=1} \frac{\rho _n F_{n-1}(x)}{\sum ^{N}_{m=1}\rho _m F_{m-1}(x)} b(x,r,n) \\&= \sum ^{N}_{n=1}w^{\rho }_n(x)b(x,r,n). \end{aligned}$$

$\square$

1.2 Proof of Proposition 2

Proof

We show that the equilibrium outcome described in Proposition 1 survives the D1 criterion. Given $\rho$, we consider ${\hat{n}}$ that satisfies $b(x,r,{\hat{n}})=b(x,r,\rho )$. Note that such an ${\hat{n}}$ may not be an integer. There exists the unique ${\hat{n}} \in [1,N]$ because

• every $\rho$ generates the unique $b(x,r,\rho )$,

• $b(x,r,\rho )$ is the weighted average of $b(x,r,{\hat{n}})$ for ${\hat{n}}=1,\ldots ,N$, and

• $b(x,r,{\hat{n}})$ increases with ${\hat{n}}$.

Thus, we can replace the set of a bidder’s best responses with the set of beliefs when considering the D1 criterion.

We define the following sets. For a reserve price r,

$$\begin{aligned}&N^0_n(r) \equiv \{ {\hat{n}} \in [1,N] | U(r,{\hat{n}},n)=U^*_n \}, \\&N_n(r) \equiv \{ {\hat{n}} \in [1,N] | U(r,{\hat{n}},n)>U^*_n \}. \end{aligned}$$

Note that each set is defined as a subset of [1, N] instead of $\{ 1,\ldots ,N \}$ and that $N^0_n(r)$ is a singleton. Moreover, $N^0_n(r)=\inf {N_n(r)}$ holds since $U(r,{\hat{n}},n)$ is a continuous and increasing function of ${\hat{n}}$.

Suppose that a type n seller can benefit from choosing r if a bidder’s belief is $n'$. Let ${\hat{n}}$ be the minimum of such $n'$s. Note that the type n seller’s payoff is greater for $n>{\hat{n}}$ since $b(x,r,{\hat{n}})$ and hence $U(r,{\hat{n}},n)$ increase with ${\hat{n}}$. Then, the D1 criterion implies that a bidder assigns a probability of one to the seller type for which the least number is the lowest. Therefore, we can replace Definition 3 with the following.

Definition 4

An equilibrium outcome survives the D1 criterion if and only if for every off-the-equilibrium reserve price r, the equilibrium can be supported with belief $\rho _{n'}=0$ whenever, for $N_n(r)\not = \emptyset$,

$$\begin{aligned} \inf {N_n}(r)<N^0_{n'}(r). \end{aligned}$$

In what follows, we show that the equilibrium outcome described in Proposition 1 passes the test given by Definition 4. We prove that $\rho _{n'}=0$ holds for any $r \in (r_n,r_{n+1})$ and $n' \not = n$ (i.e., ${\hat{n}}(r)=n$ holds). We define $U^{-1}({\tilde{U}}|r,n)={\hat{n}}$ for ${\tilde{U}}=U(r,{\hat{n}},n)$. Note that $U^{-1}(\cdot | r,n)$ is an increasing function of ${\hat{n}}$ for any $r \ge r^*$ and n since $U(r,{\hat{n}},n)$ increases with ${\hat{n}}$.

First, suppose that $n' \le n-1$. Since the profile constitutes an equilibrium, $U^*_{n'}=U(r_{n'},n',n') \ge U(r_n,n,n')$ holds. The inequality is strict for $n'<n-1$. Thus, we obtain $U^{-1}(U^*_{n'}|r_n,n') \ge n = U^{-1}(U^*_n|r_n,n)$. Note that $N^0_{n'}(r_n)=U^{-1}(U^*_{n'}|r_n,n')$ and $N^0_n(r_n)=n$.

Since $\partial U(r,{\hat{n}},n)/\partial r<0$ holds for $r>r({\hat{n}},n)$, $U^{-1}({\tilde{U}}|r,n)$ increases with r. By Lemma 3, $r_n \ge r(n,n)>r(n,n')$ holds for $n'<n$. Therefore, the single-crossing condition implies that $U^{-1}(U^*_{n'}|r,n')>U^{-1}(U^*_n|r,n)$ for $r \in (r_n,r_{n+1})$. This inequality implies that $\inf {N_n(r)}<N^0_{n'}(r)$ for $n' \le n-1$. Therefore, $\rho _{n'}=0$ should hold for $n' \le n-1$.

Second, suppose that $n' \ge n+1$. We prove that $\rho _{n'}=0$ by mathematical induction. Thus, for the first step, suppose that $n'=n+1$. Since $U^*_n=U(r_{n+1},n+1,n)$ and $U^*_{n+1}=U(r_{n+1},n+1,n+1)$ hold in equilibrium, we obtain $U^{-1}(U^*_n|r_{n+1},n)=U^{-1}(U^*_{n+1}|r_{n+1},n+1)=n+1$. Since $r_{n+1}>r_n \ge r^*$ implies that $\partial U(r,n,n)/\partial r<0$ and $\partial U(r,n+1,n+1)/\partial r<0$ hold for $r>r_n$, we obtain $\partial U^{-1}(U^*_n|r,n)/\partial r>0$ and $\partial U^{-1}(U^*_{n+1}|r,n)/\partial r>0$ for $r>r_n$. Moreover, the single-crossing condition implies that $\partial ^2 U^{-1}/\partial r \partial n<0$. Thus, for $r \in (r_n,r_{n+1})$, we obtain $U^{-1}(U^*_n|r,n)<U^{-1}(U^*_{n+1}|r,n+1)<n+1$.

Next, we suppose that, for $n'=n+k$, $U^{-1}(U^*_n|r,n)<U^{-1}(U^*_{n+k}|r,n+k)$ holds for $r \in (r_n,r_{n+1})$. Since a similar discussion implies that $U^{-1}(U^*_n|r_{n+k},n+k)=U^{-1}(U^*_{n+k+1}|r_{n+k+1},n+k+1)=n+k+1$ holds, we obtain $U^{-1}(U^*_{n+k}|r,n+k)<U^{-1}(U^*_{n+k+1}|r,n+k+1)<n+k+1$ for $r<r_{n+k+1}$. Since $r_{n+1}<r_{n+k+1}$, we obtain $U^{-1}(U^*_{n}|r,n)<U^{-1}(U^*_{n+k+1}|r,n+k+1)$ for $r \in (r_n,r_{n+1})$. Consequently, by mathematical induction, we obtain $U^{-1}(U^*_n|r,n)<U^{-1}(U^*_{n'}|r,n)$ for $r \in (r_n,r_{n+1})$. This inequality implies that $\inf {N_n(r)}<N^0_{n'}(r)$ for $n' \ge n+1$. Therefore, $\rho _{n'}=0$ should hold for $n' \ge n-1$. $\square$

1.3 Proof of the payoff equivalence for secret and public reserve prices

Proof

First, we rewrite the weight function as follows.

$$\begin{aligned} w_n(x)&= \frac{\rho ^0_n F_{n-1}(x)}{\sum ^{N}_{m=1}\rho ^0_m F_{m-1}(x)}\\&= \frac{\left( \sum ^{N}_{\ell =1}\ell g_{\ell } \right) \rho ^0_n F_{n-1}(x)}{ \left( \sum ^{N}_{\ell =1}\ell g_{\ell } \right) \sum ^{N}_{m=1}\rho ^0_m F_{m-1}(x)} \\&= \frac{\left( \rho ^0_n \sum ^{N}_{\ell =1}\ell g_{\ell } \right) F_{n-1}(x)}{\sum ^{N}_{m=1} \left( \rho ^0_m \sum ^{N}_{\ell =1}\ell g_{\ell } \right) F_{m-1}(x)} \\&= \frac{n g_n F_{n-1}(x)}{\sum ^{N}_{m=1} m g_m F_{m-1}(x)} \\&= \frac{g_n n F_{n-1}(x) f(x)}{\sum ^{N}_{m=1} g_m m F_{m-1}(x) f(x)} \\&= \frac{g_n f_n(x)}{\sum ^{N}_{m=1} g_m f_m(x)}. \end{aligned}$$

The third equality from the top follows $n g_n=\rho ^0_n \sum ^{N}_{m=1}m g_m$. Using this relation, we obtain

$$\begin{aligned} EU^{secret}(r)&= \sum ^{N}_{n=1}g_n \int ^{\bar{x}}_{r} \left( \sum ^{N}_{m=1}w_m(x)b(x,r,m) \right) f_n(x)dx \\&= \int ^{\bar{x}}_{r} \sum ^{N}_{n=1}g_n f_n(x) \left( \sum ^{N}_{m=1}w_m(x)b(x,r,m) \right) dx \\&= \int ^{\bar{x}}_{r} \sum ^{N}_{n=1}g_n f_n(x) \left( \sum ^{N}_{m=1} \frac{g_m f_m(x)}{\sum ^{N}_{\ell =1} g_{\ell } f_{\ell }(x)} b(x,r,m) \right) dx \\&= \int ^{\bar{x}}_{r} \sum ^{N}_{n=1}g_n f_n(x)b(x,r,n)dx \\&= \sum ^{N}_{n=1}g_n \int ^{\bar{x}}_{r}b(x,r,n)f_n(x)dx = EU^{public}(r). \end{aligned}$$

$\square$