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Quadratic funding with incomplete information

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Abstract

Quadratic funding is a public good provision mechanism that satisfies desirable theoretical properties, such as efficiency under complete information, and has been gaining popularity in practical applications. We evaluate this mechanism in a setting of incomplete information regarding individual preferences, and show that this result only holds under knife-edge conditions. We also estimate the inefficiency of the mechanism in a variety of settings, and characterize circumstances in which inefficiency increases with population size. We show how these findings can be used to estimate the mechanism’s inefficiency in a wide range of situations under incomplete information.

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Notes

  1. Gitcoin (https://gitcoin.co/fund) and HackerLink (https://hackerlink.io/en) are two of the main platforms that currently use QF for funding software development, and WeTrust (https://blog.wetrust.io/conclusion-of-the-first-lr-experiment-709b018b5f83) has used QF for matching donations.

  2. For an application by within the Democratic Party in the United States, see https://www.wired.com/story/colorado-quadratic-voting-experiment/

  3. While the topic of budget balance falls outside of the scope of this paper, it is worth noting that the quadratic funding mechanism runs a deficit whenever two or more individuals contribute to the public good. This implies that the mechanism faces the challenge of raising revenue for funding its deficit in ways that do not distort individual incentives when choosing a public good contribution. In practice, most applications of this mechanism have solved this issue by relying on philanthropic funding for the matching.

  4. To simplify notation, in the results and proofs in this section, we omit the type profiles from all functions.

  5. It is worth noting the role that quasi-linearity of the utility function plays in this efficiency result. It allows us to write the total welfare as \(\sum _{i = 1}^I v_i(F) - F\), so the characterization of the (interior) efficient allocation is \(\sum _{i = 1}^I v_i'(F) =1\). At the same time, the characterization of individual contributions under the funding mechanism \(\varPhi \) is given by \(v_i(\varPhi (\textbf{c})) = (\varPhi _i(\textbf{c}))^{-1}\). Therefore, satisfying the condition \(\sum _{i=1}^I\left( \varPhi _i^{\prime }(c)\right) ^{-1}=1\) (which QF satisfies) implies efficiency of the equilibrium. If the utility is not quasi-linear, it would not be possible to restore the efficiency with that mechanism, since the wealth effect of each individual and her wealth level will affect the funding mechanism outcome. For an alternative intuitive derivation of QF, see (Buterin et al. 2019) section 4.2.

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Acknowledgements

Wilfredo L. Maldonado would like to thank the Fundação Instituto de Pesquisas Econômicas - FIPE and the CNPq of Brazil 306473/2018-6, for financial support. We thank Gustav Alexandrie, Maya Eden, Loren Fryxell, Zoë Hitzig, Rossa O’Keeffe-O’Donovan, Lennart Stern, Benjamin Tereick, seminar participants at the Global Priorities Institute, and two anonymous referees for valuable comments.

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Appendices

Appendix A: Proofs

Proof

(Proposition 2.1) Let \( v: \mathbb {R} _+ \times \varTheta \rightarrow \mathbb {R} \) be defined by \( v(F;\theta ) = \sum _{i=1}^{I} v_i(F;\theta _{i}) \), and fix \( \theta \in \varTheta \). By assumption 1, we have that \( v(\cdot ; \theta ) \) is strictly concave and continuously differentiable, and that \( \lim _{F \rightarrow \infty } v'(F; \theta ) = 0\). We then have two cases: \( v'(0; \theta ) \le 1 \) or \( v'(0; \theta ) > 1 \). In the first case, it follows from the first order condition of definition 5 that \( F = 0 \) is efficient. Furthermore, since \(v(\cdot ; \theta )\) is a strictly concave function, we have that \( v'(F; \theta ) < 1 \) for all \( F > 0 \) and thus, again from definition 5, there can be no efficient provision \( F > 0 \). Thus, \( F^{e}(\theta ) = 0 \) is the unique efficient provision.

Now suppose that \( v'(0; \theta ) > 1 \). It follows that \( F = 0 \) is not efficient. Since \( \lim _{F \rightarrow \infty } v'(F; \theta ) = 0\), there exists \( A \in \mathbb {R} \) such that \( v'(A; \theta ) < 1 \). Thus, since \( v'(\cdot ; \theta ) \) is continuous, \( v'(0; \theta ) > 1 \) and \( v'(A; \theta ) < 1 \), it follows from the intermediate value theorem that there exists \( 0< B < A \) such that \( v'(B; \theta ) = 1 \). Additionally, since \( v'(\cdot ; \theta ) \) is strictly decreasing, we have that \( v'(F; \theta ) \ne 1 \) for all \( F \ne B \). Therefore, \( F^{e}(\theta ) = B \) is the unique efficient provision.

For any \( \theta \in \varTheta \), we then have that in both cases there is a unique efficient provision \( F^{e}(\theta ) \ge 0 \). Thus, we construct aunique function \( F^{e}: \varTheta \rightarrow \mathbb {R} _+ \) that maps each profile of types to its efficient funding, as desired. \(\square \)

Proof

(Proposition 2.2) We begin by proving the following claims.

Claim A.1

Suppose that, for all \( i \in \mathbb {I}\) and \( \theta _i \in \varTheta _i \), we have that \( v_i(\cdot ; \theta _i) \) is a continuously differentiable, strictly increasing and strictly concave function. Then, for any \( i \in \mathbb {I}\) and \( \textbf{c}_{-i} \in \mathbb {R} ^{I-1}_+ \), the best-response correspondence \( c_i(\cdot ; \textbf{c}_{-i}): \varTheta _i \rightarrow \mathbb {R} _+ \) is at most single valued.

Proof

(Claim A.1) Without loss of generality, let us consider \(i=1\). Let \( \theta _1 \in \varTheta _1 \) be the type of this individual, and suppose that there exist \( a,b \in \mathbb {R} _+ \), \( a < b \), best-responses to \( \textbf{c}_{-i} \). Let \( \varepsilon = b^{1/2}-a^{1/2} > 0 \). From the first order conditions, we have

$$\begin{aligned}a^{1/2}&= E \left[ v_1'\left( \left[ a^{1/2} + \sum _{i=2}^{I} (c_i(\theta _i))^{1/2} \right] ^2; \theta _i \right) \cdot \left[ a^{1/2} + \sum _{i=2}^{I} (c_i(\theta _i))^{1/2} \right] \,\vert \,\theta _i \right] \\&= a^{1/2} E \left[ v_1'\left( \left[ a^{1/2} + \sum _{i=2}^{I} (c_i(\theta _i))^{1/2} \right] ^2; \theta _i \right) \,\vert \,\theta _i \right] \\&\quad + E \left[ v_1'\left( \left[ a^{1/2} + \sum _{i=2}^{I} (c_i(\theta _i))^{1/2} \right] ^2; \theta _i \right) \cdot \left[ \sum _{i=2}^{I} (c_i(\theta _i))^{1/2} \right] \,\vert \,\theta _i \right] . \end{aligned}$$

Notice that the term in the third line above is nonnegative; thus, the equality above is satisfied only if \( E \left[ v_1'\left( \left[ a^{1/2} + \sum _{i=2}^{I} (c_i(\theta _i))^{1/2} \right] ^2; \theta _i \right) \,\vert \,\theta _i \right] \le 1 \). Thus, since \( b > a \) and \( v_1' \) is strictly decreasing, we have that

$$\begin{aligned}&E \left[ v_1'\left( \left[ b^{1/2} + \sum _{i=2}^{I} (c_i(\theta _i))^{1/2} \right] ^2; \theta _i \right) \cdot \left[ \sum _{i=2}^{I} (c_i(\theta _i))^{1/2} \right] \,\vert \,\theta _i \right] \\&< E \left[ v_1'\left( \left[ a^{1/2} + \sum _{i=2}^{I} (c_i(\theta _i))^{1/2} \right] ^2; \theta _i \right) \cdot \left[ \sum _{i=2}^{I} (c_i(\theta _i))^{1/2} \right] \,\vert \,\theta _i \right] , \end{aligned}$$

and

$$\begin{aligned}&b^{1/2} E \left[ v_1'\left( \left[ b^{1/2} + \sum _{i=2}^{I} (c_i(\theta _i))^{1/2} \right] ^2; \theta _i \right) \,\vert \,\theta _i \right] \\&< a^{1/2} E \left[ v_1'\left( \left[ a^{1/2} + \sum _{i=2}^{I} (c_i(\theta _i))^{1/2} \right] ^2; \theta _i \right) \,\vert \,\theta _i \right] + \varepsilon . \end{aligned}$$

Thus, adding the inequalities and using the fact that a and b satisfy the first order conditions, it follows that \( a^{1/2} + \varepsilon > b^{1/2} \). But this contradicts the definition of \( \varepsilon \). This contradiction completes the proof. \(\square \)

Claim A.2

Suppose that, for all \( i \in \mathbb {I}\) and \( \theta _i \in \varTheta _i \), we have that \( v_i(\cdot ; \theta _i) \) is a strictly concave and continuously differentiable function, and \( \lim _{F \rightarrow \infty } v_i'(F;\theta _i) = 0 \). Then, there exists \( A > 0 \) such that, for any \( i \in \mathbb {I}\) and \( \theta _i \in \varTheta _i \), if the contributions of all other individuals belong to [0, A] for any profile of types, then i’s best-response also belongs to this interval.

Proof

(Claim A.2) Under the given hypotheses, we can then define, for each \( i \in \mathbb {I}\), a function \( f_i: \varTheta _i \rightarrow \mathbb {R} _+ \) mapping each \( \theta _i \) to \( f(\theta _i) > 0 \) such that \( v_i'(f(\theta _i);\theta _i) < 1/I\). Now, letting \( A:= \max \{f_i(\theta _i); i \in \mathbb {I}, \theta _i \in \varTheta _i \} \), by the hypothesis of strict concavity of \(v_i\) it follows that \( v_i'(A;\theta _i) < 1/I\), for all \( i \in \mathbb {I}\) and all \( \theta _i \in \varTheta _i \). Now, suppose that, for some \( j \in \mathbb {I}\), we have that \( c_i(\theta _i) \in [0,A] \) for all \( i \ne j \) and all \( \theta _i \in \varTheta _i \), and the best-response of j with type \(\theta _j \in \varTheta _j\) is \( k > A.\) Since \( k > 0 \), the first order conditions for individual j’s problem imply that

$$\begin{aligned} \begin{aligned} k^{1/2}&= E \left[ v_j' \left( \left[ k^{1/2} + \sum _{i \ne j} (c_{i}(\theta _{i}))^{1/2} \right] ^2 ; \theta _{j}\right) \cdot \left[ k^{1/2} + \sum _{i \ne j} (c_{i}(\theta _{i}))^{1/2} \right] \,\vert \,\theta _{j} \right] \\&< \frac{1}{I} E \left[ k^{1/2} + \sum _{i \ne j} (c_{i}(\theta _{i}))^{1/2} \,\vert \,\theta _{j} \right] \\&< \frac{1}{I} I\cdot k^{1/2}, \end{aligned} \end{aligned}$$

where the first inequality follows from \( v_j'(A;\theta _j) < 1/I\) and \( v_j(\cdot ;\theta _j) \) being strictly concave, and the second inequality follows from \( c_i(\theta _i) \in [0,A] \) for all \( i \ne j \). It results that \( k < k \); this contradiction completes the proof. \(\square \)

Now, we return to the proof of the proposition. For each individual \( i \in \mathbb {I}\), let \( \vert \varTheta _i \vert = L_i \). We can restrict the domain and range of the best-response functions to the interval [0, A] , using \( A>0 \) given in Claim A.2. Thus, the problem of individual i with type \( \theta _i \) is

$$\begin{aligned} \max _{c_i \in [0,A]} E\left[ v_i\left( \varPhi ^{QF}(c_i, \textbf{c}_{-i}(\theta _{-i})); \theta _i \right) \,\vert \,\theta _i \right] - c_i. \end{aligned}$$

Notice that the function that is being maximized is continuous and the feasibility correspondence is continuous and compact valued (it is the constant interval [0, A]). Thus, by the Theorem of the Maximum (Berge 1963, ch. 6), we have that the best response correspondence for i is not empty and is upper hemicontinuous. Additionally, by Claim A.1, we can conclude that it is a continuous function. Hence, the whole game best-response function \( BR: [0,A]^{\sum _{i = 1}^{I} L_i} \rightarrow [0,A]^{\sum _{i = 1}^{I} L_i} \) is also continuous. Since \( [0,A]^{\sum _{i = 1}^{I} L_i} \) is a compact and non-empty set, the Brouwer fixed point theorem (Milnor 1965) allows us to conclude that BR has a fixed point, which is clearly an equilibrium for QF. \(\square \)

Proof

(Proposition 3.2) By proposition 2.1, we know that the hypotheses adopted here ensure a unique efficient provision \( F^{e}\ge 0 \). There are two cases: \( F^{e}= 0 \) or \( F^{e}> 0 \). First, suppose we have \( F^{e}= 0 \). We are going to show that \( \textbf{0} \) is an equilibrium for \( \varPhi ^{QF}\). Consider the problem faced by some individual \( i \in \mathbb {I}\) when all other individuals are contributing zero to the mechanism:

$$\begin{aligned} \max _{c_i \ge 0} v_i \left( \left[ c_i^{1/2} + 0 \right] ^{2} \right) - c_i. \end{aligned}$$

Which can be written as

$$\begin{aligned} \max _{c_i \ge 0} v_i \left( c_i \right) - c_i. \end{aligned}$$

Thus, the first order condition for the individual i is that \( v_i'(c_i) \le 1 \), with equality holding when \( c_i > 0 \). But note that, since \( F^{e}= 0 \), we have from definition 5 that \( \sum _{j = 1}^{I} v_j'(0) \le 1 \). In particular, since \( v_j \) is increasing for all \( j \in \mathbb {I}\), this implies that \( v_i'(0) \le 1 \). Thus, \( c_i = 0 \) satisfies the first order condition for i. But since the choice of i was arbitrary, we have that \( \textbf{0} \) is an equilibrium for \( \varPhi ^{QF}\).

Now, suppose \( F^{e}> 0 \). For all \( i \in \mathbb {I}\), let

$$\begin{aligned} c_i = \left( v_i'(F^{e}) \cdot (F^{e})^{1/2}\right) ^2. \end{aligned}$$
(6)

It is easy to check that \(F^{e}= \varPhi ^{QF}(\textbf{c}) = [\sum _{i=1}^{I} c_i^{1/2}]^2.\) Rearranging, we obtain

$$\begin{aligned} v_i'(F^{e}) = \frac{(c_{i})^{1/2}}{(F^{e})^{1/2}}, \end{aligned}$$

which is precisely the first order condition for the individual i’s optimization problem in the QF mechanism. We can thus conclude that the vector \( \textbf{c}= (c_1, \cdots , c_{I}) \) as defined by equation (6) is an equilibrium of QF and its provision is efficient.

Thus, in all cases, there exists an equilibrium allocation \( \textbf{c}^* \) such that \( \varPhi ^{QF}(\textbf{c}^*) = F^{e}\), as we wanted to show. \(\square \)

Proof

(Proposition 3.3) We begin by proving the following claim.

Claim A.3

Suppose that, for all \( i \in \mathbb {I}\), we have that \( v_i \) is continuously differentiable and strictly increasing. Then, the only possible corner equilibrium for the quadratic funding mechanism is the allocation \( \textbf{0} \in \mathbb {R}^n_+ \).

Proof

Suppose by contradiction that, for all \( i \in \mathbb {I}\), we have that \( v_i \) is continuously differentiable and strictly increasing, and that there exists an equilibrium strategy profile \( \textbf{c} \ne \textbf{0} \) such that \( c_j = 0 \) for some \( j \in \mathbb {I}\). The first order condition for j is given by

$$\begin{aligned} \frac{v_j'(\varPhi ^{QF}(\textbf{c})) \cdot \left[ \sum _{i = 1}^n c_i^{1/2} \right] }{c_j^{1/2}} \le 1. \end{aligned}$$
(7)

But note that, since the aggregate contribution is strictly positive, and since \( v_j \) is strictly increasing, \( c_j = 0 \) is the denominator with a fraction with a strictly positive numerator. Hence, the inequality in (7) cannot hold, implying that \( c_j = 0 \) does not satisfy the first order condition for j. This contradiction completes the proof. \(\square \)

Now, we proceed to the main proof. First, suppose that, at \(F = 0\), aggregate marginal utility is greater than one, that is, that \( \sum _{i = 1}^Iv_i'(0) > 1 \), and that no individual has marginal utility greater than one at a zero provision level, that is, \( v_i'(0) \le 1 \) for all \( i \in \mathbb {I}\). Now, consider the problem that some individual \( i \in \mathbb {I}\) faces when every other individual contributes zero to the public good, that is, \( c_j = 0 \) for all \( j \ne i \). This problem can be simply written as

$$\begin{aligned} \max _{c_i \ge 0} v_i \left( c_i \right) - c_i. \end{aligned}$$
(8)

The first order condition is thus that \( v_i'(c_i) \le 1 \), with equality holding when \( c_i > 0\). But since \( v_i'(0) \le 1 \), we have that \( c_i^* = 0 \) is a solution. Since the choice of i was arbitrary, we have that \( \textbf{0} \) is a Nash equilibrium of this game, which is not optimal by hypothesis.

Now, suppose by contradiction that there exists an inefficient equilibrium, and that it is not true that the conditions \( \sum _{i = 1}^Iv_i'(0) > 1 \) and \( v_i'(0) \le 1 \) for all \( i \in \mathbb {I}\) simultaneously hold. There are two cases: zero is the efficient provision level, or there exists \( i \in \mathbb {I}\) such that \( v_i'(0) > 1 \). If zero is the efficient provision, then it follows from claim A.3 that the inefficient equilibrium is interior. On the other hand, suppose that there exists \( i \in \mathbb {I}\) such that \( v_i'(0) > 1 \). Then, if all \( j \ne i \) contribute zero to the public good, the problem for i is, as in (8),

$$\begin{aligned} \max _{c_i \ge 0} v_i \left( c_i \right) - c_i. \end{aligned}$$

But note that, since \( v_i'(0) > 1 \), \( c_i = 0 \) does not satisfy the first order condition \( v_i'(c_i) \le 1 \). Hence, \( \textbf{0} \) is not an equilibrium. But then, again by claim A.3, the inefficient equilibrium must be interior. Therefore, in both cases, the inefficient equilibrium must be interior. But this contradicts proposition 3.1. This contradiction completes the proof. \(\square \)

Proof

(Proposition 4.1) First, suppose that \( F^{e}(\theta ) = 0 \) for all \( \theta \in \varTheta \). Let us prove that \( \textbf{c}(\theta ) = 0 \) for all \( \theta \) is an equilibrium for QF. Suppose that, for some \( i \in \mathbb {I}\) we have that all other individuals are playing the strategy profile \( \textbf{c}_{-i}(\theta _{-i}) = 0 \). The problem faced by the individual i with some type \( \theta _i \in \varTheta _i \) is given by

$$\begin{aligned} \max _{c_i \ge 0} E\left[ v_i\left( \varPhi ^{QF}(c_i, \textbf{c}_{-i}(\theta _{-i})); \theta _i \right) \,\vert \,\theta _i \right] - c_i. \end{aligned}$$

Substituting \( \textbf{c}_{-i}(\theta _{-i}) = 0 \) and the definition of QF, the problem above becomes

$$\begin{aligned} \max _{c_i \ge 0} v_i\left( c_i ; \theta _i \right) - c_i, \end{aligned}$$

let \(c_i(\theta _i)\) be the solution, then the first order conditions are

$$\begin{aligned} v_i'(c_i(\theta _i); \theta _i) \le 1, \end{aligned}$$

with equality holding if \( c_i (\theta _i) > 0 \). Since \( F^{e}(\theta ) = 0 \) for all \( \theta \in \varTheta \), it implies that \( \sum _{j = 1}^{I} v_j'(0;\theta _{j}) \le 1 \). It follows that \( v_i'(0;\theta _{i}) \le 1 \). Thus, \( c_i(\theta _i) = 0 \) satisfies the first order conditions for i, as we wanted to show.

Now, we prove the reciprocal using a contradiction argument. The hypothesis asserts that there exist \( \theta ' \in \varTheta \) for which \( F^{e}(\theta ') = 0 \) and that QF is efficient. Then, suppose that there is a \( \theta '' \in \varTheta , \) such that \( F^{e}(\theta '') > 0 \). Let \(\textbf{c}\) be an efficient equilibrium for QF. The first order condition for individual i’s problem for the type profile \( \theta '' \) is

$$\begin{aligned} E\left[ v_i'(\varPhi ^{QF}(\textbf{c}^*(\theta ))) ; \theta ''_i) \cdot \left( \frac{\varPhi ^{QF}(\textbf{c}^*(\theta ))}{c^*_i(\theta ''_i)}\right) ^{1/2} \,\vert \,\theta ''_{i}\right] \le 1, \end{aligned}$$
(9)

which cannot be satisfied for \( c^*_i(\theta ''_i) = 0 \), since \( \Pr (\theta '') > 0, \) \( F^*(\theta '') = F^{e}(\theta '') > 0 \), and \( v_i'(F^*(\theta ''); \theta _i'') > 0. \) Therefore, \( c^*_i(\theta ''_i) > 0.\) Since i is arbitrary, it results \( \textbf{c}^*(\theta '') \gg \textbf{0} \). Now, let us consider the first order condition of individual i’s problem when her type is \(\theta _i'\),

$$\begin{aligned} E\left[ v_i'(\varPhi ^{QF}(\textbf{c}^*(\theta ))) ; \theta '_i) \cdot \left( \frac{\varPhi ^{QF}(\textbf{c}^*(\theta ))}{c^*_i(\theta '_i)}\right) ^{1/2} \,\vert \,\theta '_{i}\right] \le 1, \end{aligned}$$
(10)

with equality holding when \( c^*_i(\theta '_i) > 0 \). Since \( \Pr (\theta '_i, \theta ''_{-i}) > 0 \), and \( \textbf{c}^*_{-i}(\theta ''_{-i}) \gg \textbf{0} \), we have that \( c^*_i(\theta '_i) > 0,\) because the numerator of the expected value when \(\theta = (\theta '_i, \theta ''_{-i})\) is strictly positive. Thus, \( c^*_i(\theta '_i) > 0 \). But then, \( \varPhi ^{QF}(\textbf{c}(\theta ')) > 0 = F^{e}(\theta ') \), which is a contradiction to the efficiency of QF. This completes the proof. \(\square \)

Proof

(Proposition 4.2) Without loss of generality, let \(j=1\). First, suppose that QF is efficient, that is, there exists an equilibrium strategy profile \( \textbf{c}^* \) such that \(\varPhi ^{QF}(\textbf{c}^*(\theta )) = F^{e}(\theta ) \). For \( \theta \in \varTheta \), the efficiency of \( F^{e}(\theta ) > 0 \) implies

$$\begin{aligned} v_1'(F^{e}(\theta ); \theta _1) + \sum _{i = 2}^{I} v_i'(F^{e}(\theta ); \theta _i^1) = 1. \end{aligned}$$

Thus,

$$\begin{aligned} \sum _{i = 2}^{I} v_i'(F^{e}(\theta ); \theta _i^1) = 1 - v_1'(F^{e}(\theta ); \theta _1). \end{aligned}$$
(11)

Since \(j=1\) has complete information regarding the other individuals’ preferences, it follows that

$$\begin{aligned} v_1'(F^{e}(\theta );\theta _1) = \frac{(c_1(\theta ))^{1/2}}{(F^{e}(\theta ))^{1/2}}. \end{aligned}$$
(12)

Thus, equations (11) and (12) imply that

$$\begin{aligned} \sum _{i = 2}^{I} v_i'(F^{e}(\theta ); \theta _i^1) = \frac{\sum _{i = 2}^{I} (c_i(\theta _i^1))^{1/2}}{(F^{e}(\theta ))^{1/2}}, \end{aligned}$$

and so letting \( \alpha = \sum _{i = 2}^{I} (c_i(\theta _i^1))^{1/2} \) yields the desired result.

To prove the converse, suppose that there exists a constant \( \alpha \in \mathbb {R} \) such that \( \sum _{i = 2}^{I} v_i'(F^{e}(\theta ); \theta _i^1) = \alpha (F^{e}(\theta ))^{-1/2} \). Define the contribution of individual \( i \in \mathbb {I}\) with type \( \theta _i \in \varTheta _i \) as

$$\begin{aligned} c_i(\theta _i) = E \left( \left[ v_i'(F^{e}(\theta )) \cdot (F^{e}(\theta ))^{1/2} \,\vert \,\theta _i^1 \right] \right) ^2, \end{aligned}$$
(13)

which is well-defined since \( F^{e}(\theta ) \) exists and is unique. Note that these contributions satisfy the first order conditions if the generated level of the public good is equal to \(F^{e}(\theta )\), so let us prove this equality.

Since each individual \(i=2,\ldots ,n\) has only one single type, the conditional expectation in equation (13) is equal to the unconditional expectation. Taking the square root of both sides and taking the sum yields

$$\begin{aligned} \sum _{i=2}^{I} (c_i(\theta _i^1))^{1/2}&= \sum _{i=2}^{I} E \left[ v_i'(F^{e}(\theta )) \cdot (F^{e}(\theta ))^{1/2} \right] \\&= E \left[ (F^{e}(\theta ))^{1/2} \sum _{i=2}^{I} v_i'(F^{e}(\theta )) \right] \\&= E \left[ (F^{e}(\theta ))^{1/2} \frac{\alpha }{(F^{e}(\theta ))^{1/2}} \right] \\&= \alpha . \end{aligned}$$

Thus, substituting the left-hand side on \( \sum _{i = 2}^{I} v_i'(F^{e}(\theta ); \theta _i^1) = \alpha (F^{e}(\theta ))^{-1/2} \) we get

$$\begin{aligned} \sum _{i = 2}^{I} v_i'(F^{e}(\theta ); \theta _i^1) = \frac{\sum _{i=2}^{I} (c_i(\theta _i^1))^{1/2}}{(F^{e}(\theta ))^{1/2}} \end{aligned}$$
(14)

On the other hand, since the individual 1 has complete information, rearranging equation (13) we get

$$\begin{aligned} v_1'(F^{e}(\theta ); \theta _1) = \frac{(c_1(\theta _1))^{1/2}}{(F^{e}(\theta ))^{1/2}}. \end{aligned}$$
(15)

Finally, adding equations (14) and (15), and using the fact that \( \sum _{i = 1}^{I} v_i'(F^{e}(\theta ); \theta _i) = 1 \) (because \(F^{e}(\theta )\) is the efficient provision), we get that \(\varPhi ^{QF}(\textbf{c}(\theta )) = F^{e}(\theta ) \). Thus, it follows that the contributions specified in equation (13) do indeed satisfy the first order conditions of an equilibrium, and that this equilibrium is efficient, as we wanted to show. \(\square \)

Proof

(Proposition 5.2) If \( \eta = 1/2 \), then we have that \( v_i'(F;\theta _i) = \beta _i(\theta _i) F^{-1/2} \). For every \( i \in \mathbb {I}\), let us define \( c_i(\theta _i) \) by

$$\begin{aligned} c_i(\theta _i) = \left( E \left[ v_i' (F^{e}(\theta ) ; \theta _i) \cdot (F^{e}(\theta ))^{1/2} \,\vert \,\theta _i \right] \right) ^2, \end{aligned}$$
(16)

which is well-defined since \( F^{e}(\theta ) \) exists and is unique. These contributions satisfy the first order conditions of the individual problem. Let us show that \( \varPhi ^{QF}(\textbf{c}(\theta )) = F^{e}(\theta ) \). Substituting \( v_i'(F;\theta _i)\) in Eq. (16) we get

$$\begin{aligned} (c_i(\theta _i))^{1/2}&= E \left[ \beta _i(\theta _i) \cdot (F^{e}(\theta ))^{-1/2} \cdot (F^{e}(\theta ))^{1/2} \,\vert \,\theta _i \right] \\&= \beta _i(\theta _i), \end{aligned}$$

therefore, we have

$$\begin{aligned} v_i'(F^{e}(\theta );\theta _i) = \frac{(c_i(\theta _i))^{1/2}}{(F^{e}(\theta ))^{1/2}}. \end{aligned}$$

So, summing this expression for all i,

$$\begin{aligned} \sum _{i = 1}^{I} v_i'(F^{e}(\theta );\theta _i) = \frac{\sum _{i = 1}^{I} (c_i(\theta _i))^{1/2} }{(F^{e}(\theta ))^{1/2}}. \end{aligned}$$

Hence, since \( F^{e}(\theta ) \) is efficient, we have that the left-hand side of the above equation is equal to one. We then have that \( \varPhi ^{QF}(\textbf{c}(\theta )) = \left( \sum _{i = 1}^{I} (c_i(\theta _i))^{1/2}\right) ^2 = F^{e}(\theta ) \), and the equilibrium \(\textbf{c}\) is efficient.

To prove the converse, suppose that QF is efficient, and that \( \eta > 1/2 \) (the case where \( \eta < 1/2 \) is analogous). By hypothesis, we know that there exists some \( j \in \mathbb {I}\) with different types \( \theta _j^k, \theta _j^\ell \in \varTheta _j \). Since \(\beta _j\) is injective by assumption, it follows that \( \beta _j(\theta _j^k) \ne \beta _j(\theta _j^\ell ) \). Without loss of generality, let \( \beta _j(\theta _j^k) > \beta _j(\theta _j^\ell ) \). Now, let \( \overline{\theta } \in \varTheta \) be the type profile such that \( \beta _i(\overline{\theta }_i) \ge \beta _i(\theta _i) \), for all \( i \in \mathbb {I}\) and all \( \theta _i \in \varTheta _i \). Let \( \textbf{c}^* \) be the efficient equilibrium. Then, the first order condition for individual i in the type profile \( \overline{\theta } \) implies that

$$\begin{aligned} (c^*_i(\overline{\theta }_i))^{1/2} = \beta _i(\overline{\theta }_i) \cdot E \left[ (F^{e}(\theta ))^{1/2-\eta } \,\vert \,\overline{\theta }_i \right] . \end{aligned}$$
(17)

Efficiency implies that, for any \( \theta \in \varTheta \),

$$ \sum _{i=1}^{I} v_i'(F^{e}(\theta ); \theta _i) = \sum _{i = 1}^{I} \beta _i(\theta _i) \cdot (F^{e}(\theta ))^{-\eta } = 1. $$

Then, as \( \beta _i(\overline{\theta }_i) \ge \beta _i(\theta _i) \) for all \( \theta _i \) and all i, it follows that

$$ \sum _{i=1}^{I} v_i'(F^{e}(\theta ); \overline{\theta }_i) = \sum _{i = 1}^{I} \beta _i(\overline{\theta }_i) \cdot (F^{e}(\theta ))^{-\eta } \ge 1, $$

for all \( \theta \in \varTheta \). From strict monotonicity of \( v_i'(\cdot ; \theta _i) \), we have that \( F^{e}(\overline{\theta }) \ge F^{e}(\theta ) \), for all \( \theta \in \varTheta \), in particular, \( F^{e}(\overline{\theta }) > F^{e}(\theta _j^\ell ,\overline{\theta }_{-j}) \). Thus, as \( 1/2 - \eta < 0 \), it follows that

$$ (F^{e}(\overline{\theta }))^{1/2-\eta } \le (F^{e}(\theta ))^{1/2- \eta } $$

for all \( \theta \in \varTheta \) and, in particular,

$$ (F^{e}(\overline{\theta }))^{1/2-\eta } < (F^{e}(\theta _j^\ell , \overline{\theta }_{-j}))^{1/2- \eta }. $$

Since \( \Pr \left( \theta _j^\ell \,\vert \,\overline{\theta }_i \right) > 0 \) for all \( i \ne j \), we then have

$$\begin{aligned} E \left[ F^{e}(\theta ))^{1/2-\eta } \,\vert \,\overline{\theta }_i \right] \ge (F^{e}(\overline{\theta }))^{1/2 - \eta } \end{aligned}$$
(18)

for all \( i \in \mathbb {I}\), with strict inequality when \( i = j \). By (17) and (18) we have that

$$\begin{aligned} (c^*_i(\overline{\theta }_i))^{1/2} \ge \beta _i(\overline{\theta }_i) \cdot (F^{e}(\overline{\theta }))^{1/2-\eta } \end{aligned}$$
(19)

for all \( i \in \mathbb {I}\), with strict inequality when \( i \ne j \). Adding (19) across all \( i \in \mathbb {I}\) yields

$$\begin{aligned} \sum _{i = 1}^{I} (c^*_i(\overline{\theta }_i))^{1/2} = (F^{e}(\overline{\theta }))^{1/2} > \sum _{i = 1}^{I} \beta _i(\overline{\theta }_i) \cdot (F^{e}(\overline{\theta }))^{1/2-\eta }. \end{aligned}$$
(20)

Dividing both sides of the inequality in (20) by \( (F^{e}(\overline{\theta }))^{1/2} \), we get

$$\begin{aligned} 1 > \sum _{i = 1}^{I} \beta _i(\overline{\theta }_i) \cdot (F^{e}(\overline{\theta }))^{-\eta } = \sum _{i = 1}^{I} v_i'(F^{e}(\overline{\theta }); \overline{\theta }_i), \end{aligned}$$
(21)

which is a contradiction to the definition of efficient provision. This completes the proof. \(\square \)

Appendix B: Figures

Fig. 4
figure 4

Absolute inefficiency (y-axis, exponential scale) as a function of population size (x-axis), for \(\eta = 0.499\)

Fig. 5
figure 5

Relative inefficiency (y-axis, exponential scale) as a function of population size (x-axis), for four different values of \(\eta \) (plots)

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Freitas, L.M., Maldonado, W.L. Quadratic funding with incomplete information. Soc Choice Welf (2024). https://doi.org/10.1007/s00355-024-01512-7

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