A public good model with lotteries in large groups

Abstract

We analyze the effect of a large group on a public goods model with lotteries. We show that as populations get large, and with preferences in which people only care about their private consumptions and the total supply of the public good, the level of contributions converges to the one given by voluntary contributions. With altruistic preferences of the warm-glow type, the contributions converge to a level strictly higher than those given by voluntary contributions, but in general they do not yield first-best levels. Our results are important to clarify why in general governments do not rely on lotteries for a large part of the revenue creation for public good provision. They are also useful to understand why lottery proceeds are earmarked to worthy causes, where warm glow is likely to be larger.

Appendix

Proof of Proposition 1

Proof

This proposition is already shown in Morgan (2000), we merely add it here for completeness.

$G^{*}$ solves

$$\begin{aligned} \sum _{i=1}^{n}h_{i}^{\prime }(G^{*})=H(G^{*})-H^{\prime }(G^{*}) \left( \sum _{i=1}^{n}\omega _{i}-G^{*}\right) . \end{aligned}$$

(7)

and we also obtain $G^{V}$ by adding first-order conditions of optimization problems for each agent $i$,

$$\begin{aligned} \sum _{i=1}^{n}h_{i}^{\prime }(G^{V})=nH(G^{V})-H^{\prime }(G^{V})\left( \sum _{i=1}^{n}\omega _{i}-G^{V}\right) . \end{aligned}$$

(8)

It is then easy to verify that $G^{V}<G^{*}$. Also $G^{L}$, solves the sum of first-order conditions.

$$\begin{aligned} \sum _{i=1}^{n}h_{i}^{\prime }(G^{L})=H(G^{L})\left( n-(n-1)\frac{R}{R+G^{L}} \right) -H^{\prime }(G^{L})\left( \sum _{i=1}^{n}\omega _{i}-G^{L}\right) \end{aligned}$$

(9)

Comparing expressions (8) and (9), Morgan (2000) remarks that the two expressions differ by the term associated with the negative externality of the lottery multiplied by $H(G)$. Thus, $G^{V}<G^{L}$. Taking the limit of (9) as $R\rightarrow \infty$, expression (9) becomes identical to (7). $\square$

Proof of Proposition 2

Proof

At an interior maximum, the first-order condition of (2) with respect to $x$,

$$\begin{aligned} -H(G)+(\omega _{i}-x_{i})H^{\prime }(G)+h_{i}^{\prime }(G)(g_{1}f\left( x_{i}\right) +1)+h_{i}(G)g_{1}f^{\prime }\left( x_{i}\right) =0 \end{aligned}$$

(10)

The equilibrium level of public good provided by voluntary contributions with the presence of warm glow giving, solves the sum of first-order conditions,

$$\begin{aligned} \sum _{i=1}^{n}h_{i}^{\prime }(G^{wg})(g_{1}f\left( x_{i}\right) +1)+\sum _{i=1}^{n}h_{i}(G^{wg})g_{1}f^{\prime }\left( x_{i}\right)= & {} nH(G^{wg}) \nonumber \\&-H^{\prime }(G^{wg})\left( \sum _{i=1}^{n}\omega _{i}-G^{wg}\right) \nonumber \end{aligned}$$

Then, we have

$$\begin{aligned} \sum _{i=1}^{n}h_{i}^{\prime }(G^{wg})= & {} nH(G^{wg})-H^{\prime }(G^{wg})\left( \sum _{i=1}^{n}\omega _{i}-G^{wg}\right) \\&-\sum _{i=1}^{n}h_{i}^{\prime }(G^{wg})g_{1}f\left( x_{i}\right) -\sum _{i=1}^{n}h_{i}(G^{wg})g_{1}f^{\prime }\left( x_{i}\right) \nonumber \end{aligned}$$

(11)

Compare expressions (8) and (11) to verify that the result holds. The two expressions differ by the term associated with the effect of the warm glow. $\square$

Proof of Proposition 3

Proof

Take the first-order conditions of (4) with respect to $x_{i}$ to find

$$\begin{aligned}&\left( R\frac{\widehat{x}-x_{i}}{\widehat{x}^{2}}-1\right) H(\widehat{x} -R)+\left( \omega _{i}-x_{i}+R\frac{x_{i}}{\widehat{x}}\right) H^{\prime }( \widehat{x}-R) \nonumber \\&\quad + \, h_{i}^{\prime }(\widehat{x}-R)(g_{1}f\left( x_{i}\right) +1)+h_{i}(\widehat{x }-R)g_{1}f^{\prime }\left( x_{i}\right) =0 \nonumber \end{aligned}$$

The public goods provision $G^{wgL}$ solves the sum of the first-order conditions,

$$\begin{aligned}&\sum _{i=1}^{n}h_{i}^{\prime }(G^{wgL})(g_{1}f\left( x_{i}\right) +1)+\sum _{i=1}^{n}h_{i}(G^{wgL})g_{1}f^{\prime }\left( x_{i}\right) \\&\quad =H(G^{wgL})\left( n-(n-1)\frac{R}{R+G^{wgL}}\right) \\&\quad \quad - \, H^{\prime }(G^{wgL})\left( \sum _{i=1}^{n}\omega _{i}-G^{wgL}\right) \end{aligned}$$

We have

$$\begin{aligned} \sum _{i=1}^{n}h_{i}^{\prime }(G^{wgL})= & {} H(G^{wgL})\left( n\!-\!(n\!-\!1)\frac{R}{ R\!+\!G^{wgL}}\right) \!-\!H^{\prime }(G^{wgL})\left( \sum _{i=1}^{n}\omega _{i}\!-\!G^{wgL}\right) \nonumber \\&-\sum _{i=1}^{n}h_{i}^{\prime }(G^{wgL})g_{1}f\left( x_{i}\right) -\sum _{i=1}^{n}h_{i}(G^{wgL})g_{1}f^{\prime }\left( x_{i}\right) \end{aligned}$$

(12)

Notice that expressions (11) and (12) differ by the term associated with the negative externality of the lottery multiplied by $H(G)$. Similarly as to the model without warm glow, the public goods provision under the lottery is greater than under voluntary contributions. That is, $G^{wg}<G^{wgL}$. $\square$

Proof of Proposition 4

Proof

To obtain the optimal level of provision, we use the first-order conditions from the problem

$$\begin{aligned} \sum _{i=1}^{n}U_{i}^{wg}=\sum _{i=1}^{n}(\omega _{i}-x_{i})H(G)+h_{i}(G)\left( g_{1}f\left( x_{i}\right) +g_{0}\right) \end{aligned}$$

which yield

$$\begin{aligned} (\omega _{i}-x_{i})H^{\prime }(G^{wg*})+h_{i}^{\prime }(G^{wg*})\left( g_{1}f\left( x_{i}\right) +g_{0}\right) -H(G^{wg*})+h_{i}(G^{wg*})g_{1}f^{\prime }\left( x_{i}\right) =0 \end{aligned}$$

(13)

which using linearity of $H\left( .\right)$ and $h_{i}\left( .\right)$ can be expressed as

$$\begin{aligned} \left( (\omega _{i}-x_{i})H+h_{i}\left( g_{1}f\left( x_{i}\right) +g_{0}\right) -HG^{wg*}+h_{i}G^{wg*}g_{1}f^{\prime }\left( x_{i}\right) \right) =0 \end{aligned}$$

We now take limits for $n$ large and assuming that $G^{wg*}$ is of $O\left( n\right)$ (something we later establish) we have

$$\begin{aligned}&\lim _{n\rightarrow \infty }\left( \frac{1}{n}\left( (\omega _{i}-x_{i})H+h_{i}\left( g_{1}f\left( x_{i}\right) +g_{0}\right) -HG^{wg*}+h_{i}G^{wg*}g_{1}f^{\prime }\left( x_{i}\right) \right) \right) =0\\&\quad \lim _{n\rightarrow \infty }\left( \frac{1}{n}\left( -HG^{wg*}+h_{i}G^{wg*}g_{1}f^{\prime }\left( x_{i}\right) \right) \right) =0 \end{aligned}$$

which means that

$$\begin{aligned} \lim _{n\rightarrow \infty }\left( f^{\prime }\left( x_{i}\right) \right) = \frac{H}{g_{1}h_{i}} \end{aligned}$$

so that

$$\begin{aligned} \lim _{n\rightarrow \infty }\frac{1}{n}G^{wg*}=\lim _{n\rightarrow \infty } \frac{1}{n}\sum _{i=1}^{n}x_{i}=\lim _{n\rightarrow \infty }\frac{1}{n} \sum _{i=1}^{n}f^{\prime -1}\left( \frac{H}{g_{1}h_{i}}\right) \end{aligned}$$

(14)

when $f\left( x\right) =\ln \left( x\right)$

$$\begin{aligned} \lim _{n\rightarrow \infty }\frac{1}{n}G^{wg*}=\lim _{n\rightarrow \infty } \frac{1}{n}\sum _{i=1}^{n}x_{i}=\lim _{n\rightarrow \infty }\frac{1}{n} \sum _{i=1}^{n}\left( \frac{g_{1}h_{i}}{H}\right) \end{aligned}$$

when $f\left( x\right) =\frac{1}{1-\alpha }\left( x\right) ^{1-\alpha }$

$$\begin{aligned} \lim _{n\rightarrow \infty }\frac{1}{n}G^{wg*}=\lim _{n\rightarrow \infty } \frac{1}{n}\sum _{i=1}^{n}x_{i}=\lim _{n\rightarrow \infty }\frac{1}{n} \sum _{i=1}^{n}\left( \frac{g_{1}h_{i}}{H}\right) ^{\frac{1}{\alpha }} \end{aligned}$$

Note that this also establishes $G^{wg*}$ is of $O\left( n\right)$.

The first-order condition of (12) with respect to $x_{i}$ is,

$$\begin{aligned}&\left( R\frac{\widehat{x}-x_{i}}{\widehat{x}^{2}}-1\right) H(\widehat{x} -R)+\left( \omega _{i}-x_{i}+R\frac{x_{i}}{\widehat{x}}\right) H^{\prime }( \widehat{x}-R)\nonumber \\&\quad + \, h_{i}^{\prime }(\widehat{x}-R)(g_{1}f\left( x_{i}\right) +1)+h_{i}(\widehat{x}-R)g_{1}f^{\prime }\left( x_{i}\right) =0 \end{aligned}$$

(15)

using linearity this is equivalent to

$$\begin{aligned}&\left( R\frac{\widehat{x}-x_{i}}{\widehat{x}^{2}}-1\right) H(\widehat{x} -R)+\left( \omega _{i}-x_{i}+R\frac{x_{i}}{\widehat{x}}\right) H+h_{i}(g_{1}f\left( x_{i}\right) +1)\\&\quad +\, h_{i}(\widehat{x}-R)g_{1}f^{\prime }\left( x_{i}\right) =0 \end{aligned}$$

We now take limits for $n$ large and assuming that $G^{wgL }$ is of $O\left( n\right)$ (something we later establish) we have

$$\begin{aligned}&\lim _{n\rightarrow \infty }\left( \left( R\frac{\widehat{x}-x_{i}}{\widehat{x }^{2}}-1\right) H(\widehat{x}-R)+\left( \omega _{i}-x_{i}+R\frac{x_{i}}{ \widehat{x}}\right) H\right. \\&\quad \left. +\, h_{i}(g_{1}f\left( x_{i}\right) +1)+h_{i}(\widehat{x} -R)g_{1}f^{\prime }\left( x_{i}\right) \right) =0 \\&\quad \lim _{n\rightarrow \infty }\left( \left( \frac{R}{\widehat{x}}-1\right) H( \widehat{x}-R)+h_{i}(\widehat{x}-R)g_{1}f^{\prime }\left( x_{i}\right) \right) =0 \end{aligned}$$

which means that

$$\begin{aligned} \lim _{n\rightarrow \infty }\left( f^{\prime }\left( x_{i}\right) \right) = \frac{H}{g_{1}h_{i}}\left( 1-\lim _{n\rightarrow \infty }\frac{R}{\widehat{x}} \right) \end{aligned}$$

when $f\left( x\right) =\ln \left( x\right)$

$$\begin{aligned} \lim _{n\rightarrow \infty }\frac{1}{x_{i}}= & {} \frac{H}{g_{1}h_{i}}\left( 1-\lim _{n\rightarrow \infty }\frac{R}{\widehat{x}}\right) \\ \lim _{n\rightarrow \infty }x_{i}= & {} \frac{g_{1}h_{i}}{H}\left( \lim _{n\rightarrow \infty }\frac{\widehat{x}}{\widehat{x}-R}\right) \\ \lim _{n\rightarrow \infty }\frac{\widehat{x}}{n}= & {} \lim _{n\rightarrow \infty }\frac{\widehat{x}}{\widehat{x}-R}\frac{1}{n}\sum _{i=1}^{n}\frac{ g_{1}h_{i}}{H} \\ \lim _{n\rightarrow \infty }\frac{\widehat{x}-R}{n}= & {} \lim _{n\rightarrow \infty }\frac{1}{n}\sum _{i=1}^{n}\frac{g_{1}h_{i}}{H} \end{aligned}$$

when $f\left( x\right) =\frac{1}{1-\alpha }\left( x\right) ^{1-\alpha }$

$$\begin{aligned} \lim _{n\rightarrow \infty }x_{i}^{-\alpha }= & {} \frac{H}{g_{1}h_{i}}\left( 1-\lim _{n\rightarrow \infty }\frac{R}{\widehat{x}}\right) \\ \lim _{n\rightarrow \infty }x_{i}^{\alpha }= & {} \frac{g_{1}h_{i}}{H}\left( \lim _{n\rightarrow \infty }\frac{\widehat{x}}{\widehat{x}-R}\right) \\ \lim _{n\rightarrow \infty }x_{i}= & {} \lim _{n\rightarrow \infty }\left( \frac{ g_{1}h_{i}}{H}\right) ^{\frac{1}{\alpha }}\left( \frac{\widehat{x}}{\widehat{ x}-R}\right) ^{\frac{1}{\alpha }} \\ \lim _{n\rightarrow \infty }\frac{\widehat{x}}{n}= & {} \lim _{n\rightarrow \infty }\left( \frac{\widehat{x}}{\widehat{x}-R}\right) ^{\frac{1}{\alpha }} \frac{1}{n}\sum _{i=1}^{n}\left( \frac{g_{1}h_{i}}{H}\right) ^{\frac{1}{ \alpha }} \\ \lim _{n\rightarrow \infty }\left( \frac{\widehat{x}-R}{n}\right) ^{\frac{1}{ _{\alpha }}}= & {} \lim _{n\rightarrow \infty }\left( \frac{\widehat{x}}{n} \right) ^{\frac{1-\alpha }{\alpha }}\frac{1}{n}\sum _{i=1}^{n}\left( \frac{ g_{1}h_{i}}{H}\right) ^{\frac{1}{\alpha }} \\ \lim _{n\rightarrow \infty }\left( \frac{G^{wgL}}{n}\right) ^{\frac{1}{a}}= & {} \lim _{n\rightarrow \infty }\left( \frac{\widehat{x}}{n}\right) ^{\frac{ 1-\alpha }{\alpha }}\frac{1}{n}\sum _{i=1}^{n}\left( \frac{g_{1}h_{i}}{H} \right) ^{\frac{1}{\alpha }} \\ \lim _{n\rightarrow \infty }\left( \frac{G^{wgL}}{n}\right) ^{\frac{1}{a}}= & {} \lim _{n\rightarrow \infty }\left( \left( \frac{\widehat{x}}{n}\right) ^{ \frac{1-\alpha }{\alpha }}\frac{G^{wg*}}{n}\right) \end{aligned}$$

Assume, by way of a contradiction, that $G^{wg*}=G^{wgL},$then

$$\begin{aligned} \lim _{n\rightarrow \infty }\left( \frac{G^{wgL}}{n}\right) ^{\frac{1}{a}}= & {} \lim _{n\rightarrow \infty }\left( \left( \frac{\widehat{x}}{n}\right) ^{ \frac{1-\alpha }{\alpha }}\frac{G^{wgL}}{n}\right) \\ \lim _{n\rightarrow \infty }\left( \frac{G^{wgL}}{n}\right) ^{\frac{1-\alpha }{a}}= & {} \lim _{n\rightarrow \infty }\left( \frac{\widehat{x}}{n}\right) ^{ \frac{1-\alpha }{\alpha }} \\ \lim _{n\rightarrow \infty }\left( \frac{G^{wgL}}{n}\right)= & {} \lim _{n\rightarrow \infty }\left( \frac{\widehat{x}-R}{n}\right) =\lim _{n\rightarrow \infty }\left( \frac{\widehat{x}}{n}\right) \end{aligned}$$

This entails a contradiction, and the result follows. $\square$

Proof of Proposition 5

Proof

The first-order condition,

$$\begin{aligned} \left( R\frac{\widehat{x}-x_{i}}{\widehat{x}^{2}}-1\right) H(\widehat{x} -R)+\left( \omega _{i}-x_{i}+R\frac{x_{i}}{\widehat{x}}\right) H^{\prime }( \widehat{x}-R) \\ + \, h_{i}^{\prime }(\widehat{x}-R)(g_{1}f\left( x_{i}\right) +1)+h_{i}(\widehat{x }-R)g_{1}f^{\prime }\left( x_{i}\right) =0 \end{aligned}$$

where $\widehat{x}=\sum x_{i}$ noting that $H^{\prime }$ and $h^{\prime }$ are bounded by concavity and assuming that $\widehat{x}-R$ is of $O\left( n\right)$ (something we later establish) as $n\rightarrow \infty$, we obtain

$$\begin{aligned}&\lim _{n\rightarrow \infty }\left( \frac{1}{n}\left( \left( R\frac{\widehat{ x}-x_{i}}{\widehat{x}^{2}}-1\right) H(\widehat{x}-R)+\left( \omega _{i}-x_{i}+R\frac{x_{i}}{\widehat{x}}\right) H^{\prime }(\widehat{x} -R)\right) \right. \\&\quad +\left. \frac{1}{n}\left( h_{i}^{\prime }(\widehat{x}-R)(g_{1}f\left( x_{i}\right) +1)+h_{i}(\widehat{x}-R)g_{1}f^{\prime }\left( x_{i}\right) \right) \right) \\&\quad =\lim _{n\rightarrow \infty }\frac{1}{n}\left( \frac{R}{\widehat{x}}H( \widehat{x}-R)-H(\widehat{x}-R)+h_{i}(\widehat{x}-R)g_{1}f^{\prime }\left( x_{i}\right) \right) =0 \\&\quad =\lim _{n\rightarrow \infty }\frac{1}{n}\left( \frac{R-\widehat{x}}{ \widehat{x}}H(\widehat{x}-R)+h_{i}(\widehat{x}-R)g_{1}f^{\prime }\left( x_{i}\right) \right) =0 \end{aligned}$$

so that

$$\begin{aligned} \lim _{n\rightarrow \infty }\frac{1}{n}\left( h_{i}(\widehat{x} -R)g_{1}f^{\prime }\left( x_{i}\right) \right)= & {} \lim _{n\rightarrow \infty } \frac{1}{n}\left( \frac{\widehat{x}-R}{\widehat{x}}H(\widehat{x}-R)\right) \\ \lim _{n\rightarrow \infty }\left( f^{\prime }\left( x_{i}\right) \right)= & {} \lim _{n\rightarrow \infty }\left( \frac{\widehat{x}-R}{\widehat{x}}\frac{ H(\widehat{x}-R)}{g_{1}h_{i}(\widehat{x}-R)}\right) \nonumber \\= & {} \lim _{n\rightarrow \infty }\left( \frac{\widehat{x}-R}{\widehat{x}}\frac{1 }{g_{1}k_{i}}\right) \nonumber \\ \lim _{n\rightarrow \infty }\left( x_{i}\right)= & {} \lim _{n\rightarrow \infty }f^{\prime -1}\left( \frac{\widehat{x}-R}{\widehat{x}}\frac{1}{g_{1}k_{i}} \right) \nonumber \end{aligned}$$

(16)

so that we have the desired result

$$\begin{aligned} \lim _{n\rightarrow \infty }\frac{\widehat{x}}{n}=\lim _{n\rightarrow \infty } \frac{1}{n}\sum _{i=1}^{n}f^{\prime -1}\left( \frac{\widehat{x}-R}{\widehat{x} }\frac{1}{g_{1}k_{i}}\right) \end{aligned}$$

Note also that from Eq. (16) if $g_{1}=0,$ it must be the case that $\lim _{n\rightarrow \infty }\left( \widehat{x}\!-\!R\right) /\widehat{x}\!=\!0$ so that $\lim _{n\rightarrow \infty }\widehat{x}/n=\rho$. $\square$