Risk aversion and bandwagon effect in the pivotal voter model

Abstract

The empirical literature on the effects of opinion polls on election outcomes has recently found substantial evidence of a bandwagon effect, defined as the phenomenon according to which the publication of opinion polls is advantageous to the candidate with the greatest support. This result is driven, in the lab experiments, by a higher turnout rate among the majority than among the minority. Such evidence is however in stark contrast with the main theoretical model of electoral participation in public choice, the pivotal voter model, which predicts that the supporters of an underdog candidate participate at a higher rate, given the higher probability of casting a pivotal vote. This paper tries to reconcile this discrepancy by showing that a bandwagon effect can be generated within the pivotal voter model by concavity in the voters’ utility function, which makes electoral participation more costly for the expected loser supporters. Given the strict relationship between concavity and risk aversion, the paper also establishes the role of risk aversion as a determinant of bandwagon.

Expressions for \(p_{1}\), \(p_{2}\) and q

I derive here the expressions for $p_{1}$, $p_{2}$ and q in system (6). For the calculation of $p_{1}$ and $p_{2}$, I draw on the analysis in Goeree and Grosser (2007). An agent is pivotal only when either the rest of the voting agents is equally split between the two candidates (turning a draw into a victory) or when the opposing candidate would win by one vote if the agent abstains (turning a loss into a draw). Given the tie-breaking rule in the case of a draw, in both cases the probability of being pivotal is divided by one-half. Clearly the first scenario can happen if the number of other voters is even, while the second if such number is odd. The two cases can be combined using the floor operator $\lfloor {\cdot }\rfloor$. Consider first a supporter of candidate 1, supposing that k other agents also prefer candidate 1, $n-k-1$ prefer candidate 2 and a total of l agents vote. Assuming that the cutoff costs $c_{b}$ and $c_{s}$ are both in (0, 1), the probability of being pivotal $p_{1}$ is equal to

$$\begin{aligned} \begin{aligned}p_{1}(c_{b},c_{s})=\frac{1}{2}\sum \limits _{l=0}^{n-1}\sum \limits _{k=\lfloor {\frac{l}{2}}\rfloor }^{n-1-\lfloor {\frac{l+1}{2}}\rfloor }\left( {\begin{array}{c}n-1\\ k\end{array}}\right) \pi ^{k}(1-\pi )^{n-k-1}\left( {\begin{array}{c}k\\ \left\lfloor {\frac{l}{2}}\right\rfloor \end{array}}\right) \left( {\begin{array}{c}n-k-1\\ \lfloor {\frac{l+1}{2}}\rfloor \end{array}}\right) \\ \times (c_{b})^{\lfloor {\frac{l}{2}}\rfloor }(1-c_{b})^{k-\lfloor {\frac{l}{2}}\rfloor }(c_{s})^{\lfloor {\frac{l+1}{2}}\rfloor }(1-c_{s})^{n-k-1-\lfloor {\frac{l+1}{2}}\rfloor } \end{aligned} \end{aligned}$$

(11)

Analogously for a supporter of candidate 2, the probability of being pivotal $p_{2}$ is equal to

$$\begin{aligned} \begin{aligned}p_{2}(c_{b},c_{s})=\frac{1}{2}\sum \limits _{l=0}^{n-1}\sum \limits _{k=\lfloor {\frac{l}{2}}\rfloor }^{n-1-\lfloor {\frac{l+1}{2}}\rfloor }\left( {\begin{array}{c}n-1\\ k\end{array}}\right) (1-\pi )^{k}\pi ^{n-k-1}\left( {\begin{array}{c}k\\ \lfloor {\frac{l}{2}}\rfloor \end{array}}\right) \left( {\begin{array}{c}n-k-1\\ \lfloor {\frac{l+1}{2}}\rfloor \end{array}}\right) \\ \times (c_{s})^{\lfloor {\frac{l}{2}}\rfloor }(1-c_{s})^{k-\lfloor {\frac{l}{2}}\rfloor }(c_{b})^{\lfloor {\frac{l+1}{2}}\rfloor }(1-c_{b})^{n-k-1-\lfloor {\frac{l+1}{2}}\rfloor } \end{aligned} \end{aligned}$$

(12)

In addition to $p_{1}$ and $p_{2}$, I need to calculate the probability q, which emerges in the agent’s decision problem if the assumption of linearity in the utility function is relaxed. To do so, notice that when l other agents vote, candidate 1 wins if at least $\frac{l}{2}+1$ agents vote for him for l even, or at least $\frac{l+1}{2}$ for l odd. The two cases can again be combined using the floor operator: if l agents vote, candidate 1 wins if at least $\lfloor {\frac{l}{2}+1}\rfloor$ vote for him. Moreover, by the assumption on the tie-breaking, candidate 1 also wins with probability one-half if the voters are equally split between the two candidates (l must be even). Hence, assuming $c_{b}$ and $c_{s}$ both in (0, 1), q is equal to

$$\begin{aligned} \begin{aligned}q(c_{b},c_{s})=\sum \limits _{l=0}^{n-1}\sum \limits _{k=v}^{n-1-l+v}\sum \limits _{v=\lfloor {\frac{l}{2}+1}\rfloor }^{l}\left( {\begin{array}{c}n-1\\ k\end{array}}\right) \pi ^{k}(1-\pi )^{n-k-1}\left( {\begin{array}{c}k\\ v\end{array}}\right) \left( {\begin{array}{c}n-k-1\\ l-v\end{array}}\right) \\ \times (c_{b})^{v}(1-c_{b})^{k-v}(c_{s})^{l-v}(1-c_{s})^{n-k-1-l+v}\\ +\,\frac{1}{2}\sum \limits _{l\text { even, }l=0}^{n-1}\sum \limits _{k=\frac{l}{2}}^{n-1-\frac{l}{2}}\left( {\begin{array}{c}n-1\\ k\end{array}}\right) \pi ^{k}(1-\pi )^{n-k-1}\left( {\begin{array}{c}k\\ \frac{l}{2}\end{array}}\right) \left( {\begin{array}{c}n-k-1\\ \frac{l}{2}\end{array}}\right) \\ \times (c_{b})^{\frac{l}{2}}(1-c_{b})^{k-\frac{l}{2}}(c_{s})^{\frac{l}{2}}(1-c_{s})^{n-k-1-\frac{l}{2}} \end{aligned} \end{aligned}$$

(13)

The probabilities $p_{1}$, $p_{2}$ and q are strictly positive also in the case that $c_{b}$ or $c_{s}$ are equal to zero (see the proof of Proposition 6 in the online supplementary material). In particular, $p_{1}=p_{2}=q=\frac{1}{2}$ if $c_{b}=c_{s}=0$.