‘Everybody’s doing it’: on the persistence of bad social norms

Abstract

We investigate how information about the preferences of others affects the persistence of ‘bad’ social norms. One view is that bad norms thrive even when people are informed of the preferences of others, since the bad norm is an equilibrium of a coordination game. The other view is based on pluralistic ignorance, in which uncertainty about others’ preferences is crucial. In an experiment, we find clear support for the pluralistic ignorance perspective . In addition, the strength of social interactions is important for a bad norm to persist. These findings help in understanding the causes of such bad norms, and in designing interventions to change them.

Appendix

1.1 Table of results

See Table 3.

Table 3 Key performance indicators by group

1.2 Proofs

1.2.1 Stage-game equilibria

It follows from the decision rule specified in Proposition 1 that, in equilibrium, we require that players prefer $\omega _i=1$ at least as much as $\omega _i=-1$ if $d_i<c^*$, that players prefer $\omega _i=-1$ at least as much as $\omega _i=1$ if $d_i>c^*$ and, in particular, that a player is exactly indifferent between $\omega _i=-1$ and 1 if she draws private values with a difference equal to the threshold $c^*$. We use this latter property of the equilibrium to endogenously calculate the threshold.

The threshold $c^*$ depends both on an individual’s beliefs about group behavior as well as the (fixed) social factor. Solving for this threshold allows us to compute a general equilibria condition that holds for any given distribution of the private shocks. Then an individual i maximizing her expected utility chooses $\omega _i=-1$ if $d_i>2Jm^e_i$. To endogenously solve for an equilibrium, we first rewrite $m^e_i$ as:

$$\begin{aligned} m^e_i=\displaystyle \frac{1}{N-1}\sum ^{N-1}_{k=0}\left( \left( {\begin{array}{c}N-1\\ k\end{array}}\right) p^k(1-p)^{(N-1-k)}(2k-N+1)\right) \end{aligned}$$

(7)

where p is the probability of a single draw of $d_i<c^*$ so that i chooses $\omega _i=1$. Then each term in the series is the expected value for each possible value of $m_i$, which can be written in the form $\frac{2k-N+1}{N-1}$ for each $k\in \{0,N-1\}$.

Letting $m_i^{e*}$ be the equilibrium expected average choice of the others in a group, corresponding to a threshold $c^*$, we can rewrite $c^*=2Jm_i^{e*}$ in (7). Then solving for an individual i drawing exactly $d_i=c^*$ with ${\mathbb{e}}V_i(-1)={\mathbb{e}}V_i(1)$ allows us to solve endogenously for the expectation $m_i^{e*}=m_j^{e*} \;\;\forall i,j$:

$$\begin{aligned} m_i^{e*} =\frac{1}{N-1}\sum \limits _{k=0}^{N-1}\genfrac(){0.0pt}0{N-1}{k}F(2Jm_i^{e*}-d)^k(1-F(2Jm_i^{e*}-d))^{(N-1-k)}(2k-N+1) \end{aligned}$$

(8)

At first sight, an individual’s expectations appears to depend on the size of the group, N. We perform the replacements $M=N-1$ and $F=F(2Jm_i^{e*}-d)$ for notational convenience to rewrite (8) as:

$$\begin{aligned} m_i^{e*} =\frac{1}{M}\sum \limits _{k=0}^{M}\genfrac(){0.0pt}0{M}{k}F^k(1-F)^{(M-k)}(2k-M) \end{aligned}$$

(9)

It can be shown that the sum of this series is independent of group size as follows: Let k be a binomially-distributed random variable with parameters $n=M,p=F$. Then $\mathbb {E}(k)=MF$ and so the right-hand side of (9) simplifies to $2F-1$.

Thus, (8) can be rewritten as $m_i^{e*} =2F(2Jm_i^{e*}-d)-1$, which notably does not depend on N. Similarly, the researcher’s prediction of the expected average choice level of the whole group solves:

$$\begin{aligned} m^* =2F(2Jm^*-d)-1 \end{aligned}$$

(10)

1.2.2 Effect of group size

While group size does not influence the stage-game equilibria, it may still affect the probability of a group switching from a bad equilibrium to a good equilibrium in a given round. Consider a scenario in which the bad norm $\omega _{it}=1$ is persistent on account of relatively large J and $m^e_{it}$, such that in the majority of rounds $\rho _{it}=0$. Ex-ante, the probability of an individual choosing $\omega _{it}=-1$ in a given round t is $\hat{\rho }_t$, regardless of the group size. Now consider the rounds in which $0<\rho _{it}<0.5$; that is, the bad norm $\omega _i=1$ is still in effect but at least one group member receives a private shock difference large enough to induce choosing $\omega _{it}=-1$. This likelihood is not the same across group sizes. The probability that at least one group member chooses $\omega _{it}=-1$ increases with N, and so we would expect a higher proportion of rounds with$\rho _{it}\ne 0$ in larger groups while the bad norm persists. However, the marginal effect of a group member choosing $\omega _{it}=-1$ on the overall group proportion $\rho _{it}$ decreases with N, and so of those rounds where $\rho _{it}\ne 0$ while the bad norm persists, we would expect that $\rho _{it}$ is higher on average for smaller groups.

Now, assume there is some ‘tipping proportion’ $\tilde{\rho }$ that, if reached after a previous equilibrium of full conformity to the bad norm ($\rho ^*\approx 0$), would result in a switch to the ‘good’ equilibrium $\rho ^*\approx 1$ with almost certainty. The tipping proportion is greater than the predicted group proportion $\hat{\rho }_t$ so that on expectation it should not be breached in a given round. Then, after a round in which $\rho _{t-1}\approx 0$, the probability of reaching the tipping proportion in round t is the probability that at least $N\tilde{\rho }$ individuals choose $\omega _{it}=-1$. From the researcher’s perspective, the number of individuals choosing $\omega _{it}=-1$ follows a binomial distribution so that $N\rho _t\sim \mathcal {B}(N,\hat{\rho }_t)$ and hence:

$$\begin{aligned} \Pr \left( \rho _t\ge \tilde{\rho }\right)&=1-\Pr \left( \rho _t<\tilde{\rho }\right) \nonumber \\&=1-\sum ^{\lfloor N\tilde{\rho }\rfloor }_{j=0}\left( {\begin{array}{c}N\\ j\end{array}}\right) \hat{\rho }_t^j(1-\hat{\rho }_t)^{N-j} \end{aligned}$$

(11)

where $\lfloor N\tilde{\rho }\rfloor$ is the largest integer less than $N\tilde{\rho }$.

This function does not change monotonically with N. However, some idea can be garnered as to how the probability is affected across general size increases. The binomial distribution can be approximated by a normal distribution with mean $N\hat{\rho }_t$ and variance $N\hat{\rho }_t(1-\hat{\rho }_t)$ when $N\hat{\rho }_t>5$. Assuming this is met, equation (11) can be approximated by:

$$\begin{aligned} \qquad \Pr \left( \rho _t\ge \tilde{\rho }\right)&=1-\Pr \bigg (\frac{N(\rho _t-\hat{\rho }_t)}{\sqrt{N\hat{\rho }_t(1-\hat{\rho }_t)}}<\frac{N(\tilde{\rho }-\hat{\rho }_t)}{\sqrt{N\hat{\rho }_t(1-\hat{\rho }_t)}}\bigg )\nonumber \\&\approx 1-\varPhi \Big (\sqrt{N}\frac{\tilde{\rho }-\hat{\rho }_t}{\sqrt{\hat{\rho }_t(1-\hat{\rho }_t)}}\Big ) \end{aligned}$$

(12)

which, for $\tilde{\rho }>\hat{\rho }_t$, is a decreasing function of N.

When a bad norm is in effect, smaller groups are thus generally more likely to breach the tipping proportion in a given round. The effect of size on persistence increases slowly and not monotonically, although comparisons can be made for sizes that are not very close together. This is due to the discrete nature of the possible proportions and hence the upper sum limit $\lfloor N\tilde{\rho }\rfloor$.