We analyze 2012 presidential election data from France, Finland and Russia. To evaluate the validity and the linkage of both digit tests, we selected these cases for multiple reasons. First and most importantly, France and Finland are assumed to be fraud free, while the Russian election was most likely manipulated. There have been no accusations of electoral fraud or anomalies in France for quite some time (com-pare Klimek et al., 2012; OSCE/ODIHR, 2012a). In the Finish presidential election, some anomalies were found in the correlation between overall turnout and True Finns party’s vote share, which can be attributed to successful voter mobilization by the controversial party (Klimek et al., 2012). Apart from this naturally caused anomaly, the OSCE characterized the election as clean (OSCE/ODIHR, 2012b). By contrast, Russian elections do not appear to be controversial over the fact that there have been strong indications of fraud since 2003 (see Myagkov, Ordeshook and Shakin, 2009; Mebane and Kalinin, 2009; Deckert, Myagkov and Ordeshook, 2011; Kalinin and Mebane, 2011; Kobak, Shpilkin and Pshenichnikov, 2012; Klimek et al., 2012).
The final election OSCE observer report assessed voting on election day as good and very good in 95% of the observed polling stations. However, due to procedu-ral irregularities during the counting process, almost one-third of observed polling stations were assessed as bad or very bad. The irregularities cited included cases of
2.3. DATA AND METHODS
group, proxy and multiple voting; improperly sealed ballot boxes and indications of buses transporting groups of voters to vote at multiple polling stations. Concerns regarding the counting procedures were mainly listed as insufficient transparency during the counting process and some ballot box stuffing (OSCE/ODIHR, 2012c, 10f). In Russia it has been shown that ballot box stuffing changes the shape of the turnout distribution and generates a high correlation between vote share and turnout (Myagkov, Ordeshook and Shakin, 2009; Kobak, Shpilkin and Pshenichnikov, 2012;
Klimek et al., 2012).
The second reason for the choice is that these countries have dispersed vote count distributions. It is known that narrow vote count distributions with limited range cause a strong deviation from the expected 2BL distribution (Shikano and Mack, 2011). Similar limitations have been addressed for LD distributions, as they “require that vote counts do not cluster within a very small range of numbers” (Beber and Scacco, 2012, 217). Electoral units in France, Finland and Russia vary more than, for example, in Germany, which should increase the fit with 2BL and LD in vote counts. Table 2.3 provides information about the main quantities of the data. It shows that mean (eligible) voters in France and Russia are similar, while those from Finland are slightly higher.6 The variation in numbers of voters across districts is substantial for all three countries, but France has a particularly large range of voters.
This is a potential drawback of the French election data, which are aggregated at the municipality level – the lowest level available in France. Vote counts of smaller municipalities are provided at the level of polling stations, while large municipalities (which contain different polling stations) report aggregated data. In general, this increases the scope of vote counts and therefore should contribute to the fit of 2BL and LD. To avoid inconsistency in our analysis, we include a control sample for the first and second rounds of the French election, excluding all municipalities that exceed the 90% quintile of voters. This excludes all aggregated data points and decreases the range and mean of the number of voters substantially;7 50% of the vote counts vary between about 100 and 500 voters, which reveals the limited scope of the control data and makes deviations more likely. This is mainly problematic if districts have very homogeneous distributions, which French elections are not particularly known for.
In the following analysis, we rely in principle on the most popular test statistic, Pearson’sχ2. For 2BL we calculate the test statistics as:
χ22BL =
9
X
i=0
(di−dqi)2
dqi , (2.1)
6In contrast to many electoral districts in many countries, those in Finland are not designed to contain a similar number of eligible voters.
7There is no other clear distinction possible between the aggregated data and vote counts from polling stations.
30
Table 2.3: Descriptive statistics of the presidential elections in 2012
N obs. N units Mean Mean Mean Range 25% of 75% of Pres. election 2012 unit size eligible voter voters voters voters voters France: first round 36,441 92 396 1,085 887 4 - 378,905 135 667
France: control 32,766 91 360 437 371 4 - 1,673 124 493
France: second round 36,441 92 396 1,085 891 6 - 382,975 136 664
France: control 32,767 91 360 437 371 6 - 1,669 125 493
Finland: first round 2,301 14 164 1,902 1,329 72 - 8,095 542 1,897 Finland: second round 2,302 14 164 1,901 1,257 37 - 8,231 497 1,824
Russia 95,566 84 1,138 1,152 752 0 - 19,711 247 1,195
where qi denotes the expected relative frequency of i in the second digit, di is the empirical frequency of second digit i in a district and d is the sum of polling stations in the district. The statistic is assumed to be distributed according to a χ2 distribution with 9 degrees of freedom. Therefore, we can evaluate the significance of the deviation of empirical data from Benford’s Law using a critical value of 16.92 at a significance level of 5%. This method is also applicable when testing the deviation of the last digit; the only difference is that the frequency of the last digit is compared with the expected frequency of the uniform distribution.8
We conduct separate test statistics for each of a country’s electoral districts, which makes it necessary to adjust the assessment of statistical significance for multiple testing. As suggested by Mebane (2006b), we use the false discovery rate (FDR) method (Benjamini and Yekutieli, 2005; Benjamini and Hochberg, 1995). When as-suming independence across tests, then lett={1, . . . , T} index theT-tested district and let St be the significant probability of the test statistic for each district. St is sorted from the smallest to the largest value, starting withS1. The FDR is controlled using the following procedure: for a chosen test level α, let dbe the smallest value such that S(d+1) > (d+ 1)α/T. d gives the number of tests rejected by the FDR criterion, which should be d= 0 if digits in vote counts distribute according to 2BL or LD.
Testing 2BL and LD with χ2 means that the power of the tests depends on the sample size – i.e., the number of polling stations in a district. Therefore, we exclude all electoral districts that contain 50 or fewer polling stations. For the French election data, this reduces the number of districts from 107 to 92 (91 for the control sample) – and from 15 to 14 districts for Finland, and 85 to 84 for Russia. If we adjust the test statistic for each candidate based on the number of tests we compute in each election according to the FDR criteria (compare N units in Table 2.3), we get an adjustedχ2 of 29.45 in France (29.42 for the control data), 24.50 in Finland and 29.22 in Russia.9
8This formula and detailed information are listed in Mebane (2008b, 179) and Shikano and Mack (2011), and on the last digit in Beber and Scacco (2012).
9Data from France and Finland are available at http://www.interieur.gouv.fr/Elections and http://pxweb2.stat.fi/Database/statfin/vaa/pvaa/pvaa_2012/pvaa_2012_en.asp.