In my last article, I refuted one mathematical proof of fraud in Russian elections. That “proof” stemmed from the observation that the distribution of votes for United Russia (UR) among election precincts was non-Gaussian. I showed that theoretically there is no reason for the distribution to be Gaussian and presented non-bell-shaped curves in American elections. There are other “proofs” out there. The Wall Street Journal, for instance, found “fingerprints of fraud” in Russian elections. The deviation from Gaussian distribution is not listed as a “fingerprint,” but the WSJ has others:
Prime Minister Vladimir Putin's ruling United Russia party captured a high share of voters—far above the 49.3% it received nationwide—in precincts where voter turnout was reported to be well above the national average, according the analysis. That dynamic suggests broad ballot-stuffing, according to experts in vote monitoring.
Figure 1. Fingerprints of fraud: (a) in Russian 2011 parliamentary elections according to WSJ analysis; (b) in British 2010 elections according to my analysis.
Apparently, The WSJ assumed that the true turnout was more or less the same in different precincts, but in some of them, Putin’s KGB agents stuffed the boxes with ballots for United Russia. Such an operation could, indeed, increase both the turnout and UR vote share in the affected precincts. This looks like a reasonable explanation, and in 2008 American election, I could not find a significant correlation between turnout and the percentage of the vote for Obama. However, Kuznetsov has found such correlations in 2010 British parliamentary elections. Following this lead, I applied the WSJ’s analysis to British elections. Results are in Figure 1.
As you can see, when you go from constituencies with turnout below 60 percent to constituencies with turnout above 60 percent, the vote share of the Conservative Party increases two fold. The share of Liberal Democrats (LD) increases by one third, while the share of Labour Party drops 40 percent and the share of all other parties drops twice. Apparently, United Russia had borrowed the ballot-stuffing technique from the Conservative-LD coalition that rules Britain.
The WSJ gives its estimate of vote fraud in Russian elections:
A comprehensive examination of the full results from Russia's nearly 100,000 voting precincts reveals statistical anomalies that experts say are consistent with widespread vote-rigging. These irregularities could cast doubt, by one rough measure, over as many as 14 million of the 65.7 million votes reportedly cast.
Unfortunately, the WSJ authors do not say how they got this estimate. Apparently, not from the deviations from the bell curve, since they do not even mention them. However, Gazeta.ru had published an elaborate article by Shpilkin, which estimates the number of votes stolen in Russian elections. The method uses the very correlation between turnout and UR vote share, which the WSJ dubbed “a fingerprint of fraud.” Here is a figure from that article.
Figure 2. 2011 Russian parliamentary elections results in Moscow. Every dot represents a precinct (several hundred voters). You can clearly see that the higher the turnout is, the more votes United Russia gets.
I made a similar picture for British elections.
Figure 3. Results of 2010 British parliamentary elections.
In the above graph, every data point represents a constituency (several tens of thousands voters). You can see that the higher the turnout is, the more votes the Conservatives get.
Figure 4. Estimating the amount of fraud in British parliamentary elections.
In the next paragraph, I will use Shpilkin’s method to estimate the number of votes stolen in British elections. I keep all his language while making the minimal necessary changes to adopt it to British elections.
Assuming that the difference in distribution of votes between the Conservative-LD coalition and the other parties depending on the attendance is the result of artificial increase of the amount of votes for the coalition, we can try to establish the size of this increase. We will try to separate from Conservative-LD votes distribution, a part proportional to the sum of votes for all the other parties. As we see in Figure 4, we could separate a part from the votes for Conservative-LD, the part, proportional to all the other voices, so that until the attendance of 62% the remainder of the UR voices after subtraction of this part practically equals 0. In our assumptions that means that there were almost no added votes. The remaining (abnormal) part of this curve should be considered an artificial increase of the amount of votes for Conservative-LD. After the division between 'normal' (proportional to the votes for the other parties) and 'abnormal' is done, we can appraise them quantitatively and try to reconstruct the "corrected" voting results without such an increase. After subtracting this data from 17.5 million votes for Conservative-LD, the normal votes count up to 8.3 million, abnormal, artificially added votes total 9.2 million.
Since we are talking about the birthplace of parliamentary democracy, the conclusion is absurd. Therefore, we should seek a different explanation of the spurious correlation between attendance and vote share. Could it be that the members of the Conservative party are more active voters than the members of Labour party? Consider the following simplified model. There are two parties: Conservative and Labour. Three-quarters of the Conservative and one-half of the Labourists take the trouble to vote. One town is 80 percent Conservative and 20 percent Labour. The attendance here will be 70 percent; 60 percen of the registered voters will vote Conservative and 10 percent, Labour. Another town is 80 percent Labour and 20 percent Conservative. The turnout will be 55 percent; 15 percent of the registered voters will vote Conservative and 40 percent, Labour. We can get something similar to Figure 3 without any ballot stuffing by MI5.
Perhaps we do not need to postulate the KGB ballot-stuffing to explain the results of Russian elections as well?