Mathematical proof of fraud in Russian elections unsound
In crowds, it is stupidity and not mother wit that is accumulated.
Gustave Le Bon, The Crowd: A Study of the Popular Mind
On Saturday 10th December thousands-strong crowds swarmed Russian cities protesting against alleged fraud in the December 4th parliamentary elections. Some of the allegations have mathematical origin. A Washington Post article states:
“Obviously, he [Putin] doesn’t agree with Gauss,” one commenter wrote, referring to pioneering mathematician Carl Friedrich Gauss, who lived 200 years ago. Disenchanted Russians argue that United Russia’s reported election results are so improbable as to violate Gauss’s groundbreaking work on statistics.
The article does not say what exactly the problem with the election result is and what work of Gauss is relevant. It only says that he lived 200 years ago. However, this may be enough to trigger an alert: the science had somehow advanced during past 200 years. I took a closer look at the allegations.
The banner says “We are for Normal distribution.” (Image from http://nl.livejournal.com/1082778.html)
In an article titled “Mathematics against Election Committee: Gauss against Churov [the head of the committee]” the blogger complained that the distribution of the percent of the vote for the United Russia party among election precincts is non-Gaussian. This, he wrote, is an evidence of an election fraud because Gaussian distribution arises
Always. In every case, when there is not one factor, but many. Whatever is measured in large quantities. Make a plot of how many millions of men in the country have the height of 165, 170, 175 centimeters and so on – and you will also get a symmetric bell-curve with the top corresponding to the most typical height in the country.
Yes, the heights of people are Gaussian-distributed. But what about incomes? They are distributed as if indeed most people were 170 centimeters tall, but often you would meet a three-meter guy. Rarely you would encounter a five-meter man, more rarely – a ten-meter one. Sometime, from a distance, you would see a hundred-meter person. And there would be several hundred-kilometer chaps in the country. This distribution is very far from Gaussian, but for some reason it does not attract the wrath of our mathematicians, of our Berezovskies. There are many non-Gaussian distributions in both nature and society [1] and there is no reason to believe that the distribution of the percent of the vote for a party among election precincts must be Gaussian. We can illustrate this using the work of the Russian mathematician Andrei Markov [2].
The culprit is that to get a Gaussian distribution the many factors must be independent. Consider an urn with one white and one black ball. Let us pull a random ball out of the urn, record its color and put the ball back. If we make a large number of such independent trials, the distribution of the fraction of the pulled white balls will be Gaussian. This is because the color of the ball we pull this time does not depend on the color of the ball we pulled out in the preceding trial: that very independence of factors. In the case of elections independence of factors means, that people chose their political views independently of their neighbours, co-workers and friends. To account for dependent events Markov [2] modified the model in the following way. The urn initially contains one white and one black ball. We pull out a random ball, then put it back and in addition add to the urn another ball of the same color. After two trials, we could pull out either two black, two white or one black and one white ball. Elementary combinatorics shows that these three combinations are equally probable. That is the number of pulled out white balls can be 0, 1, or 2, and each of these numbers has the same probability – 1/3. You can prove by induction that after N trials all numbers from 0 to N of pulled out white balls are equiprobable (you can also find the proof in Chapter 7 of Ref. [1]). Interestingly, this uniform distribution even somewhat resembles the mathematically impossible distribution presented on the banner of protest.
![The banner says “We do not trust Churov [the head of Election committee]! We trust Gauss” (Image from http://nl.livejournal.com/1082778.html)](/SpringboardWebApp/userfiles/sig/image/AbdelUpload/b25-vi.jpg)
The banner says “We do not trust Churov [the head of Election committee]! We trust Gauss” (Image from http://nl.livejournal.com/1082778.html)
What relation can the ball problem have to the elections? Let us consider the following model. In a small city, which has only one election precinct, in the beginning there are two party members. One represents the White ball party and the other – Black ball party. Each of them starts agitating for his party. When the agitator persuades someone to join his party, the new party member himself starts agitating. Let us suppose that the agitator for White ball party got lucky the first. Now there are two people agitating for White party and only one for Black party. If we suppose that each agitator has equal chances to succeed, then the probability that the new party member will join the White party is two times bigger than the probability that he joins the Black party. We have a one to one correspondence with Markov’s model. This means that the vote percentage distribution among the precincts must be not Gaussian, but uniform. Of course, the model we just considered is oversimplified. We completely neglected the influence of people living in different precincts on each other. This dependence, though lesser than the influence of neighbours, co-workers and friends still exists. That we have only two parties in the model may be not that big of a defect. The bloggers agitated to vote for anyone, but the United Russia party. So we can present the situation like a choice of either for or against United Russia. Of course, there is no reason to believe that the model I just described is close to reality. It is thus not clear what the distribution of vote percentages among precincts must be according to Science. It is, however, clear that there is no grounds for a demand for this distribution to be Gaussian.
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Mikhail Simkin
Quote:
Non-Gaussian curves in US elections you can see here
http://www.significancemagazine.org/details/webexclusive/1435463/US-elections-are-as-non-normal-as-Russian-elections.html
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bbzippo
Quote:
They most definitely do not follow normal distribution. The vote ratios for different parties are obviously linearly realted (they add up to 100%). So you cannot say "all parties are normal but one".
E.g. if you plot a simple scatter of United Russia vs. Communist P. it becomes apparent that they simply are linear transforms of each other.
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Steve Coleman
The same questions about possible election fraud were raised concerning the previous parliamentary election in 2007, which had an outcome similar to the latest. But one can largely explain the results of the 2007 election from an understanding of social conformity, which has an especially strong effect on Russian voters compared with other countries. This does not mean that no voting irregularities occurred, but they did not have much impact on the results. See the article by Stephen Coleman published on this in 2010 in the Russian language Journal of the New Economic Association, Vol 5, pp 73-90; English title is "Russian Election Reform and the Effect of Social Conformity on Voting and the Party System: 2007 and 2008." An English version is available on line at the website of the New Economic Association. The Russian version of the article is online in volume 5, "Реформа российской избирательной системы и влияние социальной конформности на голосование и партийную систему: 2007 и 2008."
reply to this comment
Alex
So your argument is that Gaussian distribution for all other parties is explained by bloggers' efforts?
Something is definitely missing in your analysis. You can't really be serious saying that 32 million people voted at random because bloggers told them so.
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Mikhail Simkin
Quote:
When they say that all election results must be Gaussian, I can take it as an erroneus, but a scientific statement. It has backing in Central Limit Theorem. However, your statement that they could be non-Gaussian but only all of them is nothing but a whim.
The bloggers agitated people to attend elections and vote for any opposition party. Suppose that the division between those supporting United Russia and those opposing it happened according to Markov's model. Suppose that those who oppose United Russia vote for randomly selected opposition party. This is reasonable, since they just came to vote against United Russia. This will lead to uniform distribution for United Russia and Gaussian for the rest of the parties.
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Alex
Now that's just brilliant. A post that defies mathematics with randomly picked stories. Surely, sociology is just as simple as your ball example.
How about other parties? Are you sure you haven't missed anything? How come the results of other parties follow normal distribution?
Yes, election results should not necessarily be Gaussian. But no mathematical argument would ever explain, why all (!) parties follow Gaussian, except one.
reply to this comment
Mikhail Simkin
Quote:
While there could be other statistical issues with the elections, the major isue was what I said. You can see the equation for Gaussian distribution on the protesting banner. You can read it in The Washington Post. Finally I found a blog in English for you
http://www.themarketfinancial.com/russian-electionsstatistical-bias-between-10000000-and-20000000-votes/128218
reply to this comment
Michael
I think the Gaussianity issue is a red herring (while they are some kind of bell curve, and one can get one from uniform distributions, they are obviously not Gaussian). Wasn't the mathematical controversy about precinct/polling station turnout data, rather than actual party votes per se as the article in Russian perhaps confusingly argues? Apparently, the empirical voter turnout distribution for all precincts for each presidential and legislative election in Russia in the past decade shows a remarkable change in shape and jaggedness in the vicinity of the right tail, and while this does not conclusively prove anything, this kind of exploratory modus operandi is pretty standard for getting clues about what the data might mean.
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Mikhail Simkin
Quote:
Protesters argued that the results of Russian elections contradict mathematics. You argue that they are different from those in the USA. A very different kind of argument.
Regarding peaks I may look into it. However, I need not just percentages, but the actual numbers of votes to make any conclusion.
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Vladimir
It would be much more convenient if you just showed non-Gaussian curves with highly regular sharp peaks at every 5 or 10% obtained during elections in the USA as for example.
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