Claim: R-Squared Coefficient of Determination as a Election Fraud Signal

Mick West

Administrator
Staff member
R-Squared is a number you can calculate from two sets of data that given an indication of if there's a correlation between them, and how tight that correlation is. You can calculate this in Excel simply by using the RSQ(range1, range2) on two ranges of data (like two columns). Here I calculate it for the straight-ticket voters vs. the individual candidate voters. I do this for both Biden and Trump.

2020-11-18_13-08-50.jpg

That looks a bit more complicated than it is. $E$2:$E$253 is just specifying column E, rows 2-253

As you can see, Biden's R-Squared (which I'm going to abbreviate as R2) is 0.141, and Trump's is 0.896. Very different numbers (R2 can only go from 0 to 1). This difference has been suggested as evidence, a detectable "signal", that indicates election fraud. For example by Dr. Shiva.
2020-11-18_13-13-41.jpg

What does this mean? Well if you do a scatter-plot of two ranges of data, with one on the X axis, and the other on the Y-axis, then if there is a high correlation, an R2 of near 1.0, then the plot will cluster around a straight line. With data used here, this means that the number of individual votes a candidate is closely proportional to the number of straight party votes they get.

Here I do a scatter-plot, with trendlines for this with Biden and Trump in Kent County,
2020-11-18_13-34-19.jpg

Each dots is one precinct, with the size of the dot (and the number next to it), being the number of votes cast in that precinct. We can see things a bit clearer if we separate out the Tump and Biden graphs.
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So obvioiusly the pattern, the relationship is different between Trump and Biden. The quesiton is: why? I suspect that there are other variables at play here (probably not fraud), but I'm not sure what they are.
 

Akton

Member
My first thought is that it would be good to test this across multiple counties/states and multiple elections in multiple years to see what the normal variation in this kind of thing is. Is it actually uncommon for candidates to vary this much?

Second, unless something is going over my head, is this sort of thing not simply a measure of ticket splitting (i.e. Biden benefitted more from ticket splitting than Trump did)?
 

Mendel

Senior Member.
The quesiton is: why? I suspect that there are other variables at play here (probably not fraud), but I'm not sure what they are.
Generally, R2 being small means your model doesn't fit the data. The assumption that this proportion should be constant is wrong. Has Dr Shiva finally made a case why this assumption should be true?

A good election to compare this with is the 1992 presidential election, when Democrat Bill Clinton defeated incumbent Republican George H. W. Bush.

Second, unless something is going over my head, is this sort of thing not simply a measure of ticket splitting (i.e. Biden benefitted more from ticket splitting than Trump did)?
Yes.

If you're looking at Democratic ticket voters over Biden vote share, R²=0.85; they're quite regular.

But the individual voters for Biden are all over the place. The reason is that Michigan flipped: the Biden campaign managed to reach voters who turned out and voted, or who switched their vote. But how well they did that would depend on the demographics of the precinct, so I'd expect that to vary quite a bit. For example, if we assumed hypothetically that half of suburban women voting Trump in 2016 now switched to Biden, we wouldn't expect that shift to affect all precincts equally.

If it affects straight-party and individual Trump voters equally, then that percentage would remain unchanged; but both groups would become individual Biden ticket voters, if we assume that an individual Trump voter is unlikely to become straight-ticket democrat, and that the former straight-ticket Republican voter is even less likely to do that. Isn't that how your model from the other thread worked, Mick?

With these assumptions, the straight/individual split could remain fairly constant for Trump, and be all over the place for Biden.
 

Mick West

Administrator
Staff member
If it affects straight-party and individual Trump voters equally, then that percentage would remain unchanged; but both groups would become individual Biden ticket voters, if we assume that an individual Trump voter is unlikely to become straight-ticket democrat, and that the former straight-ticket Republican voter is even less likely to do that. Isn't that how your model from the other thread worked, Mick?
I split the precinct vertically into Republican leaning and Democrat leaning. It's poorly defined but essentially someone who votes for the senator of that party.

Those two columns are split into straight ticket (bottom half) and individual voters (upper half)

The individual voters are then split into same-part-president, and flipped or split ticked president votes.

I then move the R/D split left to right to get the distribution.

In the model before, all precincts had the same proportion of split ticket voters as a percentage of the "leaning" column.

It's a bit of a rough approach, and ideally, I'd be tweaking underlying variables and doing a more complex monte-carlo sim. BUt I got carried away with Geogebra.
 

Mick West

Administrator
Staff member
But yeah, I think asymmetric variance in the split votes will account for most of it. My model just does not do that realistically.

I'm playing around with Jupyter Notebooks to do some better visualization - might take a few days though.
 

Z.W. Wolf

Senior Member.
It's common knowledge that a large number of Republican voters rejected Trump while continuing to support down ticket Republican candidates.

This election was a rejection of Trump, not Republicans.

The number of Democrats who voted for Trump is very small.
 
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