As you move to the right however, relative consistency should still exist. Slope possibly in either direction, or more likely a scatter plot with ridiculously inconsistent R^2 correlation. There can’t be consistency to 20 to 25 percent then a cliff.
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There is no actual cliff in most of his examples.
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And you can find "cliffs" in Biden's data too
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Slope looks to be about -1 (compared to about -0.6 for the Trump graphs). Interesting.
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That probably just depends which precinct you pick. There are 176,000 to choose from.
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?? You just grouped them. A .6D correlates to a 0.4R. There is no variation to “pick” dude. you plotted them. Perhaps you or shiva have plotted one or the other wrong
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Yeah, something is wrong here. If third party votes are negligible, the slope of the Biden graph for a particular county should be the same as the Trump graph for that county (which also makes it obvious that Dr. Ayyadurai's argument that non-zero slope implies fraud is wrong).
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No, the slopes will not be the same they will be different. You have to remember a .6D is a .4R and so on and vice versa. It’s really scalar but interrelated. There should not be a discernible slope match. For example macomb cnty for B, if 1/2
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For Trump:
y = i(x) - x
dy/dx = di/dx - 1
Add a prime for Biden vars:
y' = i'(x') - x'
dy'/dx' = di'/dx' - 1
x'+x = 1
i'(x')+i(1-x') = 1
di'/dx' + dx/dx' * di/dx = 0
-> di'/dx' = di/dx
-> dy'/dx' = di/dx - 1 = dy/dx
Where di/dx is evaluated at x=1-x'.
Did I make a mistake?
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Note that dy'/dx' at x' is equal to dy/dx evaluated at x=(1-x'). If our graph is a straight line the derivative will be the same everywhere, so that detail doesn't matter.
Example:
y = 0.1 - 0.6*x
Checking expectations at x=0 and x=1 should confirm:
y' = 0.5 - 0.6*x'
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For the sake of completeness:
y'(x') = i'(x') - x'
= 1 - i(1-x') - x'
= 1 - [y(1-x') + (1-x')] - x'
= -y(1-x')
Again i is not a function of x. You can’t make that assumption.
Y=z-x y’=z’-x’ z=1-z’ x=1-x’. Sub in all you want but z is not a function of x. You have two equations three variables so many possibilities?The question is should it be perfect and predictable. Probably no.
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These:
x' = 1 - x
y' = -y
map every possible pt in the x-y plane to the corresponding pt in the x'-y' plane based only on requiring probs to add to 1. To talk about slope you need to map a curve (fitted to data), so y'(x')=-y(1-x') is the same thing in more appropriate notation.
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