User talk:Tttfffkkk

Nomination of Divisible polynomial for deletion
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February 2021
Please refrain from using talk pages such as Talk:Benford's law for general discussion of the topic or other unrelated topics. They are for discussion related to improving the article in specific ways, based on reliable sources and the project policies and guidelines; they are not for use as a forum or chat room. If you have specific questions about certain topics, consider visiting our reference desk and asking them there instead of on article talk pages. See here for more information. Thank you. Constant314 (talk) 17:50, 15 February 2021 (UTC)
 * Now, it sounds like political bias. Tttfffkkk (talk) 23:23, 15 February 2021 (UTC)
 * It's not political bias, it's just not related to the topic. Your theory about election fraud in Milwaukee County has nothing to do with Benford's Law apart from the most tangential attempt to link the topics by saying Benford's Law might not have detected this supposed fraud. Feel free to raise your hypothesis at the Mathematics Reference Desk or a politics discussion board so that others can review your analysis and provide feedback. --Canley (talk) 11:07, 16 February 2021 (UTC)
 * Yeah, Benford's Law might or might not detect election fraud. Great hypothesis. I think I should start researching about that. Sorry, I didn't realize it. I published original work on the talk page again.Tttfffkkk (talk) 11:31, 16 February 2021 (UTC)
 * OK, no need to be sarcastic. I watched your video by the way, and I think there's some fundamental errors in the Milwaukee County analysis, happy to discuss it here if you're interested in some feedback, or at the Reference Desk... or not at all if you prefer, up to you. --Canley (talk) 11:47, 16 February 2021 (UTC)
 * Thanks for watching. Nevermind, just a coincidence. Tttfffkkk (talk) 11:57, 16 February 2021 (UTC)

Linear regression analysis of the election votes in Milwaukee County
I'm invited by Canley to raise a hypothesis here mentioned in my talk page User talk:Tttfffkkk. I made a video respecting for this linear regression analysis. Here are data sources from Milwaukee.

In conclusion, my Linear Regression model indicates that a candidate has stolen other candidates averagely 120 votes on each ward in Milwaukee

My Regression Model is Candidate's vote = Candidate's voting rate × Total votes which is no constant term expected

Form data at Nov. 4, 2020 Biden's vote = 5.3191 + 0.5821 × Total votes R^2=0.9087 Trump's vote = -6.6928 + 0.4002 × Total votes R^2=0.8255

Without constant term, the linear regression becomes: Biden's vote = 0.5877 × Total votes R^2=0.9085 Trump's vote = 0.3931 × Total votes R^2=0.8262

Form data at Nov. 5, 2020 Biden's vote = 121.6115 + 0.5629 × Total votes R^2=0.8128 Trump's vote = -120.573 + 0.4171 × Total votes R^2=0.7051 The constant term lies on the convincible interval far from 0. Here's the problem. Why does this constant term appear? I'm not an expert in Statistics. I haven't heard of any linear regression analysis about the election votes before. Anythings may surprises me.Tttfffkkk (talk) 13:04, 16 February 2021 (UTC)
 * There are several fundamental errors with this analysis:
 * You produce a linear regression model of election results at two points in time, but the variables are neither independent nor random—for a start, the November 5 results are cumulative and include the November 4 results; also as they are from two points in time so they are not random samples from the same population. You can't just apply the same linear model for one day to the next day. This is a similar error to the analysis by Charles Cicchetti, which stated that "the reported tabulations in the early and subsequent periods could not remotely plausibly be random samples from the same population" — which of course they weren't. The results over two days are different because as is well established and reported, mail-in votes which favoured Biden and the Democrats were counted at later points as they were received. Matt Parker did a video about this claim.
 * You wonder where the constant term around 120 comes from: the November 4 regressions have an intercept close to 0 — on November 4, only 418 wards had reported, which means the data has 64 (0,0) values. By November 5, all 478 wards had reported (although 3 are still 0), so a regression on the first day is going to be more weighted to the origin, but the data from later dates will reflect the support for the more popular candidate (in this case, Biden-Harris). Absolute nonsense to claim that this is evidence that "votes were stolen" and then multiply by the number of wards to produce an estimate of "57,360 votes" stolen!
 * For your interest, there is an article by Charleen Adams debunking election fraud claims using linear regression here. --Canley (talk) 21:41, 16 February 2021 (UTC)


 * I feel that you have research a lot. I appreciate your work sincerely.
 * 2.
 * I should apologize for those 64 (0,0) values included. Here's the result updated without those points:
 * Form data at Nov. 4, 2020
 * Biden's vote = 6.7388 + 0.5806 × Total votes
 * R^2=0.9003
 * Trump's vote = -8.4791 + 0.4020 × Total votes
 * R^2=0.8141
 * The constant term lies on the convincible interval including 0.
 * Without constant term, the linear regression becomes:
 * Biden's vote = 0.5877 × Total votes
 * R^2=0.9000
 * Trump's vote = 0.3931 × Total votes
 * R^2=0.8131
 * Actually, it becomes a bit far from the origin but generally pass through it.
 * 1.
 * My linear regression model assumed that there is just one voting rate. In fact, there are two different groups of votes (mail-in ballots and in-person ballots) with different voting rate. I think this would be the important concern. Suppose their votes are separated to group A and B without expected constant term:
 * $$\begin{cases}y_A=\hat{\beta_A} x_A,~\hat{\alpha_A}=\bar{y_A}-\bar{x_A}\hat{\beta_A}=0\\

y_B=\hat{\beta_B} x_B,~\hat{\alpha_B}=\bar{y_B}-\bar{x_B}\hat{\beta_B}=0\end{cases}$$
 * Now, I wonder if these two groups are mixed together to do linear regression, what exactly is the expected constant term comes out. Here's what I'm figuring:
 * $$\hat{\alpha}=\bar{y}-\bar{x}\hat{\beta}=\frac{N_A \bar{y_A}+N_B \bar{y_B}}{N_A+N_B}-\frac{N_A \bar{x_A}+N_B \bar{x_B}}{N_A+N_B}\hat{\beta}$$
 * $$=\frac{N_A}{N_A+N_B}(\bar{y_A}-\bar{x_A}\hat{\beta})

+\frac{N_B}{N_A+N_B}(\bar{y_B}-\bar{x_B}\hat{\beta})$$
 * $$=\frac{N_A \bar{x_A}}{N_A+N_B}(\hat{\beta_A}-\hat{\beta})

+\frac{N_B \bar{x_B}}{N_A+N_B}(\hat{\beta_B}-\hat{\beta})$$
 * while $$\hat{\beta}=\frac{\hat{\beta_A} N_A\sum_{x\in A} x^2-\hat{\beta_A} \left(\sum_{x\in A}x\right)^2

+\hat{\beta_B} N_B\sum_{x\in B} x^2-\hat{\beta_B} \left(\sum_{x\in B}x\right)^2 +N_A\sum_{(x,y)\in B} xy+N_B\sum_{(x,y)\in A} xy-\sum_{x\in A}x\sum_{y\in B}y-\sum_{x\in B}x\sum_{y\in A}y} {N_A\sum_{x\in A} x^2+N_A\sum_{x\in B} x^2+N_B\sum_{x\in A} x^2+N_B\sum_{x\in B} x^2-\left(\sum_{x\in A}x\right)^2-\left(\sum_{x\in B}x\right)^2-2\sum_{x\in A}x\sum_{x\in B}x}$$
 * The constant term $$\hat{\alpha}$$ becomes a mystery. I can't assert that it is expected to be 0. Tttfffkkk (talk) 04:04, 17 February 2021 (UTC)

Can't the two of you move the discussion to Tttfffkkk's talk page? I do not understand the implied model(s), if any, in the original question or in any of the further comments. As far as I can see a topic, it is not suitable for the Reference desk. --Lambiam 04:29, 17 February 2021 (UTC)
 * I've moved the conversation here, I had wondered if others would be interested to chime in but clearly not. --Canley (talk) 05:03, 17 February 2021 (UTC)

I got the constant term. $$\bar{y}=\frac{N_A \bar{y_A}+N_B \bar{y_B}}{N_A+N_B} =p_A\bar{y_A}+p_B\bar{y_B}$$ $$\bar{x}=\frac{N_A \bar{x_A}+N_B \bar{x_B}}{N_A+N_B} =p_A\bar{x_A}+p_B\bar{x_B}$$ $$\hat{\alpha_A}=\frac{\bar{y_A}E(x_A^2)-\bar{x_A}E(x_A y_A)}{D(x_A)}=0 \Rightarrow E(x_A y_A)=\hat{\beta_A} E(x_A^2)$$ $$\hat{\alpha_B}=\frac{\bar{y_B}E(x_B^2)-\bar{x_B}E(x_B y_B)}{D(x_A)}=0 \Rightarrow E(x_B y_B)=\hat{\beta_B} E(x_B^2)$$ $$\hat{\alpha}=\frac{\bar{y}E(x^2)-\bar{x}E(xy)}{D(x)}$$ $$\bar{y}E(x^2)-\bar{x}E(xy)=p_A\bar{y}E(x_A^2)+p_B\bar{y}E(x_B^2) -p_A\bar{x}E(x_A y_A)-p_B\bar{x}E(x_B y_B)$$ $$=p_A p_B\bar{y_B}E(x_A^2)+p_A p_B\bar{y_A}E(x_B^2) -p_A p_B\bar{x_B}E(x_A y_A)-p_A p_B\bar{x_A}E(x_B y_B)$$ $$=p_A p_B\hat{\beta_B}\bar{x_B}E(x_A^2)+p_A p_B\hat{\beta_A}\bar{x_A}E(x_B^2) -p_A p_B\hat{\beta_A}\bar{x_B}E(x_A^2)-p_A p_B\hat{\beta_B}\bar{x_A}E(x_B^2)$$ $$=p_A p_B(\hat{\beta_B}-\hat{\beta_A})(\bar{x_B}E(x_A^2)-\bar{x_A}E(x_B^2))$$ $$=p_A p_B \bar{x_A}\bar{x_B}(\hat{\beta_B}-\hat{\beta_A}) (\bar{x_A}-\bar{x_B}+\frac{D(x_A)}{\bar{x_A}}-\frac{D(x_B)}{\bar{x_B}})$$ $$\hat{\alpha}=\frac{p_A p_B \bar{x_A}\bar{x_B}(\hat{\beta_B}-\hat{\beta_A})} {p_A D(x_A)+p_B D(x_B)} (\bar{x_A}-\bar{x_B}+\frac{D(x_A)}{\bar{x_A}}-\frac{D(x_B)}{\bar{x_B}})$$ $$\hat{\alpha}=0$$ holds when $$E(x_A)=E(x_B)~\text{and}~D(x_A)=D(x_B)$$ which implies the nature of independent variable remains unchanged. The nature of mail-in ballots and in-person ballots possibly fluctrates the constant term. I compute it by myself but I haven't seen this kind of conclusion on text book. I have the wrong impression that this constant term remains 0. Tttfffkkk (talk) 00:19, 18 February 2021 (UTC)