Did Clyburn Help Biden in South Carolina?

By: Dr. Ikjyot Singh Kohli

The conventional wisdom by the political pundits/analysts who are seeking to explain Joe Biden’s massive win in the 2020 South Carolina primary is that Jim Clyburn’s endorsement was the sole reason why Biden won. (Here is just one article describing this.)

I wanted to analyze the data behind this and actually measure the effect of the Clyburn effect. Clyburn formally endorsed Biden on February 26, 2020.

Using extensive polling data from RealClearPolitics, I looked at Biden’s margin of victory according to various polling samples before the Clyburn endorsement. I used Kernel Density Estimation to form the following probability density function of Biden’s predicted margin of victory (as a percentage/popular vote) in the 2020 South Carolina Primary:

Assuming this probability density function has the form p(x), we notice some interesting properties:

  • The Expected Margin of Victory for Biden is given by: \int x p(x) dx. Using numerical integration, we find that this is \int x p(x) dx = 18.513 \%. The error in this prediction is given by var(x) = \int x^2 p(x) dx - (\int x p(x) dx)^2 = 107.79. This means that the predicted Biden margin of victory is 18.51 \pm 10.382. Clearly, the higher bound of this prediction is 28.89%. That is, according to the data before Clyburn’s endorsement, it was perfectly reasonable to expect that Biden’s victory in South Carolina could have been around 29%. Indeed, Biden’s final margin of victory in South Carolina was 28.5%, which is within the prediction margin. Therefore, it seems it is unlikely Jim Clyburn’s endorsement boosted Biden’s victory in South Carolina.
  • Given the density function above, we can make some more interesting calculations:
  • P(Biden win > 5%) = 1 - \int_{-\infty}^{5} f(x) dx = 0.904 = 90.4%
  • P(Biden win > 10%) = 1 - \int_{-\infty}^{10} f(x) dx = 0.799 = 79.9%
  • P(Biden win > 15%) = 1 - \int_{-\infty}^{15} f(x) dx = 0.710 = 71.0%
  • P(Biden win > 20%) = 1 - \int_{-\infty}^{20} f(x) dx = 0.567 = 56.7%

What these calculations show is that the probability that Biden would have won by more than 5% before Clyburn’s endorsement was 90.4%. The probability that Biden would have won by more than 10% before Clyburn’s endorsement was 79.9%. The probability that Biden would have won by more than 20% before Clyburn’s endorsement was 56.7%, and so on.

Given these calculations, it actually seems unlikely that Clyburn’s endorsement made a huge impact on Biden’s win in South Carolina. This analysis shows that Biden would have likely won by more 15%-20% regardless.

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ikjyotsinghkohli24

Sikh, Theoretical and Mathematical Physicist, main research in the structure and dynamics of Einstein's field equations.

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