Weighting Polls vs Bigmood

For districts with polls, we take a weighted average of the bigmood prediction and aggregate poll prediction to get to our final prediction. The weight (\(0 < weight < 1\)) of the polls depends on 1) the number of polls, 2) the grade of each poll, and 3) the age (in days before the election) of each poll. We first calculate the Grade Point Sum (GPS) of all of the polls. If a district has \(m\) polls, and the \(i\)th poll has grade \(g_{i}\) and is from \(t_{i}\) days before the election, the GPS is:

$$GPS=\sum\limits_{i=1}^m G(g_{i})e^{\frac{-t_{i}}{167}}$$

Where \(G(g_{i})\) is \(0.177\) if \(g_{i}\) is A, \(0.151\) if \(g_{i}\) is B, \(0.130\) if \(g_{i}\) is C, and \(0.077\) if \(g_{i}\) is D.

The weighting of the aggregate poll prediction (\(w_{p}\)) is then given by:

$$w_{p}=\frac{1.9}{\pi} \cdot \arctan(6.12 \cdot GPS)$$

And the weighting of the bigmood prediction is (\(1-w_{p}\)).

The constants of the arctan function were chosen so that it has an asymptote of \(0.95\) (a district with a very large GPS has a poll weight very close to \( 0.95\)) and so that a district with two B polls, each finished the day of the election, will have poll weight \(0.6\).

Blairvoyance

Blairvoyance provides a hypothesized poll result for districts without polls by interpolating on demographics of the districts that do have polls. First, Blairvoyance creates linear relationships between districts using select demographics for that district. Figure 1 is a simplified example where the three districts are plotted against two axes representing two demographics, which is effectively a representation of a relationship between districts.

  The closer they are in Figure 1, the more similar the districts by some metric formed by the demographic data. The red dots represent the districts with polling data. Blairvoyance then moves the points that represent district with polls to another dimension, which represent the poll result. Figure 2 is a representation of Figure 1 with the third axes of poll result is added, and Figure 3 shows the points being moved into the third dimension representing polls.

Keep in mind that, our model contains more than two demographics, therefore Blairvoyance utilizes more than 3-dimensions. Blairvoyance then tries to fit a curve relating demographics and poll results. The blue line in Figure 4 represents a potential fitted curve Blairvoyance might create in that simplified case. Then, in order to create a hypothesized poll result for a district without poll, it will use the poll result of the closest poll in the fitted curve, as can be seen in Figure 5.

Blairvoyance takes in polls and returns poll hypotheses. Typically, polls are only done on the districts with close races. Therefore, Blairvoyance should make poll hypotheses closer than actual polls would be; the weight of Blairvoyance should decrease as partisanship increases. Hence, the weight of Blairvoyance should be at most the best poll we can have, which is an A graded poll taken one day before the election. Let the weight of one A graded poll taken one day before election be \(w_a\). Then the weight of Blairvoyance would be \(w_a (1 + 2|bigmood - 0.5|)\). This satisfies both properties that it is at most \(w_{a}\) and decreases as partisanship increases. If some hypothetical district is completely partisan, Blairvoyance would not count toward the prediction of that district.

Calculating AUSPICE

The Agglomeration Utilizing Statistical Predictions and Inquiries Concerning Elections (AUSPICE) is the final election forecast that the ORACLE of Blair gives a district every run. AUSPICE is a combination of the distribution provided by Bigmood and the averaged polls, or Blairvoyance in the absence of polls, weighted as mentioned above. Suppose the weight of bigmood is \(w_b\) and the weight of the averaged polls for a district (or Blairvoyance) to be \(w_l\). And suppose that bigmood gives a mean of \(\mu_b\) and a standard deviation of \(\sigma_b\) for the proportion of the two-party vote going to the Democratic nominee, and the polls give a mean of \(\mu_l\) and a standard deviation of \(\sigma_l\). The AUSPICE for that district will return a mean of \(w_l \mu_l + w_b \mu_b\) and a standard deviation of \(\sqrt{(w_l \sigma_l)^2+(w_b \sigma_b)^2}\).

National Predictions

At this point, we have predicted vote shares and standard deviations of vote shares, so it might seem that calculating the overall chance that a party gets a majority of seats is rather simple. However, this is not the case.

In order to avoid excessive computations or calculating an exact number, we simulate the entire House election \(10,000,000\) times each time we run our model. The probability that each party wins a majority of seats in the House is then about the number of simulations in which this happens divided by \(10,000,000\).

A naïve approach to use here would be to simulate each district separately, using the numbers we already have. However, this approach has the implicit assumption that the district vote shares are all independent, which is certainly not the case. For one, the systematic bias of polls can often be caused by the same factors from district to district. Also, since our model is not perfect, it may be consistently off in one direction for most districts. Thus, our simulation must introduce some correlation between districts.

The way we chose to implement this is to, for each simulation, choose a number \(s\) from a normal distribution with mean \(0\) and variance \(\sigma^2\) (we will explain where \(\sigma\) comes from later). Then, for every district \(d\) with mean vote share \(\mu_d\) and variance of the vote share \(\sigma_d^2\), we choose a number \(v_d\) from a normal distribution with mean vote share \(\mu_d\) and variance \(\sigma_d^2 - \sigma^2\) and let the simulated vote share be \(v_d + s\). Since \(v_d\) and \(s\) are independent, this means that the distribution of \(v_d + s\) has mean \(\mu_d\) and variance \(\sigma_d^2\), matching our earlier computations The important part is that \(s\) is the same for all districts, meaning that a portion of the variance of the districts is shared.

It suffices to find a suitable value for \(\sigma\). There is no clear way to do this, but we decided on the following: the outcome of the House is very correlated with the House popular vote, so a good choice for \(\sigma\) is one such that the resulting distribution of the House popular vote matches our uncertainty about it. Since the individual district variation (on the order of \(\sigma \sim 5\%\)) is all independent, the contribution to the House popular vote from these sources of variation is on the order of \(5\%/\sqrt{435} \sim 0.25\%\) which is far lower than the actual uncertainty. So we should have \(\sigma\) be our uncertainty about the House popular vote.

There are many ways to forecast the House popular vote, but using the standard deviation of the national voter turnout prediction is close to the best one can do. Therefore, we use the historical standard deviation of the national generic ballot polls versus the national popular vote, which corresponds to \(\sigma \approx 1.38\%\).

Example: California’s 25th District

Please note that in this example calculations may be shown with intermediate rounding. However, the actual computations used to produce results did not use intermediate rounding. This district is currently represented by Republican Steve Knight. He is being challenged by Democrat Katie Hill. This section was updated on October 26th.

BPI

Obama received \(49.1\%\) of the two-party vote in this district in 2012. As a consequence of California’s jungle primary system, there was no Democrat on the 2014 House ballot. We therefore used the aggregate two-party Democratic performance in the jungle primary, which was \(32.8\%\); there was no incumbent running in 2014. Clinton received \(53.6\%\) of the two-party vote in this district in 2016. There was a Democratic candidate for the House seat in 2016 and he got \(46.9\%\) of the vote; there was a Republican incumbent. Since there was a Republican incumbent in 2016, we adjust the Democrat’s vote share in 2016 to be \(51.0\%\).

This district thus has a BPI of \(0.133(49.1-50) + 0.278(32.8-50) + 0.244(53.6-50)+0.345(51.0-50) = -3.59\%\). Therefore, as there is a Republican incumbent running in 2018, this district has a SEER prediction of \(50\% + 0.5(2(-3.59\%) - 8.17\%) = 42.4\%\). Since there is an incumbent, we will use a standard deviation of \(0.066\) for the SEER prediction.

Bigmood