# Research explaining mean reversion of election forecasts: why each new poll provides very very little information about the election outcome, given what we already know

by on September 19, 2012 · 6 comments

There is a huge amount of mean reversion in Nate Silver’s model right now . . . Considering that Silver’s forecast F is roughly F = λ(polls) + (1-λ)(fundamentals), either λ must be really small or truly extraordinary things have happened to Silver’s fundamentals in the past week.

I don’t know what Nate’s model is, but if he’s doing things right, then indeed the weight attached to the recent week of polls must be really small. Each week’s polls provide very very little information about the election outcome, given what we already know.

For more background, see this article (with Noah Kaplan and David Park) on the random walk and mean reversion models and this article (with Kari Lock) on Bayesian combination of state polls and election forecasts.

Andreas Moser September 19, 2012 at 12:23 pm

I usually just wait for election night. Then I find out everything that is really relevant.

Guy September 19, 2012 at 2:31 pm

I think you have it backwards, here. As each week goes by now, the survey data becomes more important as we assume it already captures whatever impact the “fundamentals” are going to have. By late October, you would want to weight the “fundamentals” essentially to zero.

Andrew Gelman September 19, 2012 at 4:24 pm

Guy:

Read carefully, I wrote, “Each week’s polls provide very very little information about the election outcome, given what we already know.”

Guy September 19, 2012 at 6:13 pm

I was reacting to the prior sentence: “I don’t know what Nate’s model is, but if he’s doing things right, indeed λ must be really small. ” I’m not sure what “really small” means, but I would think by mid-September it should be approaching 50%, no? If not, when would you say it should reach 50%?

Andrew Gelman September 19, 2012 at 8:53 pm

Guy:

Good point. I’ve clarified above.