Comments on FiveThirtyEight: Politics Done Right: Pollster Ratings v3.0

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2009-10-09T07:00:21.261-04:00

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2009-08-02T01:23:16.882-04:00

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2009-02-10T21:34:00.000-05:00

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2009-02-08T23:11:00.000-05:00

gene -- kind of figured it out after a while -- th...

2008-10-16T11:09:00.000-04:00

gene -- kind of figured it out after a while -- thanks for confirmation.

to Cedarfluteyes it's geekybest are at the top of ...

2008-10-16T10:23:00.000-04:00

to Cedarflute

yes it's geeky
best are at the top of the list

Like a previous commenter, I am puzzled about your...

2008-09-28T10:35:00.000-04:00

Like a previous commenter, I am puzzled about your equation for adding errors. Normally, errors are added as variances, so as squares. The total sum squared deviation from the true value is the sum of variance (sum squared error around the data's mean) and squared bias (squared distance between mean and true value). Am I missing something? Adding absolute values instead of squares is quite different.

This may sound sarcastic, but it's not. Are the m...

2008-07-05T09:53:00.000-04:00

This may sound sarcastic, but it's not. Are the most reliable pollsters at the top or bottom of the list?

I'm sorry to keep bugging you on this, and I'm not...

2008-04-30T00:41:00.000-04:00

I'm sorry to keep bugging you on this, and I'm not even sure you keep reading this Poblano, but I reread your original post and it seems to me you are quite explicit, pace Rasmus, in avoiding time error in weighing pollsters.

I do think you end up penalizing large samplers who otherwise are likely to be the better methodologists. Think of it this way: the best way to build a reputation with you is to keep on sampling 300 people. You make bad errors, but so does everyone else; your errors however get assigned to a sampling error.

Where your original post seems to go wrong is in assuming that time error essentially vanishes towards D-1; this is obviously false as the last minute is of course non-zero (it is captured only in part by "when did you make your decision" exit polling; last minute atmosphere has an impact at the turnout level as well). Rare is the election where this last minute effect is not of the order of magnitude of sampling error!

Anon,Yes, that exact solution occurred to me today...

2008-04-30T00:39:00.000-04:00

Anon,

Yes, that exact solution occurred to me today. In fact, I'm going to be putting an update up pronto.

Alternatively, you could just get a rough estimate...

2008-04-29T21:09:00.000-04:00

Alternatively, you could just get a rough estimate by taking the average of squared pollster errors, and then subtracting the average variance due to sample sizes. Treat the result as VAR(PIE).

If the average variance due to sample sizes is greater than the average squared pollster error, set PIE equal to zero.

Sorry, the Gaussian formula should read:P(D | VAR(...

2008-04-29T01:11:00.000-04:00

Sorry, the Gaussian formula should read:

P(D | VAR(PIE), VAR(S)) ~ PROD(exp(-(Di^2) / (2*(VAR(Si) + VAR(PIE))))

Yeah, this is a little more complicated than I tho...

2008-04-29T01:08:00.000-04:00

Yeah, this is a little more complicated than I thought.

The below approach might work:

We want to find VAR(PIE) given the actual errors for a pollster (call them D) and the sample variance VAR(S).

From Bayes' rule, we have:

P(VAR(PIE) | D, VAR(S)) ~ P(D | VAR(PIE), VAR(S)) * P(VAR(PIE) | VAR(S))

where I replace the denominator (which is independent of VAR(PIE)) with the ~ symbol to represent proportionality.

Now basically we need to come up with reasonable distributions for the following terms:

1) P(D | VAR(PIE), VAR(S))
2) P(VAR(PIE) | VAR(S))

For 1), it might make sense to use the product of independent Gaussians:

P(D | VAR(PIE), VAR(S)) = PROD(exp(-Di / (VAR(Si) + VAR(PIE)))

Where Di is the error for the ith poll, and VAR(Si) is the sample variance for ith poll.

For 2), we could use an improper uniform distribution, or more realistically an exponential distribution. All we need to chose is lambda, which can be set to some reasonable estimate.

Once the distributions are chosen, we plug in the data, find the normalization constant (wth respect to VAR(PIE)), and then find the expected value for VAR(PIE). (If we instead tried to find the most likely value, I think it would be zero.)

I haven't worked this all through, so please let me know if something seems wrong.

An alternative approach would be to explicitly use a multinomial distribution to express the probability of the data given the "true" probability and pollster error, but I think this migth be even more complicated.

Anons,OK, so what happens when the expected sampli...

2008-04-28T16:53:00.000-04:00

Anons,

OK, so what happens when the expected sampling error exceeds the actual error? For example, Quinnipiac misses Pennsylvania by just one point, when its expected S is 3?

Poblano,A couple of points:1) Anon is right about...

2008-04-28T16:29:00.000-04:00

Poblano,

A couple of points:

1) Anon is right about the sum of squares. Basically, you are modeling the pollster's prediction P as:

P = E + S + PIE

Where E is the actual results of a full election, S is the sampling error, and PIE is the pollster introduced error. S and PIE are assumed to be independent random variables. The variance of the random variable (S + PIE) is the variance of S plus the variance of PIE. The mean of (S + PIE) is the mean of S plus the mean of PIE.

2) The final unofficial PA results are in:

Clinton: 1,259,466 (54.6%)
Obama: 1,046,120 (45.4%)

http://www.electionreturns.state.pa.us/ElectionsInformation.aspx?ElectionID=27

Rasmus,One thing is that, in the older version I a...

2008-04-28T16:10:00.000-04:00

Rasmus,

One thing is that, in the older version I actually wasn't regressing to the mean enough, so taking care of that properly results in a slight decrease in SurveyUSA's relative strength.

An unknown pollster gets a PIE of +1.55, FWIW, i.e. slightly worse than average. This is actually determined via regression, since those pollsters with fewer surveys in the database performed slightly worse on balance than pollsters with more surveys. So I just extrapolated this trend toward the y-intercept and that's how I got that figure.

Anon,

Can you point me to that metadata link? I think the way I'm doing this is mathematically sound (and there a probably a number of mathematically equivalent ways of doing this) but this could be a case where my theoretical skills lag behind my applied skills, so I'd be happy to take a look.

From MetaDataI would expect PIE and Sampling error...

2008-04-28T15:47:00.000-04:00

From MetaData

I would expect PIE and Sampling error to combine via sqrt of sum of the squares, not regular addition, so you would have 3.86 instead of :
4.46 = 3.58 + 0.88

Maybe I didn't quite understand, and the PIE is an offset, not a random error.

Anon 2:There IS a time error included.The polls ha...

2008-04-28T12:28:00.000-04:00

Anon 2:
There IS a time error included.
The polls have a half-life of 30 days. And yes, large-sample size pollsters are penalized, but when large sample sizes lead to better results, they should have a better PIE= better results.

Nevertheless I like the old model better.

doktarr:
Yes, the weights in the right column are updated and
no, he doesn´t update the tracker.
We had that discussion in one of the last topics were he updated his regression model ("This Morning´s Poll, and a Regression Model Makeover").

Great stuff. I especially like the way you back y...

2008-04-28T12:14:00.000-04:00

Great stuff. I especially like the way you back your results out to an effective sample size - there are lots of ways to handle this that are mathmatically equivalent, but intuitively I really like that approach.

Have all the poll weights in the right column been updated with this approach? Also, when you make an update like this, do you go back through old data and update the "win percentage tracker"?

I probably miss the arithmetic and some of your ex...

2008-04-28T12:08:00.000-04:00

I probably miss the arithmetic and some of your explanations, but my impression is that by ignoring time error you effectively penalize large-sample pollsters (by having less of a sampling error cushion with which to absorb the time error, they end up presenting more of it to your model as a PIE).
There ought to be ways for projecting some kind of time error; I understand this would differ depending on the intensity of the last-minute campaign, but I am not sure this is not in fact some kind of fairly stable number.

No, I don´t like this new version, I´ll stick with...

2008-04-28T11:36:00.000-04:00

No, I don´t like this new version, I´ll stick with the old one.
You actually don´t "update" the Pollster Reliability, you take it away- in fact.

In CA, there is a good example too: The most recent polls of SUSA and Rasmussen get roughly the SAME weight.

SUSA is a much better pollster, but Rasmussen can make up for that just with a poll that is 4 days newer and has a sampling size of 503 compared to 500.

In my model (old 538 system with a few modifications) they had a weight of 0,96 SUSA compared to 0,86 Rasmussen- the difference is more realistic I think.

It isn´t.My numbers for AK and AL are now updated,...

2008-04-28T11:27:00.000-04:00

It isn´t.
My numbers for AK and AL are now updated, and my results for the So.Alabama polls matched those of 538 exactly, when I assumed a PIE of 1,49. In the "old" system, it was the same way, and .85 was the average. not 1, a pollster with a rating of 1 was better than average (PPP for example, but PPP is below average now).

Another question: Don´t you think the Pollster Reliability/PIE however you call it gets too less weight when it is just combined with the Sampling Error?

There is not such a big difference between

Total Expected Error = Sampling Error + Long-Run PIE
4.46 = 3.58 + 0.88

and

Total Expected Error = Sampling Error + Long-Run PIE
4.46 = 3.58 + 1,49

...

Look at AR.
The SUSA poll and the UCA poll get about the same weight- the sampling sizes are roughly the same, and the SUSA poll is only 11 days older- still the both polls get about the same weight, no matter that SUSAs PIE is smaller than UCAs.

I would think the PIE of a no-name pollster would ...

2008-04-28T10:48:00.000-04:00

I would think the PIE of a no-name pollster would be higher than average. In the "old" system, the no-name pollster got a rating of +.85, which was about midway between ARG and Gallup. That would imply that under the new system, the long-term PIE would be about +1.9

q&a:Am I right that the PIE of a no-name pollster ...

2008-04-28T10:43:00.000-04:00

q&a:
Am I right that the PIE of a no-name pollster (e.g. SO.Alabama or DWU Tiger) is 1,49?