Average probabilities, not log odds

I think it would be perhaps helpful to link to a few people advocating averaging log-odds rather than averaging probabilities, eg: 

• When pooling forecasts, use the geometric mean of the odds [? · GW]
• My current best guess on how to aggregate forecasts [EA · GW]

Personally, I see this question as being an empirical question. Which method works best?

In the cases I care about, both averaging log odds and taking a median far outperform taking a mean. (Fwiw Metaculus agrees that it's a very safe bet too)

In contrast, there are no conditions under which average log odds is the correct thing to, because violating additivity of disjoint events is never the correct thing to do (see previous paragraph).

Your example about additivity of disjoint events is somewhat contrived. Averaging log-odds respects the probability for a given event summing to 1, but if you add some additional structure it might not make sense, I agree.

Averaging log-odds is exactly a Bayesian update, so presumably you'd accept there are some conditions under which average log odds is the correct thing to do...

I did a Monte Carlo simulation for this on my own whose Python script you can find on Pastebin.

Consider the following model: there is a bounded martingale $M$ taking values in $[0,1]$ and with initial value $1/2$. The exact process I considered was a Brownian motion-like model for the log odds combined with some bias coming from Ito's lemma to make the sigmoid transformed process into a martingale. This process goes on until some time T and then the event is resolved according to the probability implied by $M\left(T\right)$. You have n "experts" who all get to observe this martingale at some idiosyncratic random time sampled uniformly from $[0,T]$, but the times themselves are unknown to them (and to you).

In this case if you knew the expert who had the most information, i.e. who had sampled the martingale at the latest time, you'd do best to just copy his forecast exactly. You don't know this in this setup, but in general you should believe on average that more extreme predictions came at later times, and so you should somehow give them more weight. Because of this, averaging the log odds in this setup does better than averaging the probabilities across a wide range of parameter settings. Because in this setup the information sets of different experts are as far as possible from being independent, there would also be no sense in extremizing the forecasts in any way.

In practice, as confirmed by the simulation, averaging log odds seems to do better than averaging the forecasts directly, and the gap in performance gets wider as the volatility of the process $M$ increases. This is the result I expected without doing any Monte Carlo to begin with, but it does hold up empirically, so there's at least one case in which averaging the log odds is a better thing to do than averaging the means. Obviously you can always come up with toy examples to make any aggregation method look good, but I think modelling different experts as taking the conditional expectations of a martingale under different sigma algebras in the same filtration is the most obvious model.

My heuristic for deciding what heuristic to use, when you're going to do something quick-n-dirty: figure out what quantity you're actually interested in, use means in its natural scale for point estimates, and transform back to your inputs.

How does this apply to some examples?

In your post, you're talking quite a lot about bets. To a first approximation marginal utility is linear in marginal wealth, so usually this means the quantity we are actually interested in is linear in probability, and the correct heuristic is "arithmetic mean" (of probability).

In SimonM's comment, we're talking about probabilities directly. Forecasting. Usually that means what we care about is calibration or a proper scoring rule, so the natural scale is [0,1] or log-odds. Now the correct heuristic is "arithmetic mean" (of log-odds of probability).

What about your difficult examples?

Expert 1 gives you a probability distribution 50% A, 25% B, 25% C, expert 2 gives you a probability distribution 25% A, 50% B, 25% C, and expert 3 gives you a probability distribution 25% A, 25% B, 50% C.

We're already doing quick-n-dirty things rather than anything rigorous. What I usually do when I want constraints and my summary statistics don't satisfy them is to just go ahead and normalize, after which of course we get 1/3, 1/3, 1/3 by symmetry after any estimation.

that depends on the prior

Once we're no longer just doing something quick-n-dirty, all, uh, bets, are off. But my heuristic is still to transform to the domain where what you care about is linear, do your somewhat-more-sophisticated dirty work, and transform back to get your point estimate.

Probably part of the intuition motivating something more like average log odds rather than average probabilities is that averaging probabilities seems to ignore extreme probabilities.

Counter-example:

9 out of 10 people give a 1:100,000,000 probability estimate of winning the lottery by picking random numbers, the last person gives a 1:10 estimate. Averaging the probabilities gives a 1:100 estimate, and you foolishly conclude these are great odds given how cheap lottery tickets are.

In your ABC example we rely on the background information that

• P(A&B)=0
• P(A&C)=0
• P(B&C)=0
• P(A or B or C)=1.

So the background information is that the events are mutually exclusive and exhaustive. But only then do probabilities need to add to one. It's not a general fact that "probabilities add to 1". So taking the geometric average does itself not violate any axioms of probability. We "just" need to update the three geometric averages on this background knowledge. Plausibly how this should be done in this case is to normalize them such that they add to one. (In the case of the arithmetic mean, updating on the background information plausibly wouldn't change anything here, but that's not the case for other possible background information.)

Of course this leads to the question: How should we perform such updates in general, i.e. for arbitrary background assumptions? I think how this is commonly done is via finding the distribution which maximizes entropy of the old distribution under the background information, or finding the distribution which minimizes the KL-divergence to the old distribution (I think these methods are equivalent). Of course this requires that we have such a distribution in the first place, rather than just three probabilities obtained by some averaging method.

But it is anyway a more general question (than the question of whether the geometric mean of the odds is better or the arithmetic mean of the probabilities): how should we "average" two or more probability distributions (rather than just two probabilities), assuming they come from equally reliable sources?

But going back to your example about extreme probabilities: Of course when we assume that sources have different reliability, then they would need to be weighted somehow differently. But the interesting case here is what the correct averaging method is for the simple case of equally reliable sources. In this simple case the geometric mean seems to make much more sense, since it doesn't generally discount extreme probabilities. (Extreme probabilities can often be even more reliable than non-extreme ones. E.g. the probability that the Earth is hollow seems extremely low.)

Also here a quick comment:

I will acknowledge that there are conditions under which averaging probabilities is also not a reasonable thing to do. For example, suppose some proposition X has prior probability 50%, and two experts collect independent evidence about X, and both of them update to assigning 25% probability to X. Since both of them acquired 3:1 evidence against X, and these sources of evidence are independent, combined this gives 9:1 evidence against X, and you should update to assigning 10% probability to X. The prior is important here;

If we assume that the prior was indeed important here then this makes sense, but if we assume that the prior was irrelevant (that they would have arrived at 25% even if their prior was e.g. 10% rather than 50%), then this doesn't make sense. (Maybe they first assumed the probability of drawing a black ball from an urn was 50%, then they each independently created a large sample, and ~25% of the balls came out black. In this case the prior was mostly irrelevant.) We would need a more general description under which circumstances the prior is indeed important in your sense and justifies the multiplicative evidence aggregation you proposed.

Lastly:

I think it helps to think about why there would be a significant gap between the worst odds you'd accept a bet at and the worst odds you'd accept the opposite bet at. One reason is that someone else's willingness to bet on something is evidence for it being true, so there should be some interval of odds in which their willingness to make the bet implies that you shouldn't, in each direction. Even if you don't think the other person has any relevant knowledge that you don't, it's not hard to be more likely to accept bets that are more favorable to you, so if the process by which you turn an intuitive sense of probability into a number is noisy, then if you're forced to set odds that you'd have to take bets on either side of, even someone who knows nothing about the subject could exploit you on average. I think the possibility that adversaries can make ambiguity resolve against you disproportionately often is a good explanation for ambiguity aversion in general, since there are many situations, not just bets, where someone might have an opportunity to profit from your loss.

This is a very interesting possible explanation for betting aversion without needing to assume, as usual, risk aversion! Or rather two explanations, one with an adversary with additional information and one without. But in the second case I don't see how a noisy process for a probability estimate would lead to being "forced to set odds that you'd have to take bets on either side of, even someone who knows nothing about the subject could exploit you on average". Though the first case seems really plausible. It is so simple I would assume you are not the first with this idea, but I have never heard of it before.

In contrast, there are no conditions under which average log odds is the correct thing to do

Taking that as a challenge, can we reverse-engineer a situation where this would be the correct thing to do?

We can first sidestep the additivity-of-disjoint-events problem by limiting the discussion to a single binary outcome.

Then we can fulfill the condition almost trivially by saying our input probabilities are produced by the procedure 'take the true log odds, add gaussian noise, convert to probability'.

Is that plausible? Well, a Bayesian update is an additive shift to the log odds. So if your forecasters each independently make a bunch of random updates (and would otherwise be accurate), that would do it. A simple model is that the forecasters all have the same prior and a good sample of the real evidence, which would make them update to the correct posterior, except that each one also accepts N bits of fake evidence, each of which has a 50/50 chance of supporting X or ~X (and the fake evidence is independent between forecasters). 

That's not a good enough toy model to convince me to use average log odds for everything, but it is good enough that I'd accept it if average log odds seemed to work in a particular domain.

The whole debate seems to be poorly founded. There are many ways to combine probability estimates that have different properties and are suited to different situations. There is no one way that works best for every purpose and context.

In some contexts, the best estimate won't even be within the range of the individual estimates. For example, suppose there is some binary question with prior 1:1 odds. Three other people have independent evidence regarding the question and give odds of 1:2, 1:3, and 1:4 based on their own evidence. What is your best estimate for the odds? (Hint: it's not anywhere between 1:2 and 1:4)