How to reconcile that Bayesian model selection considers the entire parameter space of a model?

Background example: Coin Toss

This is the standard example for Bayesian model selection (see, e.g., here). If you know this example, there is nothing new here:

We want to find out whether a coin is fair or loaded. We toss the coin a certain number of times and compare:

• $$\mathcal{M}_\text{fair}$$: a fair coin toss,
• $$\mathcal{M}_\text{loaded}$$: a coin toss with an unknown heads probability $$θ∈[0,1]$$.

Now, assuming equal priors for both models, we have:

$$p \left( \mathcal{M}_\text{loaded} \middle| \mathcal{D} \right) \propto p \left( \mathcal{D} \middle | \mathcal{M}_\text{loaded} \right) = \int_0^1 p \left( \mathcal{D} \middle | θ, \mathcal{M}_\text{loaded} \right) p \left( θ \middle | \mathcal{M}_\text{loaded} \right) \mathrm{d} θ,$$ where $$\mathcal{D}$$ is my coin-tossing data. With other words, we are averaging the likelihoods of the loaded model for all possible heads ratios.

My problem

I am generally struggling with the model selection averaging over all possible parameter combinations. As a result the outcome may strongly depend on what parameter combinations I admit for a given model (or their priors, respectively). In particular, it is usually not expected that a model aligns with a given dataset for all parameter combinations, but only a subspace thereof. By contrast, the most likely choice of parameter plays no special role.

In the coin-toss example, my problem manifests as follows: First suppose my data has a heads ratio of $$0.55$$ and the number of samples is such that Bayesian model selection prefers $$\mathcal{M}_\text{fair}$$ over $$\mathcal{M}_\text{loaded}$$. Suppose further that I consider a third model $$\mathcal{M}_\text{mildly loaded}$$, which only admits $$θ ∈ [0.3,0.7]$$. Since the above integral then excludes the least favourable $$θ$$ values, $$\mathcal{M}_\text{mildly loaded}$$ may end up being preferred over $$\mathcal{M}_\text{fair}$$.

This is a crucially different outcome: Not only do I get a different answer to the initial question (“Is the coin loaded?”), but if I also end up with a different predictor for further coin tosses, whether I choose a single model for prediction ($$\mathcal{M}_\text{mildly loaded}$$) or perform model averaging.

Another way to look at my problem is this: If I only have one model and compare two plausible parameters explaining the data in a Bayesian approach, I only need to consider the priors and Bayes factors of those two parameter values. The priors and Bayes factors of parameters that are totally implausible (given the data) don’t affect my analysis. So far so good. Now, in Bayesian model selection, the priors and Bayes factors for all parameters suddenly matter.

I consider these properties of Bayesian model selection strongly undesirable. Is my line of thought faulty or is this a known problem and if so, is there any justification or reconciliation?

My thoughts so far

• I am pretty sure I understand the math and arguments behind the equations. I am just running into contradictions and undesired behaviour.

• One possible reconciliation I considered is that since $$\mathcal{M}_\text{mildly loaded} ⊂ \mathcal{M}_\text{loaded}$$, the former should have a smaller prior, but then analogously I have $$\mathcal{M}_\text{fair} ⊂ \mathcal{M}_\text{mildly loaded} ⊂ \mathcal{M}_\text{loaded}$$ and the literature usually assumes equal model priors.

• Accounting all possible parameter sets cannot be justified by preventing overfitting. For example, $$\mathcal{M}_\text{mildly loaded}$$ and $$\mathcal{M}_\text{loaded}$$ are as prone to overfitting as long as within the overlap of $$θ$$.

• There is no clear way on how to choose the competing models. In the coin example, what keeps me from considering $$\mathcal{M}_\text{head-loaded}$$ and $$\mathcal{M}_\text{tail-loaded}$$ as separate models?

• I vaguely guess that the problem lies in the interpretation of $$p \left( \mathcal{M}_\text{loaded} \middle| \mathcal{D} \right)$$ and similar quantities, i.e., what the probability of a model means and whether it’s a desirable quantity for many applications at all.

• The Jeffreys–Lindsey paradox seems to be related and considers a similar situation. However, my issues arise without comparing to a frequentist approach.

• You may argue that ${\cal M}_\text{fair}$ has probability of zero within ${\cal M}_\text{loaded}$, so these are really different. However this does not hold for the relation between ${\cal M}_\text{mildly loaded}$ or ${\cal M}_\text{head-loaded}$; if these are true, ${\cal M}_\text{loaded}$ is true as well, so any prior probability assignment for such models will partly assign probability to the same parameter values. This doesn't look good to me for "model selection". It is not the case here that one model is wrong to the extent that another one is correct. Sep 7, 2022 at 13:36
• "the literature usually assumes equal model priors" - according to Bayesian philosophy, priors should be chosen taking into account all available information about the problem, and it is well known that assigning equal prior probabilities to models of different complexity can lead to issues and is hard to defend on "objectivity" grounds. Just assuming probabilities equal because you don't know what else to do will not lead to a proper meaningful Bayesian analysis. Sep 7, 2022 at 13:39
• @ChristianHennig: It is not the case here that one model is wrong to the extent that another one is correct. – I am not sure whether this is a critique of the scenarios I depict or of Bayesian model selection in general. However, consider the scenario where I don’t have $\mathcal{M}_\text{loaded}$ in the first place. Sep 7, 2022 at 13:54
• @ChristianHennig: it is well known that assigning equal prior probabilities to models of different complexity can lead to issues – First mind that I am only quoting the literature I seek to understand here. Also, one of the side effects of the problem I am asking about is an indirect penalty for higher-dimensional models and things like the BIC (under certain assumptions and approximations, AFAIU). Some authors even champion this as an automatic overfitting penalty. Of course there is the problem that this penalty is obfuscated, indirect, and depends on things I think it shouldn’t. Sep 7, 2022 at 13:54
• See my answer. Your observation is good in the sense that much of the literature doesn't care much about the interpretation of the prior, and in some situations this is a recipe for disaster. I think that Bayesian methods including model selection techniques are fine if the prior is chosen well, reflecting properly the situation of interest. This is often difficult to achieve, may require considerable subjective input, and in much literature you find quick and dirty shortcuts without giving any hint how a good informative prior should be chosen or how things may go wrong. Sep 7, 2022 at 14:23

A full specification of the prior would require to also specify prior probabilities for the models. Let's say we have only $${\cal M}_\text{fair}$$ and $${\cal M}_\text{loaded}$$. Let's further say you assign a prior probability of $$\frac{1}{2}$$ to both of these, and a uniform over $${\cal M}_\text{loaded}$$.

What this means is that a priori you expect there is a very substantial probability (0.5) that the coin is exactly fair, whereas you assign a very low probability to $$\theta$$ being, say, between 0.45 and 0.55 (namely $$0.5·0.1=0.05$$). This means that as long as your number of tosses is rather low, an observed relative frequency close to 0.5 will confirm the possibility that the true value is exactly 0.5, rather than favouring a model that allows it between 0.45 and 0.55 with nonzero but small probability, even if in fact the observed relative frequency is something like 0.47. If the true probability is 0.47, for a low number of observations, Bayes (with choices as indicated) may favour $${\cal M}_\text{fair}$$ over $${\cal M}_\text{loaded}$$ for this reason, however if the number of observations is large enough, also Bayes will at some point see that the observed relative frequency is so incompatible with $${\cal M}_\text{fair}$$ that it'll start favouring $${\cal M}_\text{loaded}$$.

This is all as it should be, if you realise that your prior assignment gave $$\theta=0.5$$ a strong advantage over "$$\theta\neq 0.5$$ but close". If this is backed up by reliable prior knowledge, fair enough. If you just choose your priors for the two models as ½ each because you don't know better, well this is what you get. You're free to change your prior assignments; you don't have to give ½ to $${\cal M}_\text{fair}$$, and neither do you have to choose the prior as uniform within $${\cal M}_\text{loaded}$$. In a realistic situation in which you think there's a large probability for $$\theta=0.5$$, usually you would also expect $$\theta$$ rather close to 0.5 in case of $${\cal M}_\text{loaded}$$, and your prior within $${\cal M}_\text{loaded}$$ should reflect that, with the effect that Bayes can already see that $$\theta$$ may not be equal to 0.5 but still close at a lower sample size.

The effect that $$\theta=0.5$$ gets a strong advantage may be seen as implicit penalty for the more complex model, and as such may be seen as welcome. However it isn't really clear on which grounds then the prior for $${\cal M}_\text{fair}$$ should be chosen as 0.5 rather than 0.3, or 0.6, or whatever. All these imply some kind of penalty, but there isn't really a rationale apart from understanding the underlying situation that could tell you what exactly the penalty should be. Furthermore it may be argued that this is an "un-Bayesian" consideration, as it is often not about what you believe the truth is or even how to predict future observations (which is what according to standard Bayesian philosophy the prior should be about), but rather it favours simpler models because it supposedly would be nice to have a simpler model (and people are scared of "overfitting"), so if the true $$\theta$$ is 0.508, you may be happy to select a model that has it at 0.5, even though a true Bayesian should not be happy about this.

Regarding comparing $${\cal M}_\text{mildly loaded}$$ with $${\cal M}_\text{loaded}$$, I think this is not a well defined model selection problem, as both can be true at the same time with positive probability ($${\cal M}_\text{loaded}$$ gives probability 0 to $${\cal M}_\text{fair}$$ being true, so arguably these two won't be true at the same time). Any assignment of prior probabilities within $${\cal M}_\text{mildly loaded}$$ and $${\cal M}_\text{loaded}$$ plus probabilities for these two models can be emulated by a suitably chosen prior within $${\cal M}_\text{loaded}$$ alone, without even introducing $${\cal M}_\text{mildly loaded}$$, which means that "the probability of $${\cal M}_\text{mildly loaded}$$ being true as opposed to $${\cal M}_\text{loaded}$$" is not identifiable and not meaningful. (A valid probability model is defined in this way, fair enough, so it is not forbidden to start like this, but if you look at quantities like the posterior probability of $${\cal M}_\text{loaded}$$, this won't have an interpretation that is useful for model selection).

So I wouldn't introduce $${\cal M}_\text{mildly loaded}$$ in the first place, but rather handle all $$\theta\neq \frac{1}{2}$$ by suitable prior choice within $${\cal M}_\text{loaded}$$, which gives you a posterior that can be interpreted in a straightforward manner. The same thing can of course be said regarding $${\cal M}_\text{head-loaded}$$ etc.

PS: I can imagine a situation in which the introduction of both $${\cal M}_\text{loaded}$$ and $${\cal M}_\text{mildly loaded}$$ and probabilities for both of them are meaningful, which is if you have information that there are two possibilities how your coin might have been constructed (maybe you want to guess which company did it), and for one you know that it operates according to $${\cal M}_\text{loaded}$$, the other one according to $${\cal M}_\text{mildly loaded}$$. You also then need to assume the priors within both of these. This will give you a well defined posterior probability for the source of your coin.

This, however, is not a typical situation in model selection. In model selection, there is only $$\theta$$, and $${\cal M}_\text{loaded}$$ and $${\cal M}_\text{mildly loaded}$$ are just thought constructs. This particularly means that even if you knew that the true $$\theta=0.4$$, you do not know whether this was generated from $${\cal M}_\text{loaded}$$ or $${\cal M}_\text{mildly loaded}$$, and the data do not contain information that could help to distinguish these (all information to distinguish these comes from the prior choices). This is what I mean by stating above that posterior probabilities for these are not identifiable and not meaningful.

• Thank you for your answer. While it does confirm my line of thought in many ways, I am not satisfied in one respect: I did not introduce $\mathcal{M}_\text{mildly loaded}$ to compare it with $\mathcal{M}_\text{loaded}$ but with $\mathcal{M}_\text{fair}$. Here I have the problem that $\mathcal{M}_\text{mildly loaded}$ may win over $\mathcal{M}_\text{fair}$ while $\mathcal{M}_\text{loaded}$ does not. The same happens if I choose respective priors within $\mathcal{M}_\text{loaded}$ that effectively make it $\mathcal{M}_\text{mildly loaded}$. Sep 8, 2022 at 7:54
• This becomes even more clear when splitting by direction: $\mathcal{M}_\text{head-loaded}$ may win over $\mathcal{M}_\text{fair}$, while $\mathcal{M}_\text{loaded}$ does not. In this situation, I arrive at the conclusion that the coin is more likely to be loaded than not, if and only if I split the model. While this split is admittedly somewhat artificial, it it is not in more realistic scenarios where I am coming from and, e.g., the analogues of $\mathcal{M}_\text{tail-loaded}$ and $\mathcal{M}_\text{tail-loaded}$ have different model priors. Sep 8, 2022 at 7:54
• Finally, please note that I added another attempt to illustrate my problem to my question. Sep 8, 2022 at 7:55
• @Wrzlprmft OK, I didn't realise that you were not interested in comparing ${\cal M}_\text{mildly loaded}$ and ${\cal M}_\text{loaded}$ directly. What you describe seems to refer to uniform priors, but then for example $\theta=0.47$ has a higher prior probability if you choose ${\cal M}_\text{mildly loaded}$ compared to ${\cal M}_\text{loaded}$, which explains why in a borderline situation if you observe 0.47 empirically, ${\cal M}_\text{mildly loaded}$ may "win" against ${\cal M}_\text{fair}$ where ${\cal M}_\text{loaded}$ doesn't. Sep 8, 2022 at 9:24
• Hmm, interesting point. A wonder if this is consistent when splitting the model into tail- and heads-loaded such that the combined model and parameter priors stay the same. Sep 8, 2022 at 9:28

Not a Bayesian expert but I think the term to focus on here is $$p(\theta|\mathcal{M}_\text{loaded})$$.

This is the probability distribution of how the coin might be loaded. We might suggest that this distribution is flat (all weightings equally likely) except for $$\theta=0.5$$ (a fair coin).

In your first example, the data might closely resemble a fair coin so $$\mathcal{M}_\text{fair}$$ could be preferable to a loaded coin since there is significant probability that, if the coin is loaded, it will have a value for $$\theta$$, which makes the observations unlikely.

In the second example, you are modifying the probability distribution by removing values outside $$[0.3, 0.7]$$ and so the integral is focused towards values, where the observed data is more likely.

One more point: You say that $$\mathcal{M}_\text{fair}\subset \mathcal{M}_\text{loaded}$$. But this is not correct as $$\theta=0.5$$ is not a loaded coin. Rather $$\mathcal{M}_\text{fair} \cup \mathcal{M}_\text{loaded}$$ cover the entire model space.

• Regarding your last point: From the labels, you are right, but practically it makes no difference if we include $θ=½$ in $\mathcal{M}_\text{loaded}$ because it has has measure 0. Also, in almost every application problem, the more complex model features the simpler model as a special case. Sep 6, 2022 at 14:54
• As far as I understand your answer, you mostly elaborate why the models are rank the way they do in a Bayesian model selection, when we accept this process as it is. However, my question is about why the process is the way it is. Thus, I am afraid, your answer does not resolve my problem. Also: there is significant probability that if the coin is loaded it will have a value for $𝜃$ which makes the observations unlikely. – But that applies to every set of observations (with a decent sample size). If I observe 100 tails and no heads, every $θ$ that is not close to zero is unlikely. Sep 6, 2022 at 15:05
• I feel like this might be a problem of terminology. It is not right to include $\theta=0.5$ in $M_{loaded}$ because (as I understand it) you are trying to determine the relative probabilities of these two models. If you include all possible values of $\theta$ then the probability of $M_{loaded}$ is 1 since it contains all possible coins. Sep 6, 2022 at 15:14
• Secondly, the statement 'there is significant probability that if the coin is loaded it will have a value for 𝜃 which makes the observations unlikely' is independent of the observations made. This is why I suggest focusing on interpreting $p(\theta|M_{loaded})$ as it does not contain the observations made. Sep 6, 2022 at 15:16
• If you include all possible values of $𝜃$ then the probability of $𝑀_\text{𝑙𝑜𝑎𝑑𝑒𝑑}$ is 1 since it contains all possible coins. – No, there are model priors which avoid this. Also, the literature I reference does not exclude $θ=0.5$ from the loaded model either. Finally, as already mentioned, it doesn’t affect the calculations at all. The integral from the question has the same value. Sep 6, 2022 at 15:34