Are there any examples where Bayesian credible intervals are obviously inferior to frequentist confidence intervals

A recent question on the difference between confidence and credible intervals led me to start re-reading Edwin Jaynes' article on that topic:

Jaynes, E. T., 1976. `Confidence Intervals vs Bayesian Intervals,' in Foundations of Probability Theory, Statistical Inference, and Statistical Theories of Science, W. L. Harper and C. A. Hooker (eds.), D. Reidel, Dordrecht, p. 175; (pdf)

In the abstract, Jaynes writes:

"...we exhibit the Bayesian and orthodox solutions to six common statistical problems involving confidence intervals (including significance tests based on the same reasoning). In every case, we find the situation is exactly the opposite, i.e. the Bayesian method is easier to apply and yields the same or better results. Indeed, the orthodox results are satisfactory only when they agree closely (or exactly) with the Bayesian results. NO CONTRARY EXAMPLE HAS YET BEEN PRODUCED". (the SHOUTING being mine not Jaynes')

The paper was published in 1976, so perhaps things have moved on, so my question is, are there examples where the frequentist confidence interval is clearly superior to the Bayesian credible interval (as per the challenge implicitly made by Jaynes).

Examples based on incorrect prior assumptions are not acceptable as they say nothing about the internal consistency of the different approaches.

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Under rather mild assumptions, (a) Bayesian estimation procedures are admissible and (b) all, or almost all, admissible estimators are Bayesian with respect to some prior. Thus it's no surprise that a Bayesian confidence interval "yields the same or better results." Note that my statements (a) and (b) are part of the frequentist analysis of rational decision theory. Where frequentists part company with Bayesians is not over the mathematics or even the statistical procedures, but concerns the meaning, justification, and correct use of a prior for any particular problem. – whuber Sep 3 '10 at 18:48
So, does the above comment imply that the answer to the OP's question is 'No such examples can be constructed.'? Or perhaps, some pathological example exists which violates the assumptions behind admissibility? – user28 Sep 3 '10 at 19:44
@Srikant: Good question. I think the place to begin investigating is a situation where there are non-Bayes admissible estimators--not necessarily a "pathological" one, but at least one that provides some opportunity to find a "contrary example." – whuber Sep 3 '10 at 22:29
I would add some clarity to the "incorrect prior assumptions..." by stating that the Bayesian answer and the frequentist answer must make use of the same information, otherwise you are just comparing answers to two different questions. Great question though (+1 from me) – probabilityislogic Jan 21 '11 at 12:26
pathology or not, it would probably be the first of its kind. I am very keen to see this example, for these "pathologies" usually have a good learning element to them – probabilityislogic Jan 30 '11 at 13:18

I said earlier that I would have a go at answering the question, so here goes...

Jaynes was being a little naughty in his paper in that a frequentist confidence interval isn't defined as an interval where we might expect the true value of the statistic to lie with high (specified) probability, so it isn't unduly surprising that contradictions arise if they are interpreted as if they were. The problem is that this is often the way confidence intervals are used in practice, as an interval highly likely to contain the true value (given what we can infer from our sample of data) is what we often want.

The key issue for me is that when a question is posed, it is best to have a direct answer to that question. Whether Bayesian credible intervals are worse than frequentist confidence intervals depends on what question was actually asked. If the question asked was:

(a) "Give me an interval where the true value of the statistic lies with probability p", then it appears a frequentist cannot actually answer that question directly (and this introduces the kind of problems that Jaynes discusses in his paper), but a Bayesian can, which is why a Bayesian credible interval is superior to the frequentist confidence interval in the examples given by Jaynes. But this is only becuase it is the "wrong question" for the frequentist.

(b) "Give me an interval where, were the experiment repeated a large number of times, the true value of the statistic would lie within p*100% of such intervals" then the frequentist answer is just what you want. The Bayesian may also be able to give a direct answer to this question (although it may not simply be the obvious credible interval). Whuber's comment on the question suggests this is the case.

So essentially, it is a matter of correctly specifying the question and properly intepreting the answer. If you want to ask question (a) then use a Bayesian credible interval, if you want to ask question (b) then use a frequentist confidence interval.

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Well said, especially about what question a CI actually answers. In the Jaynes' article however, he does mention that CI's (and most frequentist procedures) are designed to work well "In the long-run" (e.g. how often do you see $n \rightarrow \infty$ or "for large n the distribution is approximately..." assumptions in frequentist methods?), but there are many such procedures that can do this. I think this is where frequentist techniques (consistency,bias,convergence,etc.etc.) can be used to assess various Bayesian procedures which are difficult to decide between. – probabilityislogic Jan 21 '11 at 12:13
"Jaynes was being a little naughty in his paper..." I think the point that Jaynes was trying to make (or the point that I took from it) is that Confidence Intervals are used to answer question a) in a large number of cases (I would speculate that anyone who only has frequentist training will use CI's to answer question a) and they will think they are an appropriate frequentist answer) – probabilityislogic Jan 21 '11 at 12:19
yes, by "a little naughty" I just meant that Jaynes was making the point in a rather mischeiviously confrontational (but also entertaining) manner (or at least that is how I read it). But if he hadn't then it probably wouldn't have had any impact. – Dikran Marsupial Jan 21 '11 at 12:28

Frequentist confidence intervals bound the rate of false positives (Type I errors), and guarantee their coverage will be bounded below by the confidence parameter, even in the worst case. Bayesian credibility intervals don't.

So if the thing you care about is false positives and you need to bound them, confidence intervals are the the approach that you'll want to use.

For example, let's say you have an evil king with a court of 100 courtiers and courtesans and he wants to play a cruel statistical game with them. The king has a bag of a trillion fair coins, plus one unfair coin whose heads probability is 10%. He's going to perform the following game. First, he'll draw a coin uniformly at random from the bag.

Then the coin will be passed around a room of 100 people and each one will be forced to do an experiment on it, privately, and then each person will state a 95% uncertainty interval on what they think the coin's heads probability is.

Anybody who gives an interval that represents a false positive -- i.e. an interval that doesn't cover the true value of the heads probability -- will be beheaded.

If we wanted to express the /a posteriori/ probability distribution function of the coin's weight, then of course a credibility interval is what does that. The answer will always be the interval [0.5, 0.5] irrespective of outcome. Even if you flip zero heads or one head, you'll still say [0.5, 0.5] because it's a heck of a lot more probable that the king drew a fair coin and you had a 1/1024 day getting ten heads in a row, than that the king drew the unfair coin.

So this is not a good idea for the courtiers and courtesans to use! Because when the unfair coin is drawn, the whole room (all 100 people) will be wrong and they'll all get beheaded.

In this world where the most important thing is false positives, what we need is an absolute guarantee that the rate of false positives will be less than 5%, no matter which coin is drawn. Then we need to use a confidence interval, like Blyth-Still-Casella or Clopper-Pearson, that works and provides at least 95% coverage irrespective of the true value of the parameter, even in the worst case. If everybody uses this method instead, then no matter which coin is drawn, at the end of the day we can guarantee that the expected number of wrong people will be no more than five.

So the point is: if your criterion requires bounding false positives (or equivalently, guaranteeing coverage), you gotta go with a confidence interval. That's what they do. Credibility intervals may be a more intuitive way of expressing uncertainty, they may perform pretty well from a frequentist analysis, but they are not going to provide the guaranteed bound on false positives you'll get when you go asking for it.

(Of course if you also care about false negatives, you'll need a method that makes guarantees about those too...)

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Food for thought, however the particular example is unfair as the frequentist approach is allowed to consider the relative costs of false-positive and false-negative costs, but the Bayesian approach isn't. The correct thing to do according to Bayesian decision theory is to give an interval of [0,1] as there is no penalty associated with false-negatives. Thus in a like-for-like comparison of frameworks, none of the Bayesians would ever be beheaded either. The issue about bounding false-positives though gives me a direction in which to look for an answer to Jaynes' challenge. – Dikran Marsupial Sep 4 '10 at 9:10
Note also that if the selected coin is flipped often enough, then eventually the Bayesian confidence interval will be centered on the long run frequency of heads for the particular coin rather than on the prior. If my life depended on the interval containing the true probability of a head I wouldn't flip the coin just once! – Dikran Marsupial Sep 4 '10 at 9:57
Having though about this a bit more, this example is invalid as the criterion used to measure success is not the same as that implied by the question posed by the king. The problem is in the "no matter which coin is drawn", a clause that is designed to trip up any method that uses the prior knowledge about the rarity of the biased coin. As it happens, Bayesains can derive bounds as well (e.g. PAC bounds) and if asked would have done so, and I suspect the answer would be the same as the Clopper-Pearson interval. To be a fair test, the same information must be given to both approaches. – Dikran Marsupial Sep 6 '10 at 8:29
Dikran, there needn't be "Bayesians" and "Frequentists." They're not incompatible schools of philosophy to which one may subscribe to only one! They are mathematical tools whose efficacy can be demonstrated in the common framework of probability theory. My point is that IF the requirement is an absolute bound on false positives no matter the true value of the parameter, THEN a confidence interval is the method that accomplishes that. Of course we all agree on the same axioms of probability and the same answer can be derived many ways. – Keith Winstein Sep 7 '10 at 4:55
I agree with the first point, it is a matter of "horses for courses", but examples which show where the boundaries lie are interesting and provide insight into the "courses" best suited to each "horse". However, the examples must be fair, so that the criterion for success matches the question as posed (Jaynes is perhaps not completely immune to that criticism, which I will address in my answer which I will post later). – Dikran Marsupial Sep 7 '10 at 19:47

The problem starts with your sentence :

Examples based on incorrect prior assumptions are not acceptable as they say nothing about the internal consistency of the different approaches.

Yeah well, how do you know your prior is correct?

Take the case of Bayesian inference in phylogeny. The probability of at least one change is related to evolutionary time (branch length t) by the formula

$$P=1-e^{-\frac{4}{3}ut}$$

with u being the rate of substitution.

Now you want to make a model of the evolution, based on comparison of DNA sequences. In essence, you try to estimate a tree in which you try to model the amount of change between the DNA sequences as close as possible. The P above is the chance of at least one change on a given branch. Evolutionary models describe the chances of change between any two nucleotides, and from these evolutionary models the estimation function is derived, either with p as a parameter or with t as a parameter.

You have no sensible knowledge and you chose a flat prior for p. This inherently implies an exponentially decreasing prior for t. (It becomes even more problematic if you want to set a flat prior on t. The implied prior on p is strongly dependent on where you cut off the range of t.)

In theory, t can be infinite, but when you allow an infinite range, the area under its density function equals infinity as well, so you have to define a truncation point for the prior. Now when you chose the truncation point sufficiently large, it is not difficult to prove that both ends of the credible interval rise, and at a certain point the true value is not contained in the credible interval any more. Unless you have a very good idea about the prior, Bayesian methods are not guaranteed to be equal to or superior to other methods.

ref: Joseph Felsenstein : Inferring Phylogenies, chapter 18

On a side note, I'm getting sick of that Bayesian/Frequentist quarrel. They're both different frameworks, and neither is the Absolute Truth. The classical examples pro Bayesian methods invariantly come from probability calculation, and not one frequentist will contradict them. The classical argument against Bayesian methods invariantly involve the arbitrary choice of a prior. And sensible priors are definitely possible.

It all boils down to the correct use of either method at the right time. I've seen very few arguments/comparisons where both methods were applied correctly. Assumptions of any method are very much underrated and far too often ignored.

EDIT : to clarify, the problem lies in the fact that the estimate based on p differs from the estimate based on t in the Bayesian framework when working with uninformative priors (which is in a number of cases the only possible solution). This is not true in the ML framework for phylogenetic inference. It is not a matter of a wrong prior, it is inherent to the method.

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It is possible to be interested in the differences between Bayesian and frequentist statistics without it being a quarrel. It is important to know the flaws as well as benefits of ones preferred approach. I specifically excluded priors as that is not a problem with the framework, per se, but just a matter of GIGO. The same thing applies to frequentists statistics, for example by assuming and incorrect parametric distribution for the data. That wouldn't be a criticism of frequentist methodology, just the particular method. BTW, I have no particular problem with improper priors. – Dikran Marsupial Sep 3 '10 at 20:52
Jaynes first example: Not one statistician in his right mind will ever use an F-test and a T-test on that dataset. Apart from that, he compares a two-tailed test to P(b>a), which is not the same hypothesis tested. So his example is not fair, which he essentially admits later on. Next to that, you can't compare "the frameworks". What are we talking about then? ML, REML, LS, penalized methods,...? intervals for coefficients, statistics, predictions,...? You can as well ask whether Lutheran service is equivalent or superior to Shiite services. They talk about the same God. – Joris Meys Sep 3 '10 at 22:22
Could you clarify what is your data and what are the parameters you would be estimating in your model? I am a bit confused on this point. Also, could you please use $$instead of  to center the formula? The font size is very small right now. – user28 Sep 3 '10 at 22:23 @Srikant: The example in Felsensteins book is based on a Jukes-Cantor model for DNA evolution. Data is DNA sequences. You want to estimate the probability of change in your sequence, which is related to your branch length based on the mentioned formula. Branch lengths are defined as time of evolution : the higher the chance for changes, the more time that passed between the ancestor and the current state. Sorry, but I can't summarize the whole theory behind ML and Bayesian phylogenetic inference in just one post. Felsenstein needed half a book for that. – Joris Meys Sep 3 '10 at 22:29 I guess I just wanted you to clarify what variables in your equation was data and which ones were the parameter as it was not clear from your post especially to someone like me who is an outsider. I am still lost but I guess I would need to read the book to find out more. – user28 Sep 3 '10 at 22:44 show 36 more comments Keith Winstein, EDIT: Just to clarify, this answer describes the example given in Keith Winstein Answer on the King with the cruel statistical game. The Bayesian and Frequentist answers both use the same information, which is to ignore the information on the number of fair and unfair coins when constructing the intervals. If this information is not ignored, the the frequentist should use the integrated Beta-Binomial Likelihood as the sampling distribution in constructing the Confidence interval, in which case the Clopper-Pearson Confidence Interval is not appropriate, and needs to be modified. A similar adjustment should occur in the Bayesian solution. EDIT: I have also clarified the initial use of the clopper Pearson Interval. EDIT: alas, my alpha is the wrong way around, and my clopper pearson interval is incorrect. My humblest apologies to @whuber, who correctly pointed this out, but who I initially disagreed with and ignored. The CI Using the Clopper Pearson method is very good If you only get one observation, then the Clopper Pearson Interval can be evaluated analytically. Suppose the coin is comes up as "success" (heads) you need to choose \theta such that$$[Pr(Bi(1,\theta)\geq X)\geq\frac{\alpha}{2}] \cap [Pr(Bi(1,\theta)\leq X)\geq\frac{\alpha}{2}]$$When X=1 these probabilities are Pr(Bi(1,\theta)\geq 1)=\theta and Pr(Bi(1,\theta)\leq 1)=1, so the Clopper Pearson CI implies that \theta\geq\frac{\alpha}{2} (and the trivially always true 1\geq\frac{\alpha}{2}) when X=1. When X=0 these probabilities are Pr(Bi(1,\theta)\geq 0)=1 and Pr(Bi(1,\theta)\leq 0)=1-\theta, so the Clopper Pearson CI implies that 1-\theta \geq\frac{\alpha}{2}, or \theta\leq 1-\frac{\alpha}{2} when X=0. So for a 95% CI we get [0.025,1] when X=1, and [0,0.975] when X=0. Thus, one who uses the Clopper Pearson Confidence Interval will never ever be beheaded. Upon observing the interval, it is basically the whole parameter space. But the C-P interval is doing this by giving 100% coverage to a supposedly 95% interval! Basically, the Frequentists "cheats" by giving a 95% confidence interval more coverage than he/she was asked to give (although who wouldn't cheat in such a situation? if it were me, I'd give the whole [0,1] interval). If the king asked for an exact 95% CI, this frequentist method would fail regardless of what actually happened (perhaps a better one exists?). What about the Bayesian Interval? (specifically the Highest Posterior Desnity (HPD) Bayesian Interval) Because we know a priori that both heads and tails can come up, the uniform prior is a reasonable choice. This gives a posterior distribution of (\theta|X)\sim Beta(1+X,2-X) . Now, all we need to do now is create an interval with 95% posterior probability. Similar to the clopper pearson CI, the Cummulative Beta distribution is analytic here also, so that Pr(\theta \geq \theta^{e} | x=1) = 1-(\theta^{e})^{2} and Pr(\theta \leq \theta^{e} | x=0) = 1-(1-\theta^{e})^{2} setting these to 0.95 gives \theta^{e}=\sqrt{0.05}\approx 0.224 when X=1 and \theta^{e}= 1-\sqrt{0.05}\approx 0.776 when X=0. So the two credible intervals are (0,0.776) when X=0 and (0.224,1) when X=1 Thus the Bayesian will be beheaded for his HPD Credible interval in the case when he gets the bad coin and the Bad coin comes up tails which will occur with a chance of \frac{1}{10^{12}+1}\times\frac{1}{10}\approx 0. First observation, the Bayesian Interval is smaller than the confidence interval. Another thing is that the Bayesian would be closer to the actual coverage stated, 95%, than the frequentist. In fact, the Bayesian is just about as close to the 95% coverage as one can get in this problem. And contrary to Keith's statement, if the bad coin is chosen, 10 Bayesians out of 100 will on average lose their head (not all of them, because the bad coin must come up heads for the interval to not contain 0.1). Interestingly, if the CP-interval for 1 observation was used repeatedly (so we have N such intervals, each based on 1 observation), and the true proportion was anything between 0.025 and 0.975, then coverage of the 95% CI will always be 100%, and not 95%! This clearly depends on the true value of the parameter! So this is at least one case where repeated use of a confidence interval does not lead to the desired level of confidence. To quote a genuine 95% confidence interval, then by definition there should be some cases (i.e. at least one) of the observed interval which do not contain the true value of the parameter. Otherwise, how can one justify the 95% tag? Would it not be just a valid or invalid to call it a 90%, 50%, 20%, or even 0% interval? I do not see how simply stating "it actually means 95% or more" without a complimentary restriction is satisfactory. This is because the obvious mathematical solution is the whole parameter space, and the problem is trivial. suppose I want a 50% CI? if it only bounds the false negatives then the whole parameter space is a valid CI using only this criteria. Perhaps a better criterion is (and this is what I believe is implicit in the definition by Kieth) "as close to 95% as possible, without going below 95%". The Bayesian Interval would have a coverage closer to 95% than the frequentist (although not by much), and would not go under 95% in the coverage (\text{100%} coverage when X=0, and 100\times\frac{10^{12}+\frac{9}{10}}{10^{12}+1}\text{%} > \text{95%} coverage when X=1). In closing, it does seem a bit odd to ask for an interval of uncertainty, and then evaluate that interval by the using the true value which we were uncertain about. A "fairer" comparison, for both confidence and credible intervals, to me seems like the truth of the statement of uncertainty given with the interval. -  In your first main paragraph you seem to have confused \alpha and 1-\alpha. Where does the value of 10^12+1 come in? What do you mean by "beheaded"?? This text looks like it is need of proofreading and revision. – whuber♦ Jan 19 '11 at 18:30 10^{12} is for the trillion fair coins, and 1 is for the unfair coin. And I haven't confused \alpha and 1-\alpha the Clopper Pearson interval listed [here][1] – probabilityislogic Jan 20 '11 at 2:43 [sorry typo] 10^{12} (TeX fixed) is for the trillion fair coins, and 1 is for the unfair coin, one over this is a rough approx. to the probability of having the "bad" coin. Beheaded is the consequence of giving the wrong confidence interval. And I haven't confused \alpha and 1-\alpha the Clopper Pearson interval listed on the wiki page (search binomial proportion confidence interval). What happens is one part of the C-P interval is a tautology 1 \geq \frac{\alpha}{2} when one 1 observation. The side "flips" when X=1 to X=0, which is why there is 1-\theta and \theta. – probabilityislogic Jan 20 '11 at 2:52 Do you mean @Keith Winstein's answer? – whuber♦ Jan 20 '11 at 23:00 @whuber, yes I do mean keith winstein's answer. – probabilityislogic Jan 21 '11 at 5:39 This is a "fleshed out" example given in a book written by Larry Wasserman All of statistics on Page 216 (12.8 Strengths and Weaknesses of Bayesian Inference). I basically provide what Wasserman doesn't in his book 1) an explanation for what is actually happening, rather than a throw away line; 2) the frequentist answer to the question, which Wasserman conveniently does not give; and 3) a demonstration that the equivalent confidence calculated using the same information suffers from the same problem. In this example, he states the following situation 1. An observation, X, with a Sampling distribution: (X|\theta)\sim N(\theta,1) 2. Prior distribution of (\theta)\sim N(0,1) (he actually uses a general \tau^2 for the variance, but his diagram specialises to \tau^2=1) He then goes to show that, using a Bayesian 95% credible interval in this set-up eventually has 0% frequentist coverage when the true value of \theta becomes arbitrarily large. For instance, he provides a graph of the coverage (p218), and checking by eye, when the true value of \theta is 3, the coverage is about 35%. He then goes on to say: ...What should we conclude from all this? The important thing is to understand that frequentist and Bayesian methods are answering dierent questions. To combine prior beliefs with data in a principled way, use bayesian inference. To construct procedures with guaranteed long run performance, such as condence intervals, use frequentist methods... (p217) And then moves on without any disection or explanation of why the Bayesian method performed apparently so bad. Further, he does not give a answer from the frequentist approach, just a broad brush statement about "the long-run" - a classical political tactic (emphasise your strength + others weakness, but never compare like for like). I will show how the problem as stated \tau=1 can be formulated in frequentist/orthodox terms, and then show that the result using confidence intervals gives precisely the same answer as the Bayesian one. Thus any defect in the Bayesian (real or perceived) is not corrected by using confidence intervals. Okay, so here goes. The first question I ask is what state of knowledge is described by the prior \theta\sim N(0,1)? If one was "ignorant" about \theta, then the appropriate way to express this is p(\theta)\propto 1. Now suppose that we were ignorant, and we observed Y\sim N(\theta,1), independently of X. What would our posterior for \theta be?$$p(\theta|Y)\propto p(\theta)p(Y|\theta)\propto exp\Big(-\frac{1}{2}(Y-\theta)^2\Big)$$Thus (\theta|Y)\sim N(Y,1). This means that the prior distribution given in Wassermans example, is equivalent to having observed an iid copy of X equal to 0. Frequentist methods cannot deal with a prior, but it can be thought of as having made 2 observations from the sampling distribution, one equal to 0, and one equal to X. Both problems are entirely equivalent, and we can actually give the frequentist answer for the question. Because we are dealing with a normal distribution with known variance, the mean is a sufficient statistic for constructing a confidence interval for \theta. The mean is equal to \overline{x}=\frac{0+X}{2}=\frac{X}{2} and has a sampling distribution$$(\overline{x}|\theta)\sim N(\theta,\frac{1}{2})$$Thus an (1-\alpha)\text{%} CI is given by:$$\frac{1}{2}X\pm Z_{\alpha/2}\frac{1}{\sqrt{2}}$$But, using The results of example 12.8 for Wasserman, he shows that the posterior (1-\alpha)\text{%} credible interval for \theta is given by:$$cX\pm \sqrt{c}Z_{\alpha/2}$$. Where c=\frac{\tau^{2}}{1+\tau^{2}}. Thus, plugging in the value at \tau^{2}=1 gives c=\frac{1}{2} and the credible interval becomes:$$\frac{1}{2}X\pm Z_{\alpha/2}\frac{1}{\sqrt{2}}

Which are exactly the same as the confidence interval! So any defect in the coverage exhibited by the Bayesian method, is not corrected by using the frequentist confidence interval! [If the frequentist chooses to ignore the prior, then to be a fair comparison, the Bayesian should also ignore this prior, and use the ignorance prior $p(\theta)\propto 1$, and the two intervals will still be equal - both $X \pm Z_{\alpha/2})$].

So what the hell is going on here? The problem is basically one of non-robustness of the normal sampling distribution. because the problem is equivalent to having already observed a iid copy, $X=0$. If you have observed $0$, then this is extremely unlikely to have occurred if the true value is $\theta=4$ (probability that $X\leq 0$ when $\theta=4$ is 0.000032). This explains why the coverage is so bad for large "true values", because they effectively make the implicit observation contained in the prior an outlier. In fact you can show that this example is basically equivalent to showing that the arithmetic mean has an unbounded influence function.

Generalisation. Now some people may say "but you only considered $\tau=1$, which may be a special case". This is not true: any value of $\tau^2=\frac{1}{N}$ $(N=0,1,2,3,\dots)$ can be interpreted as observing $N$ iid copies of $X$ which were all equal to $0$, in addition to the $X$ of the question. The confidence interval will have the same "bad" coverage properties for large $\theta$. But this becomes increasingly unlikely if you keep observing values of $0$ (and no rational person would continue to worry about large $\theta$ when you keep seeing $0$).

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 Thanks for the analysis. AFAICS this is just an example of a problem caused by an incorrect (informative) prior assumption and says nothing about the internal consistency of the Bayesian approach? – Dikran Marsupial Jan 31 '11 at 8:25 Nope, the prior is not necessarily incorrect, unless one didn't actually observe a value of $0$ prior to conducting the experiment (or obtain some equivalent knowledge). It basically means that, as the true $\theta$ becomes arbitrarily large, the probability of observing these implicit observations becomes arbitrarily small (like getting a "unlucky sample"). – probabilityislogic Jan 31 '11 at 21:22 you can see by noting that the sample consists of an observation at $0$ and another one at $X$. $0$ is fixed (because it has been observed), but $X$ will be "close" to $\theta$ in most cases. So as $\theta$ becomes large, the sample average gets further and further away from both $X$ and $0$, and because the variance is fixed, the width of the CI is fixed, so it will eventually not contain either $X$ or $0$, and hence not be near either of the two likely values of $\theta$ (for one of them is an outlier when they become far apart, for fixed $\theta$) – probabilityislogic Jan 31 '11 at 21:27

are there examples where the frequentist confidence interval is clearly superior to the Bayesian credible interval (as per the challenge implicitly made by Jaynes).

Here is an example: the true $\theta$ equals $10$ but the prior on $\theta$ is concentrated about $1$. I am doing statistics for a clinical trial, and $\theta$ measures the risk to death, so the Bayesian result is a disaster, isn't it ? More seriously, what is "the" Bayesian credible interval ? In other words: what is the selected prior ? Maybe Jaynes proposed an automatic way to select a prior, I don't know !

Bernardo proposed a "reference prior" to be used as a standard for scientific communication [and even a "reference credible interval" (Bernardo - objective credible regions)]. Assuming this is "the" Bayesian approach, now the question is: when is an interval superior to another one ? The frequentist properties of the Bayesian interval are not always optimal, but neither are the Bayesian properties of "the" frequentist interval
(by the way, what is "the" frequentist interval ? )

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I am speculating, but I suspect this answer is bound to get the same treatment that others have. Someone will simply argue this is an issue of poor choice of prior and not of some inherent weakness of Bayesian procedures, which in my view partially tries to evade a valid criticism. – cardinal Apr 6 '12 at 19:42
@cardinal's comment is quite right. The prior here is off by an order of magnitude, making the criticism very weak. Prior information matters to frequentists too; what one knows a priori should determine e.g. what estimates and test statistics are used. If these choices are based on information that's wrong by an order of magnitude, poor results should be expected; being Bayesian or frequentist doesn't come into it. – guest Apr 6 '12 at 20:14
My "example" was not the important part of my answer. But what is a good choice of prior ? It is easy to imagine a prior whose support contains the true parameter but the posterior does not, so the frequentist interval is superior ? – Stéphane Laurent Apr 7 '12 at 13:14
Cardinal and guest are correct, my question explicitly included "Examples based on incorrect prior assumptions are not acceptable as they say nothing about the internal consistency of the different approaches." for a good reason. Frequentist tests can be based on incorrect assumptions as well as Bayesian ones (the Bayesian framework states the assumptions more explicitly); the question is whether the framework has weaknesses. Also if the true value was in the prior, but not the posterior, that would imply that the observations ruled out the possibility of the true value being correct! – Dikran Marsupial Apr 10 '12 at 10:33
Maybe I should edit my answer and delete my "example" - this is not the serious part of my answer. My answer mainly was about the meaning of "the" Bayesian approach. What do you call the Bayesian approach ? This approach requires the choice of a subjective prior or it uses an automatic way to select a noninformative prior ? In the second case it is important to mention the work of Bernardo. Secondly you have not defined the "superiority" relation between intervals: when do you say an interval is superior to another one ? – Stéphane Laurent Apr 10 '12 at 17:47