Tag Archives: improper

The Envelope ‘Paradox’

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At the recent IMO in Hong Kong, there were several moments where the deputy leaders had to hang around, and I spent some of these moments discussing the following problem with Stephen Mackereth, my counterpart from New Zealand. He’s a mathematically-trained philosopher, so has a similar level of skepticism to me, but for different reasons, regarding supposed paradoxes in probability. Because, as we will see shortly, I don’t think this is a paradox in even the slightest fashion, I think there’s probably too much written about this on the internet already. So I’m aware that contributing further to this oeuvre is hypocritical, but we did the thinking in HKUST’s apparently famous Einstein Cafe, so it makes sense to write down the thoughts.

[And then forget about it for eight weeks. Oops.]

The ‘Paradox’

Here’s the situation. A cryptic friend gives you an envelope containing some sum of money, and shows you a second envelope. They then inform you that one of the envelopes contains twice as much money as the other. It’s implicit in this that the choice of which is which is uniform. You have the option to switch envelopes. Should you?

The supposed paradox arises by considering the amount in your envelope, say X. In the absence of further information, it is equally likely that the other envelope contains X/2 as 2X. Therefore, the average value of the other envelope is

\frac12 \left(\frac{X}{2}+2X \right)= \frac54 X > X.

So you should switch, since on average you gain money. But this is paradoxical, since the assignment of larger and smaller sums was uniform, so switching envelope should make no difference.

Probabilistic setup

This is not supposed to be a problem on a first-year probability exercise sheet. It’s supposed to be conducive to light discussion. So saying “I won’t engage with this problem until you tell me what the probability space is” doesn’t go down terribly well. But it is important to work out what is random, and what isn’t.

There are two sources of randomness, or at least ignorance. Firstly, there is the pair of values contained in the envelopes. Secondly, there is the assignment of this pair of values to the two envelopes. The second is a source of randomness, and this problem is founded on the premise that this second stage is ‘symmetric enough’ to smooth over any complications in the first stage. If we think that probability isn’t broken (and that’s what I think), then the answer is probably that the second stage isn’t symmetric enough.

Or, that the first stage isn’t very well-defined. In what follows, I’m going to make the second stage very symmetric, at the expense of setting up the first stage in what seems to me a reasonable way using the conventional language of probability theory to record our ignorance about the values in play.

So what’s the first stage? We must have a set of possible pairs of values taken by the envelopes. Let’s call this A, so

A\subset \mathbb{A}:=\{(x,2x)\,:\, x\in (0,\infty)\}.

Maybe we know what A is, but maybe we don’t, in which we should take A=\mathbb{A}, on the grounds that any pair is possible. Suppose that your friend has chosen the pair of values according to some distribution on \mathbb{A}, which we’ll assume has a density f, which is known by you. Maybe this isn’t the actual density, but it serves perfectly well if you treat it as *your* opinion on the likelihood. Then this actually does reduce to a problem along the lines of first-year probability, whether or not you get to see the amount in your envelope.

Suppose first that you do get to see the amount, and that it is x. Then the conditional probabilities that the pair is (x/2,x) or (x,2x) are, respectively

\frac{f(x/2,x)}{f(x/2,x)+f(x,2x)},\quad \frac{f(x,2x)}{f(x/2,x)+f(x,2x)}.

So you can work out your expected gain by switching, and decide accordingly. If you don’t know the value in your envelope, you can still work out the probability that it is better (in expectation) to switch, but this isn’t really a hugely meaningful measure, unless it is zero or one.

It’s worth noting that if you can view inside your envelope, and you know A has a particular form, then the game becomes determined. For example, if

A\subset \{(n,2n), n\text{ an odd integer}\},

then life is very easy. If you open your envelope and see an odd integer, you should switch, and if you see an even integer you shouldn’t.

We’ll return at the end to discuss a case where it is always better to switch, and why this isn’t actually a paradox.

Improper prior and paradox of resampling when \mathbb{E}=\infty

For now though, let’s assume that we don’t know anything about the amounts of money in the envelopes. Earlier, we said that “in the absence of further information, it is equally likely that the other envelope contains X/2 as 2X”. In the language of a distribution on \mathbb{A}, we are taking the uniform measure. Of course this not a distribution, in the same way that there isn’t a uniform distribution on the positive reals.

However, if this is your belief about the values in the pair of envelopes, what do you think is the mean value of the content of your envelope? Well, you think all values are equally likely. So, even though this isn’t a distribution, you pretty much think the value of your envelope has infinite expectation.

[This is where the philosophy comes in I guess. Is (expressing uniform ignorance about the content of the other envelope given knowledge of your own) the same as (expressing uniform ignorance of both envelopes at the beginning)? I think it is, even though it has a different consequence here, since the former can be turned into a proper distribution, whereas the latter cannot.]

Let’s briefly consider an alternative example. It’s fairly easy to conjure up distributions which are almost surely finite but which have infinite expectation. For example \mathbb{P}(X=2^k)=2^{-k} for k=1,2,…, which is the content of the *St. Petersburg paradox*, another supposed paradox in probability, but one whose resolution is a bit more clear.

Anyway, let X and Y be independent copies of such a distribution. Now suppose your friend offers you an envelope containing amount X. You look at the value, and then you are offered the opportunity to switch to an envelope containing amount Y. Should you?

Well, if expectation is what you care about, then you definitely should. Because with probability one, you are staring at a finite value in your envelope, whereas the other unknown envelope promises infinite expectation, which is certainly larger than the value that you’re looking at.

Is this also a paradox? I definitely don’t think it is. The expectation of the content of your envelope is infinite, the expected gain is infinite with probability one, which is consistent with the expected content of the other envelope being infinite. [Note that you don’t want to be claiming that the expectation of X-Y is zero.]

An example density function

As an exercise that isn’t necessarily hugely interesting, let’s assume that f, the distribution of the smaller of the pair, is \mathrm{Exp}(\lambda). So the mean of this smaller number is 1/\lambda. Then, conditional on seeing x in my envelope, the expected value of the number in the other envelope is

\frac{\frac{x}{2} e^{-\lambda x/2} + 2x e^{-\lambda x}}{e^{-\lambda x/2}+ e^{-\lambda x}}. (*)

Some straightforward manipulation shows that this quantity is at least x (implying it’s advantageous to switch) precisely when

e^{-\lambda x/2}\ge \frac12.

That is, when x\le \frac{2\log 2}{\lambda}. The shape of this interval should fit our intuition, namely that the optimal strategy should be to switch if the value in your envelope is small enough.

The point of doing this calculation is to emphasise that it ceases to be an interesting problem, and certainly ceases to be a paradox of any kind, once we specify f concretely. It doesn’t matter whether this is some true distribution (ie the friend is genuinely sampling the values somehow at random), or rather a perceived likelihood (that happens to be normalisable).

What if you should always switch?

The statement of the paradox only really has any bite if the suggestion is that we should always switch. Earlier, we discussed potential objections to considering the uniform prior in this setting, but what about other possible distributions f which might lead to this conclusion?

As at (*), we can conclude that when f(x)+f(x/2)>0, we should switch on seeing x precisely if

f(x)\ge 2f\left(\frac{x}{2}\right).

Therefore, partitioning the support of f into a collection of geometric sequences with exponent 2, it is clear that the mean of f is infinite if everything is integer-valued. If f is real-valued, there are some complications, but so long as everything is measurable, the same conclusion will hold.

So the you-should-switch-given-x strategy can only hold for all values of x if f has infinite mean. This pretty much wraps up my feelings. If the mean isn’t infinite, the statement of the paradox no longer holds, and if it is infinite, then the paradox dissolves into a statement about trying to order various expectations, all of which are infinite.

Conclusions

Mathematical summary: it’s Bayes. Things may be exchangeable initially, but not once you condition on the value of one of them! Well, not unless you have a very specific prior.

Philosophical summary: everything in my argument depends on the premise that one can always describe the protagonist’s prior opinion on the contents of the pair of envelopes with a (possibly degenerate) distribution. I feel this is reasonable. As soon as you write down \frac12 \cdot\frac{x}{2} + \frac12 \cdot2x, you are doing a conditional expectation, and it’s got to be conditional with respect to something. Here it’s the uniform prior, or at least the uniform prior restricted to the set of values that are now possible given the revelation of your number.

Second mathematical summary: once you are working with the uniform prior, or any measure with infinite mean, there’s no reason why

\mathbb{E}\left[X|Y\right]>Y,

with probability one (in terms of Y) should be surprising, since the LHS is (almost-surely) infinite while the RHS is almost surely finite, despite having infinite mean itself.

Bayesian Inference and the Jeffreys Prior

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Last term I was tutoring for the second year statistics course in Oxford. This post is about the final quarter of the course, on the subject of Bayesian inference, and in particular on the Jeffreys prior.

There are loads and loads of articles sitting around on the web contributing the debate about the relative merits of Bayesian and frequentist methods. I do not want to continue that debate here, partly because I don’t have a strong opinion, but mainly because I don’t really understand that much about the underlying issues.

What I will say is that after a few months of working fairly intensively with various complicated stochastic processes, I am starting to feel fairly happy throwing about conditional probability rather freely. When discussing some of the more combinatorial models for example, quite often we have no desire to compute or approximate complication normalising constants, and so instead talk about ‘weights’. And a similar idea underlies Bayesian inference. As in frequentist methods we have an unknown parameter, and we observe some data. Furthermore, we know the probability that such data might have arisen under any value of the parameter. We want to make inference about the value of the parameter given the data, so it makes sense to multiply the probability that the data emerged as a result of some parameter value by some weighting on the set of parameter values.

In summary, we assign a prior distribution representing our initial beliefs about the parameter before we have seen any data, then we update this by weighting by the likelihood that the observed data might have arisen from a particular parameter. We often write this as:

\pi(\theta| x)\propto f(x|\theta)\pi(\theta),

or say that posterior = likelihood x prior. Note that in many applications it won’t be necessary to work out what the normalising constant on the distribution ought to be.

That’s the setup for Bayesian methods. I think the general feeling about the relative usefulness of such an approach is that it all depends on the prior. Once we have the prior, everything is concrete and unambiguously determined. But how should we choose the prior?

There are two cases worth thinking about. The first is where we have a lot of information about the problem already. This might well be the case in some forms of scientific research, where future analysis aims to build on work already completed. It might also be the case that we have already performed some Bayesian calculations, so our current prior is in fact the posterior from a previous set of experiments. In any case, if we have such an ‘informative prior’, it makes sense to use it in some circumstances.

Alternatively, it might be the case that for some reason we care less about the actual prior than about the mathematical convenience of manipulating it. In particular, certain likelihood functions give rise to conjugate priors, where the form of the posterior is the same as the form of the prior. For example, a normal likelihood function admits a normal conjugate prior, and a binomial likelihood function gives a Beta conjugate prior.

In general though, it is entirely possible that neither of these situations will hold but we still want to try Bayesian analysis. The ideal situation would be if the choice of prior had no effect on the analysis, but if that were true, then we couldn’t really be doing any Bayesian analysis. The Jeffreys prior is one natural candidate because it removes a specific problem with choosing a prior to express ignorance.

It sounds reasonable to say that if we have total ignorance about the parameter, then we should take the prior to be uniform on the set of possible values taken by the parameter. There are two potential objections to this. The first is that if the parameter could take any real value, then the prior will not be a distribution as the uniform distribution on the reals is not normalisable. Such a prior is called improper. This isn’t a huge problem really though. For making inference we are only interested in the posterior distribution, and so if the posterior turns out to be normalisable we are probably fine.

The second problem is more serious. Even though we want to express ignorance of the parameter, is there a canonical choice for THE parameter? An example will make this objection more clear. Suppose we know nothing about the parameter T except that it lies in [0,1]. Then the uniform distribution on [0,1] seems like the natural candidate for the prior. But what if we considered T^100 to be the parameter instead? Again if we have total ignorance we should assign T^100 the uniform distribution on its support, which is again [0,1]. But if T^100 is uniform on [0,1], then T is massively concentrated near 1, and in particular cannot also be uniformly distributed on [0,1]. So as a minimum requirement for expressing ignorance, we want a way of generating a prior that doesn’t depend on the choice of parameterisation.

The Jeffreys prior has this property. Note that there may be separate problems with making such an assumption, but this prior solves this particular objection. We define it to be \pi(\theta)\propto [I(\theta)]^{1/2} where I is the Fisher information, defined as

I(\theta)=-\mathbb{E}_\theta\Big[\frac{\partial^2 l(X_1,\theta)}{\partial \theta^2}\Big],

where the expectation is over the data X_1 for fixed \theta, and l is the log-likelihood. Proving that this has the property that it is invariant under reparameterisation requires demonstrating that the Jeffreys prior corresponding to g(\theta) is the same as applying a change of measure to the Jeffreys prior for \theta. The proof is a nice exercise in the chain rule, and I don’t want to reproduce it here.

For a Binomial likelihood function, we find that the Jeffreys prior is Beta(1/2,1/2), which has density that looks roughly like a bucket suspended above [0,1]. It is certainly worth asking why the ‘natural’ choice for prior might put lots of mass at the edge of the domain for the parameter.

I don’t have a definitive answer, but I do have an intuitive idea which comes from the meaning of the Fisher information. As the second derivative of the log-likelihood, a large Fisher information means that with high probability we will see data for which the likelihood changes substantially if we vary the parameter. In particular, this means that the posterior probability of a parameter close to 0 will be eliminated more quickly by the data if the true parameter is different.

If the variance is small, as it is for parameter near 0, then the data generated by this parameter will have the greatest effect on the posterior, since the likelihood will be small almost everywhere except near the parameter. We see the opposite effect if the variance is large. So it makes sense to compensate for this by placing extra prior mass at parameter values where the data has the strongest effect. Note that in the previous example, the Jeffreys prior is in fact exactly inversely proportional to the standard deviation. For the above argument to make sense, we need it to be monotonic with respect to SD, and it just happens that in this case, being 1/SD is precisely the form required to be invariant under reparameterisation.

Anyway, I thought that was reasonably interesting, as indeed was the whole course. I feel reassured that I can justify having my work address as the Department of Statistics since I now know at least epsilon about statistics!

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