信封交换悖论

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英文解析

This has been a very popular question. Assume that neither of you peek into your envelopes. Assume that you have $X$in your envelope, where$X$ has a fifty- fifty chance of being either $m$or$2m$ . This means that your opponent's envelope has a fifty- fifty chance of containing $2X$or$12X$ . The expected value of switching is

$$\left(\frac{1}{2}\times \mathbb{S}2X\right) + \left(\frac{1}{2}\times \mathbb{S}\frac{1}{2} X\right) = \mathbb{S}1.25X.$$

The expected benefit of switching is, therefore, $0.25X$. On this basis, it looks as though you should switch envelopes. Of course, if your opponent does not peek, and she has$Y$ in her envelope, exactly the same argument shows that she has an expected benefit to switching of $0.25Y$. So, it looks as though she should switch also. This is the first part of the "Exchange Paradox": it seems that you both benefit

from switching.

Now, suppose that neither of you peek and that you do switch envelopes once. If you still do not peek, then a repeat of exactly the same argument suggests an expected benefit of 0.25 of the contents of your envelope if you switch again. The same applies to your opponent. This is the second part of the "Exchange Paradox": it seems that you could happily switch forever (like a dog chasing its own tail). The foregoing is the naive answer.

The problem is twofold: First, you are assuming that value is expected payoff (this is so only if you are genuinely risk- neutral); 10 second, your "prior" beliefs are that you have a fifty- fifty chance of having either $m$or$2m$ . The first problem is a function of your individual risk preferences and is difficult to address. The second problem can be tackled using two approaches: the first approach is to reconsider the nature of your prior; the second approach is to "update" your prior probability assessment (this is "Bayesian" statistics as opposed to "classical" statistics).

The first approach is to reconsider the nature of your priors. Our previous (paradoxical) calculation yielded $1.25X$as the expected payoff to switching. However, this assumes that for any given$X$, it is equally likely that your opponent has$2X$ or $\frac{1}{2} X$. If you do not peek, then you are assuming a "diffuse level prior" because you assume this equality of likelihood for any$X$. Your prior is, therefore, not a valid probability density function (pdf) because the probabilities - across$X$- do not sum to 1. However, for any particular$m$, it is equally likely that you received one of$m$ or $2m$. Thus, for any particular$m$, your priors are a pdf and any paradoxes should disappear. The expected value of switching should be zero. This is easily demonstrated. Let$P(\mathbb{S}m)$denote the probability that you got$m$ (the lower amount); let

$E(V)$ denote the expected value to switching; then $E(V)$ is given by

$$\begin{array}{l}{E(V) = [E(V|\S m)\times P(\S m)] + [E(V|\S 2m)\times P(\S 2m)]}\\ {= \left(+\S m\times \frac{1}{2}\right) + \left(-\S m\times \frac{1}{2}\right)}\\ {= \S 0.} \end{array}$$

The expected value is zero, and you are thus indifferent- resolving the paradox.11 Note that $E(V\mid \S m) = +\S m$ because, conditional on your having been given the envelope containing only $\mathbb{S}m$ , you gain $\mathbb{S}m$ by switching.

The second approach is to update your prior. To update your prior, you need information. The most obvious source of information is to peek into your envelope. So, assume that both you and your opponent peek into your envelopes. Now it gets subjective. If you see an amount that seems very high, then you update your prior probabilities: the probability that you have the high- value envelope increases, and the probability that you have the low- value envelope decreases. You no longer see value in switching envelopes.12 If you see an amount that seems very low, then you see value in switching. The problem now is that you must subjectively assess the amount in the envelope as being either "low" or "high." The "Bayesian Resolution of the Exchange Paradox" is covered in detail in Christensen and Utts (1992).

If you have both peeked, and you do switch, then you will not switch again. This is because one of you gained, and that person will not want to lose by switching back. A similar question (but with an upper bound on the quantities possible) appears in Dixit and Nalebuff (1991, Chapter 13). The Dixit and Nalebuff book on strategic thinking is well worth a look.