Solving the two envelopes problem

But we know from basic probability theory that for two independent random variables, E(X/Y) > 1 does actually imply E(X) > E(Y).

This does not follow; a counterexample:

Suppose X and Y are independent random variables, with X taking the values {2,100}, and Y the values {1,150}, each with equal probability (i.e., each of X and Y follows the Bernoulli distribution with p = 0.5). Then we have
E(X/Y) = (2/1 + 100/1 + 2/150 + 100/150) / 4 = 25.67 > 1,
but
E(X) = 51 < 75.5 = E(Y).

Keep in mind that the equation E(1/Y) = 1/E(Y) does not hold in general, because taking the inverse is not a linear transformation. To evaluate the expectation after a nonlinear transformation, one requires not just the orignal expectation, but also the pdf of the distribution. (I can't be sure this is what you did, but misapplying this gives: E(X/Y) = E(X)E(1/Y) = E(X)/E(Y). The first equality holds if X and Y are independent or by definition if X and 1/Y are uncorrelated, but the second equality does not hold in general.)

Another commentary I once read regarding the "two envelopes paradox":

In order to make any sense out of this problem, you have to assume some prior probability distribution over the amount of money in the two envelopes. For many of these possible distributions, once you open one of the envelopes and learn how much money is inside, you now know more about whether the other envelope has more or less money than the one you opened. For example, if you assume that neither envelope has more than $1,000 and then open an envelope with $800, the other envelope has to have $400, so, contrary to the line of reasoning in the "paradox", switching would be bad.

On the other hand, perhaps you only want to think about distributions for which it seems the paradox still holds: ones in which that, regardless of how much money you find in envelope A, envelope B still has an equal chance of being twice as much or half as much. Well, it turns out that you can prove that this criterion also implies that the expected value of the amount of money in envelope A is infinity. This makes the paradox seem much less paradoxical: first, when your expected value is infinity, any specific finite result is disappointing (which is why switching is always correct), and second, any finite number multiplied by infinity is still infinity (which explains how each envelope can have an expected value of 1.25 times the other).

(Apparently, my original source was David Chalmers, of all people.)

Generally, people do take R and B to be fixed amounts of money - then, it's worth pointing out that the expected value of the blue envelope is not .5*.5R+.5*2R, because R there stands for 2 different amounts of money.

If the amounts of money are $5 and $10, then R in the first case stood for $10 and R in the second case stood for $5, so it should really be .5*.5($10)+.5*2($5), and the expected value is just $7.50.

One of the envelopes contains 2 glort's worth of dollars, and the other envelope contains 4 glorts' worth of dollars. I don't know the exchange ratio of glorts to dollars, except that it is positive.

Now it is clear that the expected value is 3 glorts' worth of dollars, regardless of the envelope picked. It's still possible that the game master is using some variant of "if the value of the glort is low for the duration of this trial, put the larger amount of money in envelope B; if the value of the glort is high, put the larger amount of money in envelope a" that favors one envelope over the other.

It is aimed at people with only basic command of probabilities

Such people have probably not heard of the two envelopes problem and thus the title is not informative for them.* It seems to me that it would be useful to put something in the title to indicate that this post is about probabilities and maybe that it is fairly introductory. I could be wrong about how people read this site. Maybe everyone clicks through to the article and reads as far as the italicized section, but it seems low cost to give it a more informative title.

* Also, I doubt that people who do associate "two envelopes" with probability are not very likely to think of this particular problem, for what it's worth.

Added: "this site" = discussion. If the plan is to move this to main, the structure makes more sense. But I still think the title could be better.

There is another very cool puzzle that can be considered a followup which is:

There are two envelopes in which I, the host of the game, put two different natural numbers, chosen by any distribution I like, that you don't have access. The two envelopes are indistinguishable. You pick one of them (and since they are indistinguishable, this can be considered a fair coin flip). After that you open the envelope and see the number. You have a chance to switch your number for the hidden number. Then, this number is revealed and if you choose the greater you win, let's say a dollar, otherwise you pay a dollar.

Now, before everything I said happens, you must devise a strategy that guarantees that you have a greater than 1/2 chance of winning.

Some notes:

1- the problem may be extended for rational, or any set of constructive numbers. But if you want to think only in probabilities this is irrelevant, just an over formalism.

2- This may seem uncorrelated to the two envelopes puzzle at first, but it isn't.

3- I saw this problem first on EDITthis post on xkcd blag. Thanks for Vaniver for pointing out.

It is easy to prove that this implication is false, by considering a simple counter-example courtesy of VincentYu.

Even simpler counter example: X and Y take values 1 and 100, independently and with equal probability. Then E(X/Y) = 1/4 ( 1/1 + 1/100 + 100/100 + 100/1), which is basically 25.5. Ditto for E(Y/X).

Expectation is a mean, and in means, large terms dominate - for E(X/Y), the (X=100,Y=1) situation dominates, while for E(Y/X), the (X=1,Y=100) situation dominates.

In fact, if X and Y are independent, identically distributed, strictly positive and non-trivial (ie not just constants), then I think that we always have E(X/Y) > 1.

Nice.

I think you reached the wrong conclusion in your final paragraphs. Can you show how the expected value calculation could be different for "the amount of money contained in the other envelope expressed in terms of the amount of money in this envelope [which is in dollars, BTW]" versus "the amount of money in the other envelope expressed in dollars"?

I see that Wikipedia says that there is no generally accepted solution to the paradox. That almost certainly means people are interpreting it differently. I'll give my opinion. But let me recast the problem in more pointed terms.

Mr Moneybags writes a check for some real positive amount of money and puts it in an envelope. Then he writes a check for double the amount of the previous check and puts it in another envelope. He shuffles the envelopes. You pick one and open it, and you find $N. He offers to let you switch to the other just this once. Should you?

On first glance it appears that the other envelope could have $N/2 or $N*2, and that you are entirely ignorant about which is the case, so your ignorance priors are 1/2 for each possibility. Then the expected value calculation comes up all wrong! Call the hypothesis that you opened the lower value envelope L and the hypothesis that you opened the higher value envelope H.

E(switching)=P(H)*(N/2)+P(L)*(N*2)=(1/2)*(N/2)+(1/2)*(N*2)=(5/4)*N.  

But that's nonsense. My opinion as to what went wrong was that it's false that you are entirely ignorant after opening the envelope, because you can now do an update on your ignorance prior.

We'll start, of course, with Bayes' formula P(H|N)/P(L|N) = P(N|H)/P(N|L) * P(H)/P(L) * P(N)/P(N). The last ratio cancels. The second-to-last ratio also cancels because the shuffling of the envelopes made it equally likely you'd open either one. We know basically nothing about the distribution that the lower value was first drawn from, so a convenient ignorance prior on it is that it was some uniform distribution over the interval $A to $B. Consequently, the higher value would be described by a uniform distribution over the interval 2*$A to 2*$B. The distributions may or may not overlap. I'll take each case separately.

Common start: A=0.

P(N=k|H)={1/3 for 0<k<=B; 1 for B<k<=2B}
P(N=k|L)={2/3 for 0<k<=B; 0 for B<k<=2B}
Integrate each with respect to k from 0 to 2B, then plug in
P(H|N)/P(H|L) = (4/3*B)/(2/3*B) = 2
and convert back to probabilities
P(H|N)=2/3, P(L|N)=1/3.

Separate start: A>0 and 2A<B.

P(N=k|H)={0 for A<k<2A; 1/3 for 2A<=k<=B; 1 for B<k<=2B}
P(N=k|L)={1 for A<k<2A; 2/3 for 2A<=k<=B; 0 for B<k<=2B}
Integrate each with respect to k from A to 2B, then plug in
P(H|N)/P(H|L) = P(N|H)/P(N|L) = (4/3*B-2/3*A)/(2/3*B-1/3*A) = 2
and convert back to probabilities
P(H|N)=2/3, P(L|N)=1/3.

No overlap: B<2A.

P(N=k|H)={0 for A<=k<=B; 1 for 2A<=k<=2B}
P(N=k|L)={1 for A<=k<=B; 0 for 2A<=k<=2B}
Integrate each with respect to k from A to 2B, then plug in
P(H|N)/P(H|L) = P(N|H)/P(N|L) = (2B-2A)/(B-A) = 2
and convert back to probabilities
P(H|N)=2/3, P(L|N)=1/3.

So viewing your prize, you suddenly feel it's twice as likely that you got the higher amount. Bizarre, huh? And your expected value calculation becomes

E(switching)=P(H|N)*(N/2)+P(L|N)*(N*2)=(2/3)*(N/2)+(1/3)*(N*2)=N,

so there's no point switching after all. Even weirder is that, before you open the envelope, this is a situation where you know for sure what you will believe in the future, but you nevertheless can't make that update till then.

This is a really good exposition of the two envelopes problem. I recall reading a lot about that when I first heard it, and didn't feel that anything I read satisfactorily resolved it, which this does. I particularly liked the more precise recasting of the problem at the beginning.

(It sounds like some credit is also due to VincentYu.)

The flaw in the argument is simply that it assumes E(X/Y) > 1 implies that E(X) > E(Y).

I didn’t understand this sentence very well at first, because the inequality on the right is two steps removed from the one on the left. I find this version clearer:

The flaw in the argument is simply that it assumes E(X/Y) > 1 implies that E(X) / E(Y) > 1. (If E(X) / E(Y) > 1, that would imply that E(X) > E(Y).)

Here's a solution to a more general version of the problem:

Let's say that the red envelope contains N times as much money as the blue envelope with probability p, and the blue envelope contains N times as much money as the red envelope with probability (1 - p).

Without loss of generality, N is at least 1.

If N = 1, both envelopes contain the same amount, and there is no point in switching.

If N > 1, let the variable s represent the smaller amount of money between the amounts of the two envelopes. So one envelope contains s, and the other envelope contains Ns.

Scenario 1: The blue envelope contains s, and the red envelope contains Ns. This scenario occurs with probability p.

Scenario 2: The red envelope contains s, and the blue envelope contains Ns. This scenario occurs with probability (1 - p).

Assume s is not less than 0.

The expected amount of money in the blue envelope, E(B) = sp + Ns(1 - p) = s(p + N - Np).

The expected amount of money in the red envelope, E(R) = s(1 - p) + Nsp = s(1 - p + Np).

If s = 0, both envelopes contain the same amount, and there is no point in switching.

If s > 0, compare the expectations of the amounts of money in the envelopes in terms of s:

E(B) > E(R) when

s(p + N - Np) > s(1 - p + Np)

p + N - Np > 1 - p + Np

2p - 2Np > 1 - N

2p(1 - N) > 1 - N

2p < 1, because (1 - N) < 0

p < 1/2

Similarly, E(B) < E(R) when p > 1/2, and E(B) = E(R) when p = 1/2.

The ratios of the expected amounts of money in each envelope depend only on p when s > 0 and N > 1.

Both envelopes contain the same amount of money when s = 0, N = 1, or both.

So if your degree of belief that the red envelope contains more money than the blue envelope is the same as your degree of belief that the blue envelope contains more money than the red envelope, don't bother switching, unless you need to kill time (which you already knew intuitively, but Q.E.D.)

I got the idea of using the variable s to represent the smaller of the amounts in the two envelopes from R Falk 2008: "The Unrelenting Exchange Paradox".