The Envelope Paradox: What Does Maths Say?

[matt grime]

Wow, you managed to make the probabilities add up to more than 1 in a really trivial way (n=0...).

Now, I seem to have read recently that it is provable that there is no prior probability distribution where it is *always* preferable to switch. I think you should work out those posteriors again, after making it a probability distribution in the first place, though.

 

[gel]

Wow, you managed to make the probabilities add up to more than 1 in a really trivial way (n=0...).

Sorry, that was just a typo. I fixed my post - try re-reading it now.

 

[Count Iblis]

Ok, we know this intuitively. It seems that the key is to find the flaw in the logic of the OP's analysis. Why does his math seem to show that switching will on average yield a better result?

Because one assumes a uniform prior (which is not really possible as pointed out earlier). But if it were then the expected amount of money in the envelope would have to be infinite. So, if you only find a finite amount of money in the envelope, you should switch.

 

[quadraphonics]

Because one assumes a uniform prior (which is not really possible as pointed out earlier). But if it were then the expected amount of money in the envelope would have to be infinite. So, if you only find a finite amount of money in the envelope, you should switch.

I think that this is a more satisfying way of stating the resolution than to simply point out that a uniform distribution with infinite support is not technically allowed. I.e., if the expected return under not switching is infinity, then the fact that switching provides 1.25 times the expected return does not actually recommend switching, since 1.25*infinity is still just infinity, so either strategy has the same expected return.

 

[Hurkyl]

If I open an envelope and see 100$, my expected return from swapping is between 50$ and 200$ -- i.e. not infinite.

 

[matt grime]

Sorry, that was just a typo. I fixed my post - try re-reading it now.

I believe you now got the version where you're taking the difference between two random variables with infinite means.

 

[gel]

I believe you now got the version where you're taking the difference between two random variables with infinite means.

The variables X and Y in my post do have infinite means. However, the conditional distributions have finite means. So, once you open one envelope it makes sense to ask what the expected return from opening the other envelope is, and E(X|Y) >= 2Y, E(Y|X) >= 2X.

 

[matt grime]

As far as I can tell, the explanation of this is that since the expected amount you get is infinite (before peeking), then this means you shouldn't apply the rules of conditional probability, as things demonstrably go funky. Although there are examples where it will work, and some will find that explanation troublesome.

 

[gel]

I don't think it matters if the distributions have infinite mean. You can still use conditional probabilities just fine.
Here, we have E(Y|X)>= 2X and E(X|Y) >= 2Y, suggesting that it is always preferable to switch. If X and Y had finite mean then you could use the tower law to get E(Y) >= 2E(X) and E(X) >= 2E(Y) >= 4E(X). The only way is if E(X) and E(Y) are both infinite (or zero).

So the paradox can only occur with infinite means for the variables. The conditional expectations do still exist though.

And I wouldn't say that you shouldn't use conditional expectations of variations of variables with infinite mean. It can be very useful in many cases. What you have to be careful about is taking expectations of variables with infinite mean.

 

[uart]

So the paradox can only occur with infinite means for the variables

Well I suppose if you were playing a gme with an expected outcome of infinity then you'd be pretty pissed if you got any finite amount /joke. Does that example prove that some people are never satisfied?