So anyway I solved the problem above. And the idea that she doesn't get new information is valid, but only because the new information she gets which increases her credence of tails from 1/2 to 2/3 (I am waking and being asked credence therefore it isn't TH and I am in the experiment), demands mathematically by the theorem Jason posted that there was also a possibility of information decreasing her credence of tails, which would happen on TH when she realizes she isn't "woken" (poked) and realizes that it can't be Monday and her credence of tails is 0%-except it can't be done in this experiment design because she is asleep and has no credence, and the SIA is invalidated. We can reverse this by saying the 3 waking days, her credence of tails would increase to 2/3 as she doesn't get waken (sampled) on Tuesday and TH would equal 0, but sleeping beauty is asleep and doesn't have a credence. Gone. So we must rely on the SSA, which provides the answer of 1/2. Note that if the experiment was changed and the TH communication happened, there really would be no new information at all from the act of waking and the odds would be 1/2 from the SIA as well. If we then add a discriminating cue such as allowing betting some rounds, like she gets asked to bet every day but TH, then the SIA would hold up and she should wager on tails (2 to 1 fav) when she is asked to bet and answer 2/3 for her credence for tails, and when she is not asked to bet, she knows it is TH and can also answer that her credence is 0 for tails. That cue does not exist in the original SBP and as I've proven the answer is then 1/2. In this model if we were to ask her to bet on TH as well as the other days, there is no new information and it's 50/50 once again.

or at the end, its okay that she doesn't get sampled and she can know before the experiment that her credence for TH the 1/2 time it happens would be 0, and 2/3 is correct using SIA after all? I don't know the SIA or the maths (jason's theory) good enough to say. Hah. Seriously, eagerly awaiting response on this.

She would answer 1 for TH which satisfies the theorem and gives 1/2 but the experiment is over and she can't be sampled. I was saying "asleep" in my earlier posts but that was incorrect, the experiment is just over- same thing. Does it still get "counted" to satisfy the theorem you posted seems to determine whether 1/3 is a valid answer to this question yeah? Seems like clearly no and my logic above is correct

I suppose it depends on exactly what you mean by "the expected holding". To make things clear, let the "Small Envelope" amount be chosen from some random distribution like the one BruceZ gave in his post (uniform on {1,2,...100}) and let the "Large Envelope" be 2*(Small Envelope). Shuffle the two envelopes to get a "First Envelope" and "Second Envelope". Then,

P( [Second Envelope] = 2*[First Envelope]) = 1/2
P( [Second Envelope] = .5*[First Envelope]) = 1/2

Then it is NOT the case that,

E[Second Envelope] = 1.25*[First Envelope]

That would make no sense because E[Second Envelope] is constant and in BruceZ's example equals 75, while "First Envelope" remains a random varibiable.

What is true is that the expected value of the ratio is 1.25. ie.

E[(Second Env)/(First Env)] = E[(First Env)/(Second Env)] = 1.25

Since the expected value of the two ratios above are equal it's hard to see how a case can be made for an advantage in switching based on the expected value of the ratio being 1.25. The expected value of the respective ratio when not switching is also 1.25.




I don't understand this example.


PairTheBoard

No. I had the same confusion at first. E[Second Envelope] is in units of [First Envelope], or X in Aaron's notation, not dollars. I think that's what's confusing everyone, including the people on wiki, and the people who think they've solved the paradox by finding an error of equivocation. Then they keep finding new ways to interpret it, and they then find a different error. They assume there must be an error because switching is ridiculous, so they interpret it so as to find an error. The only error is in their arguments. The fact is, you interpret it this way, and there is no error. You really should switch when your numeraire is X, but not when it is dollars, and there is nothing ridiculous about different numeraires resulting in different decisions.


I was confused by that too, and the book is clearer than the post. The key is that every other bidder has an envelope with the same scrip that you start out with. Now ONLY YOU are allowed to switch with some outside person who offers you an envelope which has twice your scrip with probability 1/2, and half your scrip with probability 1/2. Now you really do switch, which is pretty obvious, and it is because the numeraire is the value of the envelope that everyone starts with. You wouldn't then swap back, because you wouldn't change numeraires to your new envelope. The original envelope that you started with that everyone else has is the correct numeraire here.

Can we avoid turning this into a thread about the envelope problem? I'm sure there are a good half dozen existing envelope problem threads that you guys can discuss "numeraires" in.

Oh. I think I see what you're saying. Treating the "First Envelope" as a numeraire, then the ratio (Second Env)/(First Env) is expressing the "Second Envelope" in units of the numeraire "First Envelope". I think it's still improper for you to say "the expected holding is 1.25*X". But I see what you're getting at. More properly I think would be,

"The EV of the Second Envelope expressed in units of the First Envelope is 1.25."

Which is just another way of saying,

E[(Second Env)/(First Env)] = 1.25

But with the respective EV, E[(First Env)/(Second Env)] = 1.25 as well, you would need something more involved to establish some priority for treating the contents of the First Envelope with priority, thus breaking the above symmetry. That's what you do in the bolded of your example above.

To specify your example a little more, suppose the envelopes are filled as I describe above with the Small Envelope from a uniform distribution on {1,2,...100} and the envelopes shuffled. Suppose then that whatever the amount turns out to be, say A, in the First Envelope, that amount is given to everybody. Suppose further that A/100 is established as the base unit of currency for everybody on a loaf of bread standard. That is, A/100 buys 1 loaf of bread. Then it is advantageous for you to switch to the Second Envelope because the expected value of the Second Envelope is 125 loafs of bread while the sure value of the numeraire First Envelope is 100 loafs of bread.

That's interesting.


PairTheBoard

Right. Yea, I saw it and started my post while you were doing yours. That's an interesting insight on "two envelopes" which I think is new to 2+2.

PairTheBoard

I think this concept of "numeraire" deserves some scrutiny. AB claims it applies to Sleeping Beauty. I'm not sure what to make of it but seeing first how he applies it to "two envelopes" seems like a reasonable place to start for understanding what he's talking about.


PairTheBoard

Except it's

E(Second Env)/(First Env) = 1.25

and

E(First Env)/(Second Env) = 1.25

Other than that, I think you've got it.

He could also be the first person in 60 years to solve this correctly. Would *that* deserve a place here ya think? And then recogniziing that the rest of the people putting out papers of nonsense every year were wasting their time trying to find an error when there was none? How 'bout that? And yes, the analogy to SB is excellent. I'll take credit for that one.

Actually, we're both right because they are equal:

E(Second Env)/(First Env) = E[(Second Env)/(First Env)] = 1.25

It's just that in the left hand expression, E(Second Env) is in units of First Env, not dollars. But you can see that the second expression is

0.5*(1/2) + 0.5*(2) = 1.25

But it is NOT

E(Second Env)/E(First Env) = 1

Hence the statement about explaining this in terms of the expected value of a ratio being different from a ratio of expected values.

Getting back to SB, this sentence on p. 336 of Aaron Brown's book Red Blooded Risk says it all even though it was not written with SB specifically in mind:

"Practical statistical analysis must specify a range of application - what the numeraire is and what it can buy. Otherwise there is no logic to connect the frequency with degree of belief."

Actually, I think my notation is the correct one.

Your notation, E(Second Env)/(First Env) does not make it clear that you're computing the expected value of the random variable (Second Env)/(First Env) ratio. Technically, your notation indicates (the expected value of the numerator) divided by the random variable in the denominator. So rather than 1.25, E(Second Env)/(First Env) is technically the random variable 75/(First Env) - via your original example. That's not what we want.

We don't want [E(Second Env)] / (First Env)

We want E[ (Second Env)/(First Env) ]



Also, that's why AB's description, "So the expected holding is 1.25*X", is not so good if X is the First Envelope and "holding" is the Second Envelope. If the "holding" is understood as the number of First Envelope units contained in the Second Envelope then what's correct is that the expected "holding" is 1.25.




PairTheBoard

So you're in the "both answers can be correct" camp?

I don't understand any practical information that possibly could be displayed by the 1/2 answer.

I think that translates to, there's no way the numeraire can be defined such that 1/2 is the answer.



We've shown that when wagering 1/2 doesn't work.

We've shown that when measuring "time spent in each world", 1/2 gives an incorrect ratio.

What question w/ a practical application can be answered with 1/2? (I.e. wagering has a clear practical application, as does the measurement of time - especially in iterated versions of this).

True, quants work with mathematical tools, and some probabilists may study questions arising from finance. But quants are not mathematicians any more than mathematicians are quants. We can ask Aaron. Aaron, are you a professional mathematician?

As for the rest, I don't think you actually know what you're talking about, but w/e. Good for you.

Edit - and since I know effectively nothing about probability theory, I'm sure I would learn a lot from the book.

Mine is fine as long as you compute E(Second Env) in units of (First Env) instead of dollars. Our ratios are the same number. Mine makes it clear that we are using (First Env) as the numeraire which is the key to understanding this.

Aaron Brown posted an example for 1/2. A rather humorous one. Only 1 answer is correct once a given numeraire is defined. Different numeraires produce different answers. Not only in this problem, but for all problems.

You know nothing about probability theory, but you don't think *I* know what I'm talking about. I suggest you read the posts since last night because everything is settled now to those who have the capacity to understand the solution. It is also very possible, IMO, that Aaron Brown has succeeded in resolving the envelope paradox where many of those who publish on it each year have failed. That's because the numeraire idea is the correct one, and that is normally used only by financial people. But not all financial people understand the link between this principle and fundamental probability issues. Aaron is in a unique position to understand both, and why his book is very unique. As for its applicability to SB, which was the initial point, there isn't really even a debate about that. If you don't think I know what I'm talking about, most of Aaron's book will go right over your head.

The way to express the Second Envelope in units of the First Envelope is by the ratio (Second Env)/(First Env). To make it clear that that's what you're taking the expectation of, the standard notation is

E[(Second Env)/(First Envelope)]


PairTheBoard

You can convert to units of the first envelope either before or after the average. The normal argument starts with the amount in your envelope X, and takes the expected value in terms of X. It is correct to say that the average amount in the second envelope is 1.25 times the amount in the first envelope. You can't make a similar statement your way. On the other hand, your way resolves the problem in terms of the difference between the expected value of a ratio and the ratio of expected values. Both insights are useful. They are equivalent.

This is the exact same as the money scenario, except if you ask her, "do you want money tomorrow?", she says, "I don't care."

Even if she won't want more money tomorrow, she should still bet on tails. Nothing has actually changed. I don't understand how you are concluding a probability of 1/2 here.

Even if all she cares about is today, today is still 1/3 MH, 1/3 TH and 1/3 TT.

I don't understand how this is redefining the numeraire at all.

that's the pdf of a laplace distribution, right?

That's a cop-out.

The envelope problem only has one answer. No you should not switch envelopes. The problem is clear enough that there's no reasonable argument for a different numeraire. Yes, you've shown that you can identify the value of one envelope as averaging out to 1.25X the amount in the other envelope. However, the same is true for the other envelope, so they are both equal. There's no way to define the numeraire as highest value of X or anything similar. Both are equal no matter what definitions you use. Even if you are able to come up with some kind of convoluted mathematical argument here, the problem makes it pretty clear (even if it's implicit rather than explicit), that the person wants money.

I think the SB problem is clear enough in the same way. MH, TH, and TT are all equally likely. The people disagreeing on that aren't redefining the numeraire in any acceptable way. They are just wrong.