Interesting Mathematical Paradox?

[jason1990]

[ QUOTE ]
Suppose there are two such groups. The First deals with the information by ignoring it (applying the indifference principle?).

[/ QUOTE ]
I would not call this the indifference principle. There are three Stages:
<ul type="square">
[1] The Information. I see amount A.
[2] The Effect. ???
[3] The Decision. Switch or not. Call off bet or not.[/list]The indifference principle assumes that there is no effect. This produces a mathematical contradiction. Hence, there must be some effect, though it is unknown.

The people in your First group may very well realize this. They simply decide to not call off their bet, regardless of the information.

[ QUOTE ]
The Second deals with the information by saying upon opening the envelope that the amount changes the experiment but they don't know how so they call off the $100 wager.

[/ QUOTE ]
Why should ignorance in Stage 2 lead one to think that calling off the bet in Stage 3 is the best decision?

What about the hypothetical group of Random Switchers who use their pocket calculator to generate an exponentially distributed random variable, and switch envelopes when A is less than the number they generated? They do better on average than the Always Switchers and the Never Switchers. This, by itself, should be convincing enough evidence that The Effect exists.

[PairTheBoard]

[ QUOTE ]
[ QUOTE ]
Suppose there are two such groups. The First deals with the information by ignoring it (applying the indifference principle?).

[/ QUOTE ]
I would not call this the indifference principle. There are three Stages:
<ul type="square">
[1] The Information. I see amount A.
[2] The Effect. ???
[3] The Decision. Switch or not. Call off bet or not.[/list]The indifference principle assumes that there is no effect.

[/ QUOTE ]

Is that correct? This Indifference Principle is new to me. But from your description of the Bent Coin I don't see that it assumes there is no effect from Bending the Coin. I see it as saying that since as far as we know, the Bend can bias Heads just as well as Tails, for the purposes of making a Bet on the First Coin Flip we may as well figure the chances are still 50-50 even though we know they are probably not.

[ QUOTE ]
What about the hypothetical group of Random Switchers who use their pocket calculator to generate an exponentially distributed random variable, and switch envelopes when A is less than the number they generated? They do better on average than the Always Switchers and the Never Switchers. This, by itself, should be convincing enough evidence that The Effect exists.

[/ QUOTE ]

I'm not arguing that there is no effect. Certainly this shows that there is. This raises an interesting question though. The Decision by Calculator improves the results of the Switchers. But if the $100 bettors use the same method to decide whether to continue their bets or call them off, does it improve their results? Will they also always get better results than by just ignoring the amount A, and always continuing the bet?

PairTheBoard

[jason1990]

[ QUOTE ]
[ QUOTE ]
The indifference principle assumes that there is no effect.

[/ QUOTE ]

Is that correct?

[/ QUOTE ]
Let E be the event that we have the smaller envelope. Unconditioned, P(E) = 1/2. If we apply the indifference principle after looking at the value in the envelope, A, then we would conclude that P(E|A) = 1/2. Notice that this implies

P(A = x and E) = P(A = x)P(E | A = x) = P(A = x)P(E).

That is, under the indifference principle, A and E are independent. This is what I mean by "no effect."

[jason1990]

I was a little dissatisfied with my Random Switchers, since I was supposed to be replying to your idea about calling off the fixed $100 bet. So here is an idea about Random Call-Offers.

There is a sequence of Groups. Group n decides whether or not to call off the bet in this way. They generate a random number U, uniform on (0,1), and call off their bet if

A &gt; sqrt{n|ln(U)|}.

Their EV is

50(2e^{-y^2/n} - e^{-4y^2/n}),

where the values in the envelopes are y and 2y. For small n, Group n will not do very well, since they will be calling off their bet too often. But the EV will increase monotonically with n. At about n = 1.64y^2, the EV of Group n will be about the same as the Never Call-Offers, $50. But the EV will continue to increase, reaching a maximum at about n = (3y^2)/ln(2). This Group will have an EV of about $59. After that, the EV will decrease monotonically, with the limit being $50. In other words, all Groups with n sufficiently large will outperform the Never Call-Offers.

[PairTheBoard]

[ QUOTE ]
I was a little dissatisfied with my Random Switchers, since I was supposed to be replying to your idea about calling off the fixed $100 bet. So here is an idea about Random Call-Offers.

There is a sequence of Groups. Group n decides whether or not to call off the bet in this way. They generate a random number U, uniform on (0,1), and call off their bet if

A &gt; sqrt{n|ln(U)|}.

Their EV is

50(2e^{-y^2/n} - e^{-4y^2/n}),

where the values in the envelopes are y and 2y. For small n, Group n will not do very well, since they will be calling off their bet too often. But the EV will increase monotonically with n. At about n = 1.64y^2, the EV of Group n will be about the same as the Never Call-Offers, $50. But the EV will continue to increase, reaching a maximum at about n = (3y^2)/ln(2). This Group will have an EV of about $59. After that, the EV will decrease monotonically, with the limit being $50. In other words, all Groups with n sufficiently large will outperform the Never Call-Offers.

[/ QUOTE ]


That's remarkable and I think an interesting new addition to the archive of 2+2 Two Envelope Facts, Results, and Implications.

I'm a little suprised because I thought the 2-1 odds were so great as to put any improvement out of reach. I guess it should be noted that for any fixed envelope amounts y,2y picking n large enough improves on the Never-Call-Offers. However, no matter how large n is chosen, if the envelope amounts y,2y are large enough the Never-Call-Offers still do better.

But wait a minute. This is actually amazing isn't it? Why can't you just wait until you see the envelope amount before you choose which n to use for your random choice? You won't know whether you're looking at y or 2y. But just choose n large enough to work for either. That way you can always insure an improved decision over just Not Calling-Off. Are you sure this is right? If so I think its a significant advance.

Wouldn't this varying n method also work for asssumed prior distributions for Y (other than the point mass)?

PairTheBoard

[jason1990]

[ QUOTE ]
But wait a minute. This is actually amazing isn't it? Why can't you just wait until you see the envelope amount before you choose which n to use for your random choice? You won't know whether you're looking at y or 2y. But just choose n large enough to work for either.

[/ QUOTE ]
Suppose you see A. You know that either y = A or y = A/2. The worst case for you is y = A. So you could choose n = CA^2 for some suitable large C. The problem is this. The smaller n is, the more likely you are to call off your bet. With this strategy, in the case that A = y, you will be more likely to call off your bet than in the case A = 2y. This is exactly the opposite of what you want to be doing. You will screw yourself with this strategy and actually end up doing worse than the Never Call-Offers no matter how large C is.

[PairTheBoard]

[ QUOTE ]
[ QUOTE ]
But wait a minute. This is actually amazing isn't it? Why can't you just wait until you see the envelope amount before you choose which n to use for your random choice? You won't know whether you're looking at y or 2y. But just choose n large enough to work for either.

[/ QUOTE ]
Suppose you see A. You know that either y = A or y = A/2. The worst case for you is y = A. So you could choose n = CA^2 for some suitable large C. The problem is this. The smaller n is, the more likely you are to call off your bet. With this strategy, in the case that A = y, you will be more likely to call off your bet than in the case A = 2y. This is exactly the opposite of what you want to be doing. You will screw yourself with this strategy and actually end up doing worse than the Never Call-Offers no matter how large C is.

[/ QUOTE ]

I thought my idea sounded too good to be true. But I guess I'm not understanding how this is working. You said,

[ QUOTE ]
For small n, Group n will not do very well, since they will be calling off their bet too often. But the EV will increase monotonically with n. At about n = 1.64y^2, the EV of Group n will be about the same as the Never Call-Offers, $50. But the EV will continue to increase, reaching a maximum at about n = (3y^2)/ln(2). This Group will have an EV of about $59. After that, the EV will decrease monotonically, with the limit being $50. In other words, all Groups with n sufficiently large will outperform the Never Call-Offers.



[/ QUOTE ]

I understand you get screwed with small n. But in my idea you'd be choosing n large. I understand you can't insure it's the optimum n. But if you choose n larger than 1.64A^2 and A=y then you've chosen n greater than 1.64y^2 and you improve over Never-Calling-Off. And if A=2y, then you've still chosen n greater than 1.64y^2 and you still improve over Never-Calling-Off. In either case you've improved over Never-Calling-Off. What am I missing here?

PairTheBoard

[jason1990]

I think what you may be missing here is that my EV calculations are based on the fact that A and n are independent. If they are dependent, then that EV formula and subsequent analysis goes out the window. If you look back at the Call-Off criterion, you will see that if n is of the form CA^2, then you will just be calling off your bet with some fixed probability that does not depend on A. Your EV will then simply be $50*(prob. of not calling off).

Let me throw something else out there. Suppose you have a function f taking values between 0 and 1. You can use this function to generate a strategy for calling off. Namely, if you see A, then you call off your bet with probability 1 - f(A). Conversely, whatever strategy you might employ will correspond to some f. For example, the Never Call-Off strategy corresponds to f(A) = 1 for all A. The Group n strategy corresponds to f(A) = e^{-A^2/n}.

If your strategy corresponds to f, then your EV will be 50(2f(y) - f(2y)). In order to outperform the Never Call-Offers, you need this to be greater than 50. Let us call a strategy "uniformly dominating" if

2f(y) - f(2y) &gt;= 1 for all y.

I claim that the only uniformly dominating strategy is the Never Call-Off strategy. I will leave it as a puzzle to prove this result.

[PairTheBoard]

[ QUOTE ]
I think what you may be missing here is that my EV calculations are based on the fact that A and n are independent.

[/ QUOTE ]

ok. I think I just fell into another version of the Two Envelope Paradox flawed thinking. n is not a random variable. But if I choose n based on A I turn it into a random variable. In fact, I do so in a way that actually worsen my results, as you point out here:

[ QUOTE ]
Suppose you see A. You know that either y = A or y = A/2. The worst case for you is y = A. So you could choose n = CA^2 for some suitable large C. The problem is this. The smaller n is, the more likely you are to call off your bet. With this strategy, in the case that A = y, you will be more likely to call off your bet than in the case A = 2y. This is exactly the opposite of what you want to be doing. You will screw yourself with this strategy and actually end up doing worse than the Never Call-Offers no matter how large C is.


[/ QUOTE ]

Another way to look at it is in an example. Say y=10. Say I look in the envelope and see 10. Say I then choose the random decision based on n=200. Will my results be better than Never-Calling-Off? Well, no. If I repeat the experiment for this fixed y and look at all the times I see 10 in the envelope, I will be calling off the bet some of the time which is poor because the Never-Calling-Off strategy wins every time I see 10 in the envelope. This even though the n=200 Decision strategy works better than the Never-Calling-Off strategy for y=10. That's because when you repeat the experiment for that EV you get to see A=20 half the time.

Introducing the function f is a nifty way to express the decision and generalize.

It looks like

2f(y) - f(2y) &gt;= 1 for all y

would imply

f(2^n *y) &lt;= 2^n(f(y)-1) +1

with right hand side going negative for large n, unless
f(y)=1

I'm going to have to think about all this a little bit. I think it has shed a little more light on the Two Envelope Paradox.

PairTheBoard