Statistics Question

[AaronBrown]

This is one of those weird things about infinity that defies common sense. I understand how uncomfortable people are with saying you can have negative expected value from making fair bets, but it is true. A simpler example is the standard martingale strategy of doubling your bet every time you lose in roulette. If you can bet unlimited amounts and guarantee to finish in finite time (as by making each spin take half as long as the previous one) you can have a positive expected value for the strategy although each bet individually is negative expected value. That's no argument for actually playing the thing, it's just a consequence of the mathematical definitions.

If you like this sort of thing, remember the infinite hotel. It's all full when the infinite bus pulls up with an infinite number of guests. "No problem," says the clerk, "I'll just tell all the existing guests to move to the room twice the number of their current room. That will free up an infinite number of odd numbered rooms, so we can accomodate everyone." It's no use arguing that the hotel was full before so it can't accomodate new guests, with infinity that kind of argument doesn't work.

Another fun one is there are more irrational numbers than rational numbers, but there are an infinite number of rational numbers between any two irrational numbers.

[mosdef]

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Another fun one is there are more irrational numbers than rational numbers, but there are an infinite number of rational numbers between any two irrational numbers.

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well, more mind-boggling is how many MORE irrational numbers there are than rational numbers.

also, the number of rational numbers in the interval (0,1) is the same as the number of rational numbers in the interval (0,2) and is the same as the number of integers.

also, most of the irrational numbers are trascendental, but we only know of about 25 of them.

these things provide great food for thought up to the point where you go insane. after that you get tenure and no one ever sees you again so it doesn't matter.

by the way, do you think the EV of the game is 0 or -1? initially i thought it was 0 (see my previous post), however you've got me doubting myself. However, i think if you write the winnings as a function of n, where n is the last toss, then sum all of the winning over all n weighted by the probability of n being the last toss, you get 0, not -1.

[AaronBrown]

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by the way, do you think the EV of the game is 0 or -1? initially i thought it was 0 (see my previous post), however you've got me doubting myself. However, i think if you write the winnings as a function of n, where n is the last toss, then sum all of the winning over all n weighted by the probability of n being the last toss, you get 0, not -1.

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It's -1, just as the roulette strategy is +1.

The problem with your logic is the probability of n being the last toss is not indpendent of whether the result is 0 or 1. You're computing the unconditional distribution of wealth at points in time and averaging them. The strategy determines a distribution of wealth conditional on stopping.

[mosdef]

Hmm... I think I see it now.

Let N be the random variable equal to the number of the last toss. N will take on values 1,2,3,... with probabilities 1/2, 1/4, etc.

Let X be your profit when the game stops. X is a random variable, written as a function of the random variable N it is X(n)=-1 for all n. So if you take the E[X] you just get -1. Correct?

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you can have a positive expected value for the strategy although each bet individually is negative expected value.

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You are right, to suggest this makes me very uncomfortable.

I have to ask you another question:

You bet $1 on a coinflip. If you lose, you stop. If you win, you bet again, but your next bet you get 2:1 odds (+EV), so you bet $2, if you lose you stop, but if you win, you are now stuck with a profit of $5 ($1 from first bet, $4 from second bet)

So your next bet is $6, and now you get 4:1. If you lose you stop, if you win you now have $29 profit.

And on your next bet you get 8:1 odds. You bet $30. if you lose you stop, if you win you have a profit of $269.

Your next bet you get 16:1 odds, etc....

Same as before, each bet takes half as long, so it's over in a second. Do you still think the EV is -$1? I think it is now positve infinity, quite a big discrepency.

[alThor]

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I think it is now positve infinity, quite a big discrepency.

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Yes; the coin flip problem as you described it is just the St. Petersburg Paradox, but with an (irrelevent) "time" story. The EV involves an infinite sum. By the standard way of calculating an infinite sum, the EV is infinite.

Yet in addition, the probability of going broke is also 100%; that's the paradox!. It's just yet another "infinity" paradox.

Now regarding the book (which I have not read), I don't know what assumptions they are using. For instance, if you bound the wealth of your opponent in the St Petersburg game, you cannot get infinite EV. The value of the world's capital is finite. Therefore one cannot apply the coin example directly to the stock market. Hence I cannot say whether there is a mistake in the book.

But in summary, you are correct that there would be a flaw in any reasoning that said, in your coin example, that "since P(ruin) = 1, EV must be -1". Again, I can't tell if that's what the book is trying to say.

alThor

[krubban]

First visit to this part of the forums for me. Feels like being in mathclass [img]/images/graemlins/smile.gif[/img]
I don't agree on your last example though that the estimated value is positive infinity. If you do an infinite amount of coinflips you will lose eventually no matter what odds you get. Unless you meant that the profit from each flip is somehow saved up and not used to bet on your next flip.

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you can have a positive expected value for the strategy although each bet individually is negative expected value.

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You are right, to suggest this makes me very uncomfortable.

I have to ask you another question:

You bet $1 on a coinflip. If you lose, you stop. If you win, you bet again, but your next bet you get 2:1 odds (+EV), so you bet $2, if you lose you stop, but if you win, you are now stuck with a profit of $5 ($1 from first bet, $4 from second bet)

So your next bet is $6, and now you get 4:1. If you lose you stop, if you win you now have $29 profit.

And on your next bet you get 8:1 odds. You bet $30. if you lose you stop, if you win you have a profit of $269.

Your next bet you get 16:1 odds, etc....

Same as before, each bet takes half as long, so it's over in a second. Do you still think the EV is -$1? I think it is now positve infinity, quite a big discrepency.

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I would like to change it a bit more

Repeat this new process (which took no more than 1 second) but do it in no more than 1/2 a second, then do the whole process again in no more than 1/4 a second

Now according to your logic the EV should be negative infinity, and I say it's positive infinity, an even bigger discrepency