Statistics Question

This was posted in the Science, Math and Philosphy forum, but it doesn't look like I'm going to get an answer there, so I posted here (apologies to the wrong-forum nits)

So I'm reading one of my textbooks, and I read something that if I understand correctly basically says:

Suppose that you have $1. You bet some one on a coin flip, even money. If you lose, you stop. If you win, 1/2 a second later you bet $2. If you lose you stop. If you win, 1/4 a second later you bet $4. If you lose you stop. If you win, 1/8 of a second you bet $8. etc...

So, the authors of the book conclude the probability that you will go broke after 1 second is 1. I agree.

However they conclude that your EV after 1 second is -$1. I disagree, I say it's $0.

What do you guys think? It may just be me not understanding a definition some where, I dunno

Anyways, here's some of the book:


Quote:
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Example 3.1 The following example of a suicide strategy is borrowed from Harrison and Pliska (1981). It can be modified easily to provide an example of an arbitrage opportunity in an unconstrained Black-Scholes setting. For simplicity, we take r = 0, T=1, and So=1. For a strictly positive constant b>0, we consider the following trading strategy:

y1(t) = {-b 0 =< t <= r(b), 0 otherwise}

y2(t) = {1+b, 0 =< t <= r(b), 0 otherwise)

where

r(b) = inf{t: St = 1 + 1/b} = inf{t:V(y,t) = 0}


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here V(y,t) is defined as:

y1(t)*St + y2(t)*Bt

St = Stock Price
Bt = Bond Price (risk free investment, like money in a savings account (means Bt>0) or loaned from a bank (means Bt<0))


Quote:
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In financial interpretation, an investor starts with one dollar of wealth, sell b shares of stock short, and buys 1+b bonds. Then he holds the portfolio until the terminal date T=1, or goes bankrupt, whichever comes first. The probability of bankruptcy under this strategy is equal to p(b) = P{r(b)<1}, so that it increases from zero to one as b increases from zero to infinity. By selling short a very large amount of stock, the investor makes his failure almost certain, but he will probably make a great deal of money if he survives. The chance of survival can be completely eliminated, however, by escalating the amount of stock sold short in the following way.

To show this, we shall modify the strategy as follows. On the time interval [0,1/2], we follow the strategy above with b = 1. The probability of bankruptcy during [0,1/2] thus equals p = P{r() =< 1/2}. If r(1) >1/2, the amount of stock sold short is adjusted to a new level b1 at time 1/2. Simultaneously, the number of bonds held is revised in a self-financing fashion. The number b1 is chosen so as to make the conditional probability of ruin during the time interval (1/2,3/4], given that we have survived up to time 1/2, equal to p again.

In general, if at any time t(n) = 1 - (1/2)^n we still have positive wealth, then we readjust (typically increase) the amount of stock sold short, so that the conditional probability of bankruptcy during (t(n), t(n+1)] is always p. To keep the strategy self-financing, the amount of bonds held must be adjusted at each time t(n) as well. The probability of survival until time t(n) is then (1-p)^n, and it vanishes as n tends to 0 (so that t(n) tends to 1). We thus have an example of a piecewise constant, self-financing strategy, (y1,y2) say, with V(y,t) = y1(0)So + y2(o) = 1,

V(y,t) = y1(t)St + y2(t)>=0, for any t in [0,1)


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by So they mean St at time t=0


Quote:
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and V(y,1) = 0. To get a reliable model of a security market we need, of course, to exclude such examples of doubling strategies from the market model.


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So basically I say V(y,1) = 1, not 0.

This is from "Martingale Methods in Financial Modelling" by Marek Musiela and Marek Rutkowski

I typed this (yes the whole damn thing) so there many or may not be some errors, also, I changed a few symbols (like y instead of psi)

[LetYouDown]

Well, you're betting $1...you're going to lose that dollar. I guess it's a matter of semantics whether it's -1 or 0. I'd side with -1. How would you qualify the EV of a bet where you bet $1 and get $1 returned to you no matter what?

Basically I am thinking that there is an infintesimal chance that you win an infinite amount of money, which is why on average you will not lose or win any money

I think a similar question is the doubling system:

You bet $1, you win you stop. If you lose you bet $2, if you win you stop. If you lose you bet $4 etc....

What do you think the EV is of this game?

And does the answer change if you are playing red on roulette (house edge involved)?

Anyways, I probably picked a bad example, as my discrepancy with the book is definatly not a matter of semantics

[LetYouDown]

No, I think I misunderstood what you meant. You're saying that the amount you win will still be inversely proportional to the chance that you win, regardless of the fact that the chance you will win approaches 0 for all practical intents and purposes?

[ QUOTE ]
No, I think I misunderstood what you meant. You're saying that the amount you win will still be inversely proportional to the chance that you win, regardless of the fact that the chance you will win approaches 0 for all practical intents and purposes?

[/ QUOTE ]

Yeah pretty much

But to tell the truth I don't know the exact mathematical definition of EV, so that might be a factor

So what do you reckon the EV of the situation in the OP is? 0 (what I think) or -1 (what the authors of the book think)?

[LetYouDown]

Practically, -1. In reality, 0. Although I'd say this qualifies as a paradox...and then again, I'm not certain that it does. All the more reason to think it's a paradox, lol.

[mosdef]

Interesting question.

You're right by the way. The expected value of your winnings is 0, not -1.

The reason for the apparent paradox is that just because the prob of your winnings equalling -1 approaches 100%, that doesn't make the EV -1. Consider the following thought exercise. Play the exact same game, but everytime you win the coin toss you actually win nothing. Now, according to the authors' logic, since prob of of your winnings equalling -1 approaches 100%, then EV -1. But this new game can't have the same EV, can it? Of course not.

Return to the original example. The reason that their logic breaks down is that they are making an incorrect inference. In particular, let X_n be the r.v. representing your winnings in the game. Let Y equal r.v. equal to -1 with probability 1. Prob(X_n=-1)->1 as n-> infinity. However, the r.v. X_n does NOT approach the r.v. Y as n-> infinity. For that to be true, the values of X_n and Y would need to fit within an arbitrarily small distance of each other over the whole range of possible outcomes. This is clearly not the case since some of the values become X_n are extremely large as n-> infinity.

This is very hard to explain without a chalk board and pictures. Please fly to Toronto with a chalk board and I will explain it to you in person.

[ QUOTE ]
Please fly to Toronto with a chalk board and I will explain it to you in person.

[/ QUOTE ]
[img]/images/graemlins/grin.gif[/img]


What does r.v. stand for?

[mosdef]

random variable

[ QUOTE ]
random variable

[/ QUOTE ]
[img]/images/graemlins/blush.gif[/img]

ok, it makes sense now, thanks, I was beginning to think I seriously misunderstood a very basic concept some where since this was in a textbook

In the past, when I have had disagreements with textbooks, it was profitable to bet on the textbook, not me