# One can earn nothing on the Brownian motion, true ?

Consider any discrete time stochastic process $p(n)$ (price) with independent increments $\xi_k$ and $E(\xi_k)=0$. E.g. Brownian motion (i.e. $\xi_k = N(0,1)$).

Consider some "trading strategy" which means - we can "buy" at some moments $t_k$, keep for some time, and then sell. The profit is difference between the price we sell and we buy. Assume we can buy only one asset, not more. It is slightly informal definition, but hope it is clear.

Question/Conjecture Is it true that E(profit) = 0 for any trading strategy, under assumption above that price is described by the independent increment process ? (May be I need also assume that distributions are symmetric and/or identical ).

Any information on specific processes like $\xi_k= \pm 1$, or $\xi_k = N(0,1)$ is highly welcome.

Motivation: since we cannot predict the price, we should not be able to earn something, so $E(profit)<=0$, however if there exist a strategy, such that $E(profit) < 0$ (strictly less than zero) , we can try to consider "inverse" (not sure it is well-defined) strategy and get $E > 0$. So it seems the only choice is to have $E(profit) = 0$.

Details: I assume that we are trading for some time n=0...N, and at last moment "N" we MUST sell on the price p(N) (if we have an asset).

Example: Consider the simple case $\xi_k = +1$ or $-1$ with probabilities $1/2$. Consider the trading only for n=0,1,2. I assume that for n=0, p(0) = 0, and we buy asset for this price.

It seems the conjecture is true in this case. It seems, there are just 4 strategies, and for all of them E(profit) = 0. Look:

Strategy 1. "As soon price>0, sell it, and then do nothing" - it seems good strategy and we should have positive profit - but no way - the trouble is if the price goes down p(1)=-1, p(2) = -2 , we did not sell for n=1, so we must sell at n=2, so we get big loss = -2. It compensates possible profits.

Strategy 2. "Keep untill the end" - obviously E(profit) = 0.

Strategy 3. "Sell at n=1", - obviously E(profit) = 0.

Strategy 4. " if p(1) =-1, then sell, otherwise keep until the end". It seems it is crazy strategy we sell when price lowered down - so we for sure have a loss, but it is compensated by the fact that if p(1) =1, p(2) = 2 we will get profit = 2.

(It is somewhat opposite strategy to strategy 1).

Combinatorial question By the way how many trading strategies are there for such binomial distribution $\xi_k= \pm 1$ ?

PS

The question seems to me well-defined mathematical question, if something is unclear, please tell me I'll explain. Please avoid discussion whether it is realistic model for price or not, or something like, that. Please treat it as a mathematical problem. If one proposes a trading strategy with $E(profit) \ne 0$ it can be easily checked by MatLab or Excel or whatever.

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Related: mathoverflow.net/questions/132430/… "How much one can earn on white noise ?" –  Alexander Chervov Jun 2 '13 at 15:33
Have a look at martingale theory and in particular Doob's optional stopping theorem (or optional sampling theorem), see e.g. en.wikipedia.org/wiki/Optional_stopping_theorem. –  Uwe Franz Jun 2 '13 at 16:45
This question is not research level, but it would be appropriate for math.stackexchange.com. –  Nate Eldredge Jun 2 '13 at 19:20
Since you deal with the finite-time horizon, the price is just a martingale even if the increments would be dependent: you just need that the conditional expectation of the increment (given previous history) is zero. As @Uwe suggested, OST would tell you that whenever you buy an asset, the expected gain is profit is zero no matter which selling time you choose. Although optimal stopping theory often considers just a single stopping time, if you take care of details and apply this fact sequentially, you get that the total expected profit for several trades is zero as well. –  Ilya Jun 3 '13 at 15:01
[continued] An alternative way would be to apply stochastic optimal control as I replied to another question of yours. The situation there becomes trivial since the value function will not depend on time. –  Ilya Jun 3 '13 at 15:03

This is OK if the "time you keep it" is a bounded random variable, for example. See the suggested topic of "martingale" for more on this.

But what if we allow an unbounded stopping time? Say the independent increments are $\pm 1$, we buy at time $0$ and then sell as soon as we are $1$ dollar ahead. The event that we are eventually $1$ dollar ahead happens with probability one.

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in the example Strategy 1 is exactly - keep until it is +1. But since I stop at fixed Time n=N. We may never get +1. –  Alexander Chervov Jun 2 '13 at 20:57

This is a specific case of Doob's optional sampling theorem

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Unless I'm missing something, discrete time and the lack of transaction costs mean that your question boils down to whether there is any strategy such that, after 1 time unit only, one can expect a profit. (A simple induction argument suffices to show this.) If that's true, then a symmetric distribution, where the probability of loss equals the probability of profit after 1 time unit, implies that expected profit will always be 0.

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Well, that was meant to be a comment. Can't see how to change it, though. –  abo Jun 2 '13 at 19:11
It might be you are right. Let me think. Thank you. –  Alexander Chervov Jun 2 '13 at 20:52