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.