My question is related to this thread, but I'm interested in a special case. Suppose that the price of an asset starts at 100 USD, and changes according to a geometric random walk; one step of 1% either up or down at each second. You start with an equity of 100 USD, and you put 100 USD at each trade.

The strategy is a very simple one-sided trade. Two orders are set repeatedly: A buy order at $x_{buy}=(100-1)^{n1}$ and then a sell order at $x_{sell} = (100+1)^{n_2}$, where $n_1$ and $n_2$ are positive integers.

We know that the price will hit $x_{buy}$ and $x_{sell}$ an infinite number of times, so this strategy is determined to be profitable (however small) in the long term. The problem is, it's an extremely slow strategy.

Suppose that you want to maximize the expected profit over a 1 year period. What values for $n_1$ and $n_2$ are optimal?

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    $\begingroup$ Hi: the probable reason that you're not getting any replies is that it's not at all clear why that strategy is profitable. You said that it's profitable but I'm not 100 % sure. So, like myself, people probably read the strategy a few times and are too confused to possibly deal with optimizing such a strategy. If your argument is that those prices are always eventually hit, that doesn't matter. You still have to be more on the correct side of the trade than not. How is that guaranteed ? $\endgroup$
    – mark leeds
    May 19 at 3:44
  • $\begingroup$ Ahhhh, you're right, I need to edit the question: You use 100 USD (not 100% of equity) at each trade. Now I think it's obvious that after a long enough time (which I can't calculate the expected value for), your profit will exceed 100 USD, and you will be permanently in profit afterwards. Is it better now? $\endgroup$
    – asmani
    May 19 at 4:22
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    $\begingroup$ Hi Asmani: I'm not trying to be difficult but I truly don't understand why it's profitable. Could you explain that ? If you do, maybe someone can explain why your explanation is not correct. Note that one should be short when they expect price to decrease and long when they expect price to increase. You haven't explained how you create your expectation. $\endgroup$
    – mark leeds
    May 19 at 18:04
  • $\begingroup$ Thank you Mark. I admit I didn't make it clear (I edited now), but what I have in mind is a one-sided strategy (otherwise your short position would definitely get liqudated at some point). You buy at 99 USD and sell at 101 USD, making a profit of 2.02 USD each time. after some time (which I don't know the expected value for), you'll have traded 50 times and have made over 100 USD profit. $\endgroup$
    – asmani
    May 19 at 19:04
  • $\begingroup$ The aforementioned example is for the case $n_1=n_2=1$. $\endgroup$
    – asmani
    May 19 at 19:11

2 Answers 2


If you assume $S_t$ is a geometric brownian motion, there is 100% probability the stock goes to 0. It's why there's a theory of stocks always going to 0. This can be proved using markov chains and stopping times, but also proved using martingales, i.e. All non-negative martingales converge, in this case, it converges to 0.

Intuitively, you can think of it like; if you will hit every possible value, then given enough time, you will - thus stopping at 0.

In your case, yes the stock will hit $x_{buy}$ and $x_{sell}$ infinitely many times, IF the stock can go negative. But in reality, once $S_t=0$, the game stops, which means you may leave with less equity than you started


interesting proposal, but if your movement up or down is random, in the long run it will be 50% probability up or down, and P&L from "up" is the same as P&L from "down" movement, so you have expected payoff of -1 * 0.5 + 1 * 0.5 = 0.

And as someone mentioned in the comments, you might be picking growing your capital (randomly) until you enter a trade and face a streak of 100 moves in the opposite direction and go bust.


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