Why a Random-Walk Trading Strategy May Not Earn Expected Profit
Summary
The document considers a one-sided strategy that repeatedly places buy and sell orders at price levels determined by step counts in a geometric random walk. The question asks how to choose those levels to maximize expected profit over a fixed horizon, but the answers do not derive an optimum. Instead, they challenge the premise that eventual visits to both levels guarantee gains. One response argues that under the stated idealized dynamics the price can converge to zero, ending trading with capital losses; another notes that symmetric up and down moves have zero expected directional payoff before costs.
The responses also point to ruin risk from a long adverse streak. Their conclusions depend on the assumed process, how prices are prevented from becoming negative, and the precise financing, order, and stopping rules. The document therefore serves as a warning about confusing repeated level crossings with positive expected return, rather than a complete strategy evaluation or quantitative optimization.
Key ideas
- Repeatedly reaching buy and sell levels does not by itself establish positive expected profit.
- Under symmetric moves, the expected directional payoff can be zero before transaction costs.
- A price process that can converge to zero may terminate the strategy with losses.
- Long adverse streaks can create substantial ruin risk for a leveraged or fully committed strategy.
- The document does not calculate optimal order-level parameters for the stated horizon.
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# Maximum profit from trading on a random walk with a specific strategy
# Maximum profit from trading on a random walk with a specific strategy
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?
## Answer by THATS MY QUANT MY QUANTITATIVE (score 1)
https://quant.stackexchange.com/a/77344
If you assume the dynamics of $S_t$ is a geometric Brownian motion (or a discrete simple symmetric random walk), there is 100% probability the stock goes 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
## Answer by John (score 0)
https://quant.stackexchange.com/a/75886
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.Shown in full with attribution under the source's licence. Licence: CC BY-SA 4.0 (Stack Exchange)
This summary was written by Stratmill's research agent from the original; it is not a copy of the source.