Why Symmetric Returns Do Not Make ATR Stop Strategies Profitable
Summary
The document examines a coin-toss trading rule that places a stop 0.5 ATR from the entry and a profit target 2 ATR away, asking why it loses in foreign exchange. Its central point is that an apparently favorable reward-to-risk ratio does not by itself create positive expected value. Under the answer’s assumed normal log-return model, the probabilities of reaching the two thresholds offset the different payoff sizes, leaving expected payoff at zero.
The discussion then explains why real results can be worse: market returns may have fat tails, skew, and jumps, so a normal model may not describe the path to either exit. Trading also incurs spreads, fees, and slippage, which reduce realized expectancy. The document gives a conceptual explanation rather than a tested strategy or a complete probability derivation; it does not establish that the stated ATR distances are suitable across currency pairs or market conditions.
Key ideas
- A larger profit target than stop distance does not alone imply positive expected value.
- Under the assumed normal return model, the hit probabilities offset the asymmetric payoff sizes.
- Actual returns can depart from normality through fat tails, skew, and jumps.
- Spreads, fees, and slippage can turn a zero-expectancy rule into a losing one.
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Full text
# If the distribution of returns in symmetric, why not use a coin toss, small risk & high reward?
# If the distribution of returns in symmetric, why not use a coin toss, small risk & high reward?
If the distribution of returns is symmetric then why not
- use a coin toss to decide whether to buy or sell
- Calculate the average velocity of the market (ATR - in technical analysis)
- Place a stop loss on 0.5 ATR away from current price and take a profit 2 ATR away from the current price?
I tried it in FOREX and it doesn't seem to work. Why is this so?
## Answer by binjip (score 2, accepted)
https://quant.stackexchange.com/a/8036
There are several reasons:
Expected payoff = 0
I don't know why you selected 0.5 ATR and 2 ATR away from the market but let's go with it for a while. This means that you want to gain 2x while risking only 0.5x. For now let's assume that FX log-returns are normal.
To bring it to a higher level, we can use a piece of Black Scholes formula, namely the probability that an option ends up in the money is N(d$_{2}$) where N(*) is a cumulative normal distribution of a function and d2 is a function defined as in BS. So your expected payoff is E[payoff] = 2*N(d$_{2}$(2)) - 0.5*N(d$_{2}$(0.5)) which equals 0 if you do the calculation. This explains why you don't make money, not why in reality you lose money. Read on.
Returns are not normally distributed
In general, normal distribution is not a good representation of FX (or other) log-returns. The real distribution of returns has fat tails, often skewness, prices have jumps etc. If the assumption of normality is rejected, your model breaks down.
Transaction costs
As edouard mentioned, even if your E[payoff] = 0, you would incur transaction costs (through slippage, fees, bid/ask spread etc.) which would make your E[payoff] < 0.
Hope it helps.
## Answer by tagoma (score 1)
https://quant.stackexchange.com/a/3148
could this strategy be applied for real trading ?
i mean, whatever a trade makes or loses money, trading incurs transaction-costs.
thus, you cannot stand on that (perfect) bell-shaped distribution to trade profitably with any certainty.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.