Validate a trading strategy without hindsight · 2 / 5
Handle a candle that touches both stop and target
A candle can contain both your stop and target without revealing which came first. Choosing the winning path silently changes an uncertain result into an invented fill.
Athenum7 minUpdated:
Specify the position and exit assumptions
Assume a one-unit long already filled at 100 before the interval, with a stop at 98 and target at 104. The exits are one-cancels-other: filling either cancels the other. For this exercise only, assume continuous prices, adequate liquidity, no gaps and exact barrier fills. Initial risk is 2 price units, defined as 1R.
An interval with open 100, high 105, low 97 and close 101 permits both target-first and stop-first paths. OHLC alone does not identify the actual path. Lower-timeframe observations help only if their sequencing and coverage resolve the ambiguity; a smaller candle can still touch both exits.
Bounds are not probabilities
Report the results under each admissible ordering. The range is conditional on the fill assumptions, not a probability distribution. Its midpoint is not expected profit unless a defensible model supplies the path probabilities. Keep ambiguous observations in the research record instead of deleting them because they are inconvenient.
A gap through the stop can make losses worse than the stop-first bound. A touched limit target may not fill because other orders are ahead in the queue. Treat gaps, partial fills and queue position as separate execution questions rather than hiding them inside an optimistic bar-ordering convention.
The same ten trades produce net bounds of 0R to 6R
For the ambiguous candle, 100 → 105 → 97 → 101 hits the target first: profit 4, or +2R. The path 100 → 97 → 105 → 101 stops out first: loss 2, or −1R. Both fit the four candle values.
Suppose eight unambiguous trades total +3R and two ambiguous trades each permit −1R or +2R. The ten-trade gross bounds are +1R and +7R. Charge a specified 0.1R per trade: total cost is 1R, leaving net bounds 0R and +6R. Reporting +6R alone conceals the unresolved ordering.
| Path | First exit | Gross result |
|---|---|---|
| 100 → 105 → 97 → 101 | Target 104 | +2R |
| 100 → 97 → 105 → 101 | Stop 98 | −1R |
- Stop-first bound
- 0 R
- Target-first bound
- 6 R
A conservative convention is not a historical observation
Always assigning stop-first can be a predeclared stress case, but do not label it the observed intrabar sequence. Conversely, a simulator's chosen path is not proof that the exchange executed that path. Preserve the distinction in both results and lesson captions.
Before acting
- Confirm the position existed before the interval.
- Specify exit cancellation and fill rules.
- Retain both admissible orderings.
- Deduct costs from the whole sample.
- Test gaps and queue uncertainty separately.
Check your understanding
Four unambiguous trades total −1R. One ambiguous trade permits −1R or +2R. Each of the five trades costs 0.1R. What are the net bounds?
Show the explained answer
The lower bound is −1 − 1 − 0.5 = −2.5R. The upper bound is −1 + 2 − 0.5 = +0.5R. The sign is unresolved, so this sample does not establish positive net expectancy.