4A.10 Time stops, and why widening a stop is negative-expectancy behaviour

4A.10.1Time stops#

A time stop exits after a fixed holding period $T$ if neither stop nor target has been hit: the "vertical barrier" of the triple-barrier method (López de Prado 2018, ch. 3; M04). Its justification is that most trading signals have a decaying edge.

Proposition 4A.13time stop for a decaying edge

Suppose the drift after entry is $\mu_t=\mu\,\mathbf 1\{t\le h\}$ with $\mu>0$ and constant volatility $\sigma$ , and the trader holds one unit until a deterministic time $T$ . Then the P&L has mean $\mu\min(T,h)$ and variance $\sigma^2T$ , and its Sharpe ratio $\mu\min(T,h)/(\sigma\sqrt T)$ is maximised at $T=h$ .

Proof

$G=\int_0^T\mu_tdt+\sigma W_T$ . For $T\le h$ the ratio is $\mu\sqrt T/\sigma$ , increasing; for $T\ge h$ it is $\mu h/(\sigma\sqrt T)$ , decreasing.∎

After the signal's horizon, holding adds variance and no mean. M04's event studies estimate $h$ (the horizon over which the conditional mean return differs from zero); the time stop should be set from that estimate, not from patience.

4A.10.2Widening the stop#

A trader enters with a stop at −1R. The price falls to −0.8R. Instead of accepting the stop, she moves it to −2R "to give the trade room". This is common, and SEBI's studies of loss-making retail traders (M01 Case 1) and the disposition-effect evidence (M14 · 14.9.2) describe its consequences. Proposition 4A.9 explains the arithmetic.

Proposition 4A.14widening a stop scales the expected loss with the expected time

Let the conditional drift after the adverse move be $\mu\le0$ . From the current point, keeping the original stop at distance $b_1$ or widening it to $b_2>b_1$ , with a target at distance $a$ , the expected P&L from now on is $\mu\,\mathbb E\tau(a,b_i)$ , and $\mathbb E\tau(a,b)$ is increasing in $b$ . So if $\mu<0$ , widening strictly lowers the expected P&L; if $\mu=0$ it leaves it at zero and raises the variance of the outcome.

Proof

The first statement is (4A.8). Monotonicity: the exit time from $(-b_2,a)$ is at least the exit time from $(-b_1,a)$ pathwise, and strictly larger with positive probability, so $\mathbb E\tau(a,b_2)>\mathbb E\tau(a,b_1)$ . For $\mu=0$ , $\mathbb EX_\tau^2=\sigma^2\mathbb E\tau=ab$ increases with $b$ .∎

Worked example 4A.9

Units of R. Current P&L −0.8R; original stop −1R (0.2R away), widened stop −2R (1.2R away), target +2R (2.8R away). Daily volatility 0.25R. Suppose the adverse move carries mild information: conditional drift −0.05R a day.

Original stop: target probability 0.3%, expected exit time 3.8 days, expected final P&L −0.99R.
Widened stop: target probability 1.0%, expected exit time 23.2 days, expected final P&L −1.96R.

With no drift, both have expected final P&L −0.8R, but the widened version has a much wider spread of outcomes and ties up capital six times longer. Widening the stop is never an improvement in expectancy; it is a purchase of hope (a slightly higher chance of getting back to the target) paid for with a larger expected loss whenever the move against you means anything.

Why traders do it anyway: prospect theory In prospect theory (M14 · 14.9.1) the value function is convex over losses. At a paper loss, a gamble that might recover to breakeven is valued above the certain loss of the same expected value, which is risk-seeking in losses. M14 Proposition 14.15 shows this produces the disposition effect: holding losers, selling winners. Moving a stop wider is the disposition effect at the scale of one trade, and the reason the rule "stops move only in the direction of the trade" is written into trading plans (M14 · 14.10.2). R-multiple accounting (M03 · 3.1) also breaks: once stops move, 1R no longer measures risk, and the system's measured expectancy is computed on a quantity the trader no longer controls.