4A.8 Trailing stops · M04A Reading markets

Motivation

"Cut losses short and let profits run" is usually implemented as a trailing stop: a stop that follows the best price reached. What does it give back, how long does it hold, and does it improve expectancy?

Reading guide

The rule itself (trail the best price by a distance $d$ set from ATR as in 4A.7, as in the chandelier exit), Proposition 4A.11's statement, "What it means" and Worked example 4A.7 are for every reader. The proof of Proposition 4A.11 and the discrete-monitoring paragraph are advanced and can be skipped on a first reading.

4A.8.1The law of the peak#

Setup

Long position, $X_t=x_0+\mu t+\sigma W_t$ , running maximum $M_t=\max_{s\le t}X_s$ , trailing distance $d>0$ . Exit at

\tau=\inf\{t: M_t-X_t\ge d\},

the first time the drawdown from the peak reaches $d$ . Exit P&L is $X_\tau-x_0=(M_\tau-x_0)-d$ under continuous monitoring.

Proposition 4A.11law of the peak; Taylor 1975, Lehoczky 1977

$M_\tau-x_0$ is exponentially distributed with mean

m(d)=\frac{e^{\gamma d}-1}{\gamma},\quad\gamma=\frac{2\mu}{\sigma^2}\qquad(m=d\text{ if }\mu=0). \tag{4A.14}

Consequently: the expected P&L is $m-d$ ; the probability of a loss is $P(M_\tau-x_0<d)=1-e^{-d/m}$ ; the give-back $M_\tau-X_\tau$ equals $d$ ; and the expected holding time is

\mathbb E\tau=\frac{m-d}{\mu}\quad(\mu\ne0),\qquad \mathbb E\tau=\frac{d^2}{\sigma^2}\quad(\mu=0). \tag{4A.15}

Proof

Let $Y=M_\tau-x_0$ . Suppose the maximum has reached level $x_0+y$ . By the strong Markov property at the hitting time of $x_0+y$ , the process restarts there, and $Y>y+\epsilon$ happens iff, starting from $x_0+y$ , it reaches $x_0+y+\epsilon$ before $x_0+y-d$ (the stop does not move until a new maximum is made, and the drawdown cannot reach $d$ while the path is above $x_0+y-d$ ). By Proposition 4A.7 (shifted to start at the level), this probability is

\frac{s(0)-s(-d)}{s(\epsilon)-s(-d)}=\frac{1-e^{\gamma d}}{e^{-\gamma\epsilon}-e^{\gamma d}}=1-\frac{\gamma\epsilon}{e^{\gamma d}-1}+O(\epsilon^2),

independent of $y$ . Hence $P(Y>y+\epsilon\mid Y>y)=1-\lambda\epsilon+O(\epsilon^2)$ with $\lambda=\gamma/(e^{\gamma d}-1)$ , so $P(Y>y)$ satisfies $G'(y)=-\lambda G(y)$ , $G(0)=1$ (the maximum starts at $x_0$ ), and $Y\sim\mathrm{Exp}(\lambda)$ with mean $1/\lambda=m(d)$ . As $\mu\to0$ , $(e^{\gamma d}-1)/\gamma\to d$ (the driftless calculation gives $\frac{\epsilon}{\epsilon+d}$ directly). Continuity of paths gives $M_\tau-X_\tau=d$ exactly. The expected P&L is $\mathbb EY-d$ . For the holding time apply Proposition 4A.9: $\mathbb E[X_\tau-x_0]=\mu\mathbb E\tau$ , so $\mathbb E\tau=(m-d)/\mu$ for $\mu\ne0$ ; letting $\mu\to0$ with $m-d=\gamma d^2/2+O(\gamma^2)$ and $\mu=\gamma\sigma^2/2$ gives $d^2/\sigma^2$ . (Integrability of $\tau$ : $\tau$ is the first time the drawdown $M_t-X_t$ , a Brownian motion with drift $-\mu$ reflected at 0, reaches $d$ ; such hitting times have finite mean, see Taylor 1975.)∎

4A.8.2Skew, give-back and drift#

What it means

With no drift, the trailing stop has zero expectancy (as it must, by Proposition 4A.9), loses money 63.2% of the time ( $1-e^{-1}$ ), loses at most $d$ , and has an exponential right tail of winners. It manufactures positive skew from nothing: many small losses, a few large wins. That is the "let profits run" payoff profile, and on noise it is worth exactly zero.
With positive drift the peak's mean grows faster than linearly in $d$ ( $m\approx d+\gamma d^2/2$ ), so wider trails capture more of a trend, at the cost of longer holding time and a larger give-back.
The give-back from the peak is always $d$ under continuous monitoring. A trader who "gave back too much" set $d$ too wide for her taste; the give-back is a choice, not bad luck.

4A.8.3Close-only monitoring#

Discrete monitoring Checked once a day on closes, the drawdown overshoots $d$ . Simulation (m04a_trailing.simulate_trailing) shows that the mean give-back is $d+\beta\sigma\sqrt{\Delta t}$ with $\beta=0.5826$ to within Monte Carlo error, and that the expected P&L and loss probability are well approximated by Proposition 4A.11 with $d$ replaced by $d+\beta\sigma\sqrt{\Delta t}$ . This is the Broadie–Glasserman–Kou correction (4A.10) applied to the drawdown process; we use it as an approximation verified numerically, not as a theorem. (The corrected formula underpredicts the expected holding time, because the overshoot itself has variance.)

Worked example 4A.7a Nifty futures trail

Nifty at 24,000, daily volatility 1% (240 points). Trailing stop 300 points.

No drift, intraday monitoring: expected peak gain 300 points, expected P&L 0, $P(\text{loss})=63.2\%$ , expected holding time $300^2/240^2=1.56$ trading days.
Drift +12 points a day (0.05% a day, a little above Nifty's 2011–2026 average of about 0.04% a day): $\gamma d=2\times12\times300/240^2=0.125$ , $m=319.6$ points, expected P&L $+19.6$ points (₹1,271 per lot of 65), $P(\text{loss})=60.9\%$ , $\mathbb E\tau=1.63$ days.
Same trail checked only at the close: effective distance $300+0.5826\times240=439.8$ points. Expected give-back about 440 points, expected P&L about +43 points with drift, and the trade lasts longer.

A 300-point trail on Nifty is a 1.6-day trade. To ride a multi-week trend, the trail must be several daily standard deviations wide; then each exit gives back several days' worth of range. There is no free lunch in the parameter.