4A.6 Stops and targets as barrier problems

Motivation

Every swing system (M04 · 4.2) has a stop and many have a target. The questions a trader asks ("How often will this stop be hit? How long will the trade last? Does a 2:1 target improve my system?") have exact answers in the simplest model of prices. The answers are sobering and useful.

4A.6.1Hit probabilities of a bracket#

Setup

Measure P&L per unit from entry: $X_t=\mu t+\sigma W_t$ , $X_0=0$ , with $t$ in trading days and $X$ in rupees (or, for percentage stops, in log-return units). A bracket has a target at $+a$ and a stop at $-b$ , $a,b>0$ . Let

\tau=\inf\{t\ge0: X_t\notin(-b,a)\},\qquad p=P(X_\tau=a).

Brownian motion with drift is the model of M02 and M08 for log prices over horizons where the drift is roughly constant. Every result below assumes continuous monitoring; discrete monitoring is treated at the end of the section. $\tau<\infty$ almost surely, and indeed $\mathbb E\tau<\infty$ (a Brownian motion leaves any bounded interval in finite expected time; the formulas below show it).

Proposition 4A.6driftless bracket

If $\mu=0$ , then $p=\dfrac{b}{a+b}$ .

Proof

$X_t=\sigma W_t$ is a martingale and $|X_{t\wedge\tau}|\le\max(a,b)$ . By optional stopping (bounded stopped martingale, $\tau<\infty$ a.s.), $0=\mathbb EX_\tau=ap-b(1-p)$ .∎

This is the continuous version of the gambler's ruin of M03 · 3.6, and it already contains the main lesson: in a driftless market, the hit probability is fixed by the geometry. A target twice as far as the stop is hit one time in three, whatever the chart looked like at entry.

Proposition 4A.7bracket with drift

For $\mu\ne0$ put $\gamma=2\mu/\sigma^2$ . Then

p=\frac{1-e^{\gamma b}}{e^{-\gamma a}-e^{\gamma b}}. \tag{4A.6}

As $\mu\to0$ , $p\to b/(a+b)$ . For a symmetric bracket $a=b$ , $p=e^{\gamma b}/(1+e^{\gamma b})$ .

Proof

Let $s(x)=e^{-\gamma x}$ (the scale function). By Itô's lemma, $ds(X_t)=s'(X_t)\,dX_t+\tfrac12s''(X_t)\sigma^2dt=s(X_t)\big(-\gamma\mu+\tfrac12\gamma^2\sigma^2\big)dt-\gamma s(X_t)\sigma\,dW_t$ . The drift term vanishes because $\tfrac12\gamma^2\sigma^2=\gamma\mu$ when $\gamma=2\mu/\sigma^2$ . So $s(X_{t\wedge\tau})$ is a bounded martingale, and optional stopping gives $s(0)=p\,s(a)+(1-p)s(-b)$ , i.e. $1=pe^{-\gamma a}+(1-p)e^{\gamma b}$ . Solve for $p$ . For the limit, expand numerator and denominator to first order in $\gamma$ : $\frac{-\gamma b}{-\gamma a-\gamma b}$ . For $a=b$ : $\frac{1-e^{\gamma b}}{e^{-\gamma b}-e^{\gamma b}}=\frac{e^{\gamma b}(e^{\gamma b}-1)}{(e^{\gamma b}-1)(e^{\gamma b}+1)}$ .∎

4A.6.2Exit time and expectancy#

Proposition 4A.8expected time to exit

If $\mu=0$ , $\mathbb E\tau=ab/\sigma^2$ . If $\mu\ne0$ ,

\mathbb E\tau=\frac{ap-b(1-p)}{\mu}. \tag{4A.7}

Proof

For $\mu=0$ : $M_t=X_t^2-\sigma^2t$ is a martingale. Optional stopping at $\tau\wedge n$ gives $\sigma^2\mathbb E[\tau\wedge n]=\mathbb E[X_{\tau\wedge n}^2]\le\max(a,b)^2$ , so by monotone convergence $\mathbb E\tau<\infty$ , and then bounded convergence on the left gives $\sigma^2\mathbb E\tau=\mathbb EX_\tau^2=a^2p+b^2(1-p)$ . With $p=b/(a+b)$ this is $\frac{a^2b+b^2a}{a+b}=ab$ . For $\mu\ne0$ : $X_t-\mu t$ is a martingale; optional stopping at $\tau\wedge n$ gives $\mu\mathbb E[\tau\wedge n]=\mathbb E[X_{\tau\wedge n}]\in[-b,a]$ , so $\mathbb E[\tau\wedge n]\le\max(a,b)/|\mu|$ and $\mathbb E\tau<\infty$ by monotone convergence; bounded convergence then gives $\mathbb EX_\tau=\mu\mathbb E\tau$ , and $\mathbb EX_\tau=ap-b(1-p)$ .∎

Proposition 4A.9exits do not create expectancy

Let $\tau$ be any stopping time with $\mathbb E\tau<\infty$ (a bracket, a trailing stop, a time stop, a discretionary exit that uses only past prices). Then

\mathbb E[X_\tau]=\mu\,\mathbb E[\tau]. \tag{4A.8}

In particular, if $\mu=0$ every exit rule has zero expectancy, and the R-multiple expectancy of a bracket with $b=1R$ and $a=kR$ is $kp-(1-p)=0$ .

Proof

$X_t-\mu t=\sigma W_t$ is a martingale. For bounded $\tau$ , optional stopping gives $\mathbb E[X_\tau]=\mu\mathbb E\tau$ . For general $\tau$ with $\mathbb E\tau<\infty$ , apply it to $\tau\wedge n$ and let $n\to\infty$ : $\mathbb E[W_{\tau\wedge n}^2]=\mathbb E[\tau\wedge n]\le\mathbb E\tau$ , so $\{W_{\tau\wedge n}\}$ is bounded in $L^2$ , hence uniformly integrable, and $\mathbb EW_{\tau\wedge n}\to\mathbb EW_\tau$ .∎

Read (4A.8) as: expectancy = drift × expected time in the trade. The exit rule chooses the shape of the P&L distribution (win rate, payoff ratio, skew) and the time spent exposed. It cannot change the sign of the expectancy, and it changes the size only through $\mathbb E\tau$ . A "2:1 reward-to-risk" rule is not an edge; with no drift, it buys a 2:1 payoff with a 1-in-3 win rate.

Worked example 4A.5a swing trade, in rupees

A stock at ₹1,000 with daily volatility 2% (₹20 a day). Stop ₹960 ( $b=40$ , so 1R = ₹40 per share), target ₹1,080 ( $a=80$ , 2R).

No drift: $p=40/120=1/3$ ; $\mathbb E\tau=80\times40/20^2=8$ trading days; expectancy 0.
Drift $\mu=+\text{₹}1$ a day (0.1% a day, about 25% a year: a strong momentum edge): $\gamma=2/400=0.005$ , $p=(1-e^{0.2})/(e^{-0.4}-e^{0.2})=0.4018$ ; $\mathbb E\tau=8.21$ days; expectancy $\mu\mathbb E\tau=\text{₹}8.21$ per share $=0.205R$ .
Drift $\mu=-\text{₹}1$ a day: $p=0.2693$ ; $\mathbb E\tau=7.68$ days; expectancy $-\text{₹}7.68=-0.192R$ .

A large edge moves the win rate of a 2:1 bracket from 33% to 40%. Over the 7–8 days of a typical trade, Indian delivery costs of about 24 bp round trip (M01: ₹237.82 on ₹1 lakh) are ₹2.4 per share here, a third of the expectancy. Win rates in the high 30s with average winners near 2R, the profile of the swing traders in M04 · 4.2.5, are what a real but modest drift looks like through a bracket.

4A.6.3Stop-outs over a horizon#

Proposition 4A.10probability of being stopped out over a horizon

For a stop at $-b$ and no target,

P\Big(\min_{t\le T}X_t\le-b\Big)=\Phi\!\Big(\frac{-b-\mu T}{\sigma\sqrt T}\Big)+e^{-2\mu b/\sigma^2}\,\Phi\!\Big(\frac{-b+\mu T}{\sigma\sqrt T}\Big). \tag{4A.9}

For $\mu=0$ this is $2\Phi(-b/(\sigma\sqrt T))$ . For $\mu>0$ and $T\to\infty$ it tends to $e^{-2\mu b/\sigma^2}<1$ ; for $\mu\le0$ it tends to 1.

Proof

For $\mu=0$ this is the reflection principle. For $\mu\ne0$ , change measure with Girsanov's theorem: under $\mathbb Q$ with $d\mathbb Q/d\mathbb P|_{\mathcal F_T}=\exp(-(\mu/\sigma)W_T-\tfrac12(\mu/\sigma)^2T)$ , $X/\sigma$ is a standard Brownian motion, and the joint density of $(\min_{t\le T}X_t, X_T)$ under $\mathbb Q$ follows from reflection. Multiplying by the likelihood ratio $\exp(\mu X_T/\sigma^2-\mu^2T/(2\sigma^2))$ and integrating over $\{\min\le-b\}$ gives (4A.9); the computation is in Karlin and Taylor (1975), ch. 7, and Shreve, Stochastic Calculus for Finance II, 7.2. The limits follow from $\Phi(\pm\infty)$ .∎

Table 4A.1 (driftless stop-out probabilities) Stop distance in multiples $k$ of the daily standard deviation, horizon $n$ trading days; entries $2\Phi(-k/\sqrt n)$ .

$k$ \ $n$	1	5	10	20	60
0.5	0.617	0.823	0.874	0.911	0.949
1	0.317	0.655	0.752	0.823	0.897
1.5	0.134	0.502	0.635	0.737	0.846
2	0.046	0.371	0.527	0.655	0.796
3	0.003	0.180	0.343	0.502	0.699

A stop one daily standard deviation away is hit within a month (20 days) 82% of the time on pure noise. "My stops keep getting hit" is usually this table, not stop-hunting. The table also shows the trade-off: a tighter stop allows a larger position for the same rupee risk (M03 · 3.9), but buys a much higher probability of being shaken out before the thesis has time to work.

4A.6.4Closing-basis stops#

Discrete monitoring A closing-basis stop (exit if the day closes below the level) or any stop checked on a grid of step $\Delta t$ misses intraday touches. Broadie, Glasserman and Kou ("A continuity correction for discrete barrier options", Mathematical Finance 7(4), 1997, 325–349) showed that a barrier monitored every $\Delta t$ is approximated, to $o(\sqrt{\Delta t})$ , by a continuously monitored barrier moved away from the start by

\beta\,\sigma\sqrt{\Delta t},\qquad \beta=-\frac{\zeta(1/2)}{\sqrt{2\pi}}\approx0.5826. \tag{4A.10}

A closing-basis stop with daily volatility $\sigma$ therefore behaves like an intraday stop placed $0.58\sigma$ further away. The same constant gives the expected overshoot: a discretely monitored barrier is crossed, on average, about $0.58\sigma\sqrt{\Delta t}$ beyond the level. That is the expected gap-through slippage of a closing-basis stop. The code (m04a_barriers) checks the corrected formulas against simulation; with $\Delta t=1/400$ the corrected two-barrier probability matches the simulated one within Monte Carlo error.

Practitioner notes

Choose the stop where the trade idea is wrong, then set position size from it (M03 · 3.9). Do not choose the stop from the position size you would like.
Targets are optional. By (4A.8) a target only shortens $\mathbb E\tau$ and so cuts expectancy when $\mu>0$ ; its use is to lock in the payoff distribution you want (for example, a fixed 2R to keep variance bounded) and to free capital.
Intraday stops on NSE: the pre-open does not accept stop-loss orders (M01 · 1.2), so a stop entered overnight becomes active only in the continuous session, after the open has printed.