Barrier options and first passage, financial engineering for quants lesson

A down-and-out call pays like an ordinary call unless the stock falls to a barrier at some moment before expiry, in which case it pays nothing. Its value depends on a question that the law of the terminal price cannot answer. How likely is the path to touch the barrier, and how does the answer change when the barrier is checked only on certain dates? This lesson defines barrier payoffs through hitting times, derives the reflection principle for driftless Brownian motion, shows why matching knock-in and knock-out prices add to the vanilla price, and derives the Brownian bridge crossing probability that links observations on dates to continuous monitoring.

Key terms

Barrier option

Definition A barrier option has a European payoff $g(S_T) \ge 0$ at maturity $T$ that is switched on or off by a path event. The contract fixes a barrier level $H > 0$ , a monitoring set $\mathcal{M} \subseteq [0, T]$ , which is the whole interval for continuous monitoring or a finite set of dates $t_1 < \dots < t_n$ , a hit convention such as $S_t \le H$ for a lower barrier, the settlement, and any rebate. With the hit event

A = \bigcup_{t \in \mathcal{M}} \{ S_t \le H \},

a down-and-out option pays $g(S_T)\mathbf{1}_{A^c}$ at $T$ and a down-and-in option pays $g(S_T)\mathbf{1}_{A}$ at $T$ , with no rebate unless the contract specifies one. Up-and-out and up-and-in options use an upper barrier with the event $S_t \ge H$ . A down barrier with $S_0 \le H$ and $0 \in \mathcal{M}$ is hit at once. Under a pricing measure $\mathbb{Q}$ with constant rate $r$ , the price of the down-and-out at time $0$ is $e^{-rT}\mathbb{E}^{\mathbb{Q}}[g(S_T)\mathbf{1}_{A^c}]$ , which depends on the joint law of the path and $S_T$ , not only on the law of $S_T$ .

In plain words A barrier option is an ordinary option with a switch. The switch depends on whether the price touches a level during the life of the contract, observed in the way the contract specifies.

Example Take a down-and-out call and a down-and-in call with strike $K = 100$ , barrier $H = 90$ and continuous monitoring. A path that ends at $120$ without falling to $90$ pays $20$ from the knock-out and $0$ from the knock-in. A path that ends at $120$ after touching $89$ pays $0$ from the knock-out and $20$ from the knock-in. A path that ends at $80$ pays nothing from either. In every case the two payoffs add to the vanilla payoff $(S_T - 100)^{+}$ .

Hitting time

Definition For a lower barrier $H$ and a monitoring set $\mathcal{M}$ , the hitting time, or first passage time, is

\tau_H = \inf\{ t \in \mathcal{M} : S_t \le H \},

with $\inf \varnothing = \infty$ . The barrier is hit by $T$ exactly when $\tau_H \le T$ , so the down-and-out payoff is $g(S_T)\mathbf{1}_{\{\tau_H > T\}}$ . For a continuous adapted price and continuous monitoring, $\tau_H$ is a stopping time, because the event $\{\tau_H \le t\}$ equals $\{\min_{0 \le s \le t} S_s \le H\}$ , which is determined by the path up to $t$ since the set of levels at most $H$ is closed. With the strict condition $S_t < H$ the hitting time is a stopping time once the filtration satisfies the usual conditions. For discrete monitoring, $\tau_H$ takes values among the dates and is a stopping time for any adapted price. In the continuous models of this lesson the conventions $S_t \le H$ and $S_t < H$ give the same probabilities, but the contract must still fix one.

In plain words The hitting time is the first moment the price is seen at or below the barrier. At each time one can tell from the observed path whether it has already happened, which is what makes it a stopping time.

Example In the constructed model of the next card, with $S_0 = 100$ , $H = 90$ and a driftless log price with $\sigma = 0.2$ , the hitting time has the distribution function

\begin{aligned} &\mathbb{Q}(\tau_{90} \le t) \\[0.4em] &\quad = 2\,\Phi\Bigl(-\frac{\ln(100/90)}{0.2\sqrt{t}}\Bigr), \end{aligned}

which is about $0.2921$ at $t = 0.25$ , $0.4563$ at $t = 0.5$ and $0.5983$ at $t = 1$ . It tends to $1$ as $t \to \infty$ , so the driftless path reaches the barrier eventually with probability one, while the probability of a hit before a fixed maturity is less than one.

Reflection principle

Definition Let $W$ be a standard Brownian motion, $a > 0$ and $T > 0$ . Then for every $y \le a$

\begin{aligned} &\mathbb{P}\Bigl(\max_{0 \le t \le T} W_t \ge a, \\[0.4em] &\qquad W_T \le y\Bigr) \\[0.4em] &\quad = \mathbb{P}(W_T \ge 2a - y), \end{aligned}

and in particular

\begin{aligned} &\mathbb{P}\Bigl(\max_{0 \le t \le T} W_t \ge a\Bigr) \\[0.4em] &\quad = 2\,\mathbb{P}(W_T \ge a) \\[0.4em] &\quad = 2\Bigl(1 - \Phi\bigl(a/\sqrt{T}\bigr)\Bigr). \end{aligned}

The proof reflects each path after its first hitting time $\tau_a$ of the level $a$ , setting $\widetilde{W}_t = W_t$ for $t \le \tau_a$ and $\widetilde{W}_t = 2a - W_t$ for $t > \tau_a$ . By the strong Markov property the path after $\tau_a$ is a Brownian motion started at $a$ and independent of the path before, and by its symmetry $\widetilde{W}$ has the same law as $W$ . A path that reaches $a$ and ends at or below $y$ is paired with a reflected path that ends at or above $2a - y \ge a$ , and every continuous path that ends at or above $a$ has reached $a$ . Since $\mathbb{P}(W_T = a) = 0$ , the hit probability is twice the upper tail. By symmetry the same formula gives $\mathbb{P}(\min_{0 \le t \le T} W_t \le -a)$ . If under $\mathbb{Q}$ the log price is $\ln S_t = \ln S_0 + \sigma W^{\mathbb{Q}}_t$ , which holds for geometric Brownian motion with $r - q = \tfrac12\sigma^2$ , then for $0 < H < S_0$

\begin{aligned} &\mathbb{Q}(\tau_H \le T) \\[0.4em] &\quad = 2\,\Phi\Bigl(-\frac{\ln(S_0/H)}{\sigma\sqrt{T}}\Bigr). \end{aligned}

The zero drift is essential. When the log drift $r - q - \tfrac12\sigma^2$ is not zero, this formula gives the wrong probability, and a drifted first passage formula is needed, which this lesson does not derive.

In plain words Every path that touches the level and finishes below it has a mirror image that finishes above it, and the two are equally likely. Paths that finish above the level must have touched it, so the probability of a touch is twice the probability of finishing above.

Example Take $S_0 = 100$ , $H = 90$ , $\sigma = 0.2$ , $T = 1$ and $r - q = 0.02 = \tfrac12\sigma^2$ , so that the log price has zero drift. The distance to the barrier in Brownian units is $\ln(100/90)/0.2 \approx 0.5268$ , and

\begin{aligned} &\mathbb{Q}(\tau_{90} \le 1) \\[0.4em] &\quad = 2\bigl(1 - \Phi(0.5268)\bigr) \\[0.4em] &\quad \approx 0.5983 . \end{aligned}

If instead $r - q = 0.05$ , the log drift is $0.03$ per year away from the barrier. The drifted formula, confirmed by a simulation, then gives about $0.5508$ , so the driftless formula would overstate the hit probability by about $0.048$ .

Knock-in knock-out parity

Definition Let a knock-in and a knock-out option have the same payoff $g(S_T)$ , maturity, barrier, hit convention, monitoring set and settlement, and no rebates. On every path exactly one of the events $A$ and $A^c$ occurs, so

\begin{aligned} &g(S_T)\mathbf{1}_{A} + g(S_T)\mathbf{1}_{A^c} \\[0.4em] &\quad = g(S_T), \end{aligned}

and by linearity of the pricing expectation, or by the law of one price for the static portfolio that holds both options,

V_{\text{in}} + V_{\text{out}} = V_{\text{vanilla}} .

The identity is model-free. It assumes nothing about the dynamics of the price, only that the contracts match. A rebate on knock-out, a different monitoring calendar or a different hit convention in one of the two contracts breaks the partition, and the identity then needs correction terms.

In plain words Holding both the knock-in and the knock-out is the same as holding the ordinary option, because whatever the path does, exactly one of them is alive at expiry.

Example In the zero-log-drift model with $r = 0.02$ , $q = 0$ , $\sigma = 0.2$ , $S_0 = 100$ , $T = 1$ , $K = 100$ and $H = 90$ , the vanilla call is worth about $8.916$ . Reflection at the level of terminal densities shows that, at every final log price above $\ln H$ , the density of paths that hit $H$ and end there equals the density of driftless paths started at $\ln(H^2/S_0)$ . Since $K \ge H$ , only such final values pay, so the down-and-in call is a vanilla call in the same model on a stock starting at $H^2/S_0 = 81$ , worth about $1.616$ . Parity then gives the down-and-out call

8.916 - 1.616 = 7.300,

which a simulation confirms. With the payoff $g = 1$ , the knock-in pays $1$ at $T$ after a hit and is worth $e^{-0.02} \cdot 0.5983 \approx 0.5865$ , the knock-out is worth $e^{-0.02} \cdot 0.4017 \approx 0.3937$ , and the two add to the zero-coupon bond price $e^{-0.02} \approx 0.9802$ .

Brownian bridge crossing probability

Definition On an interval of length $h > 0$ , let the log price $X_t = \ln S_t$ be $X_t = X_0 + mt + \sigma W^{\mathbb{Q}}_t$ with constants $m$ and $\sigma > 0$ and a $\mathbb{Q}$ -Brownian motion $W^{\mathbb{Q}}$ . Let $0 < H < \min(s, z)$ , and write $x = \ln(s/H)$ and $y = \ln(z/H)$ for the distances of the two endpoints above the barrier in log units. Then

\begin{aligned} &\mathbb{Q}\Bigl(\min_{0 \le t \le h} S_t \le H \Bigm\vert \\[0.4em] &\qquad S_0 = s,\ S_h = z \Bigr) \\[0.4em] &\quad = \exp\Bigl(-\frac{2xy}{\sigma^2 h}\Bigr). \end{aligned}

When $s \le H$ or $z \le H$ the conditional probability is $1$ . The derivation has two steps. First, the path splits as

\begin{aligned} X_t &= X_0 + \tfrac{t}{h}(X_h - X_0) \\[0.4em] &\quad + \sigma\bigl(W^{\mathbb{Q}}_t - \tfrac{t}{h}W^{\mathbb{Q}}_h\bigr), \end{aligned}

and the bridge $W^{\mathbb{Q}}_t - \tfrac{t}{h}W^{\mathbb{Q}}_h$ is Gaussian and uncorrelated with $W^{\mathbb{Q}}_h$ , hence independent of it, so given both endpoints the law of the path does not involve $m$ . Second, take the driftless case. By reflection, the density of paths that reach the barrier and end at distance $y$ is the normal transition density with variance $\sigma^2 h$ over the distance $x + y$ , while the density of all paths ending there uses the distance $y - x$ . Their ratio is

\begin{aligned} &\exp\Bigl(-\frac{(x + y)^2 - (y - x)^2}{2\sigma^2 h}\Bigr) \\[0.4em] &\quad = \exp\Bigl(-\frac{2xy}{\sigma^2 h}\Bigr). \end{aligned}

The formula needs constant coefficients on the interval. Jumps or a varying volatility change it, and general diffusion bridges need approximations with their own error analysis.

In plain words Once both endpoints are known, the drift no longer matters, and the crossing probability depends only on how far the endpoints are above the barrier compared with the variance accumulated in between. Endpoints close to the barrier or a long interval make a crossing likely.

Example With $s = 100$ , $z = 105$ , $H = 90$ , $\sigma = 0.2$ and $h = 1$ , the distances are $\ln(100/90) \approx 0.10536$ and $\ln(105/90) \approx 0.15415$ , so the conditional crossing probability is

\begin{aligned} &\exp\Bigl(-\frac{2 \cdot 0.10536 \cdot 0.15415}{0.04}\Bigr) \\[0.4em] &\quad \approx e^{-0.8121} \approx 0.4439 . \end{aligned}

This is a different question from the unconditional $0.5983$ of the reflection card, since it also uses the ending price, and an ending price of $105$ makes the paths that went down to $90$ less likely. Simulated bridges on finer and finer time grids approach the same value from below. Conditioning on the ending price is a device for computing probabilities and prices. A trader at an earlier time does not know that price.

Barrier monitoring

Definition Barrier monitoring is the rule in the contract that says when the barrier is checked, continuously on $[0, T]$ or on a finite set of dates. If $\mathcal{M}_1 \subseteq \mathcal{M}_2$ , a hit on $\mathcal{M}_1$ is also a hit on $\mathcal{M}_2$ , so for a nonnegative payoff with no rebate the knock-out payoff monitored on $\mathcal{M}_2$ is at most the one monitored on $\mathcal{M}_1$ on every path, and by no arbitrage the same order holds for their prices. By parity the knock-in prices are ordered the other way. Continuous monitoring contains every set of dates, so a continuously monitored knock-out is worth at most any matching discretely monitored one. In a constant-coefficient model, exact simulation of the price at the monitoring dates, with a hit recorded only when a sampled value is at or below $H$ , prices the discretely monitored contract. For the continuously monitored one, simulate the price exactly at dates $0 = t_0 < t_1 < \dots < t_n = T$ and weight the payoff of each path by the survival probability

\prod_{i=1}^{n} (1 - p_i),

where $\Delta t_i = t_i - t_{i-1}$ , $\ell_i = \ln(S_{t_i}/H)$ and

p_i = \exp\Bigl(-\frac{2\,\ell_{i-1}\ell_i}{\sigma^2 \Delta t_i}\Bigr)

when $\ell_{i-1} > 0$ and $\ell_i > 0$ , and $p_i = 1$ when either endpoint is at or below $H$ . Given the sampled values, the bridges on separate intervals are independent by the Markov property, so the weight is the exact conditional survival probability and the weighted estimator is unbiased. Jumps can cross a barrier between dates in a way that no Brownian bridge describes. The laws of the terminal price do not determine barrier prices either, because two path models with the same law of $S_T$ at every maturity can assign different hitting probabilities.

In plain words Removing monitoring dates can only remove hits, never add them, so adding dates can only lower the value of a knock-out or leave it unchanged. The bridge probability accounts for what the path may do between the observed dates.

Example In the zero-log-drift model with $S_0 = 100$ , $H = 90$ , $\sigma = 0.2$ and $T = 1$ , the probability of a recorded hit is about $0.4917$ with $12$ monthly dates and about $0.5733$ with $252$ equally spaced daily dates, against $0.5983$ for continuous monitoring. The monthly dates are among the daily dates, and both lie in $[0, 1]$ , so the three probabilities increase as the inclusion argument requires. A simulation that samples only the $12$ monthly values and uses the bridge weights gives $0.5983$ with a standard error of about $0.0007$ .

Check yourself

In the constructed model with $K = 100$ and $H = 90$ , the vanilla call is worth $8.916$ and the matching continuously monitored down-and-in call is worth $1.616$ . What is the matching down-and-out call worth? If the knock-out instead paid a rebate of $1$ at $T$ whenever it is knocked out, what would the knock-in and the knock-out be worth together, with $r = 0.02$ and the hit probability $0.5983$ ?

Show the answer

Parity for matching contracts without rebates gives $8.916 - 1.616 = 7.300$ for the down-and-out call. A rebate of $1$ paid at $T$ on the hit event adds the value $e^{-0.02} \cdot 0.5983 \approx 0.5865$ to the knock-out, so the two contracts are then worth

\begin{aligned} &1.616 + 7.300 + 0.5865 \\[0.4em] &\quad \approx 9.503, \end{aligned}

more than the vanilla $8.916$ . With the rebate, the two payoffs no longer add to the vanilla payoff on paths that hit the barrier, so the simple parity does not apply.

In the zero-log-drift model with $S_0 = 100$ , $\sigma = 0.2$ and $T = 1$ , move the barrier to $H = 95$ . Find the probability of a hit by $T$ under continuous monitoring and the crossing probability given $S_1 = 105$ , and compare both with $H = 90$ .

Show the answer

The distance in Brownian units is $\ln(100/95)/0.2 \approx 0.2565$ , so the hit probability is

\begin{aligned} &2\bigl(1 - \Phi(0.2565)\bigr) \\[0.4em] &\quad \approx 0.7976, \end{aligned}

compared with $0.5983$ for $H = 90$ . Both endpoints are above $95$ , so the bridge formula applies with $\ln(100/95) \approx 0.051293$ and $\ln(105/95) \approx 0.100083$ , giving

\begin{aligned} &\exp\Bigl(-\frac{2 \cdot 0.051293 \cdot 0.100083}{0.04}\Bigr) \\[0.4em] &\quad \approx e^{-0.2567} \approx 0.7736, \end{aligned}

compared with $0.4439$ for $H = 90$ . Moving the barrier toward the starting price shrinks both distances and raises both probabilities.

A path simulated on monthly dates, with $h = 1/12$ , is at $95$ on one date and at $93$ on the next, with $H = 90$ and $\sigma = 0.2$ in a constant-coefficient model. A classmate records no hit on that interval and uses such paths to price a continuously monitored knock-out. What is the probability that the continuous path crossed $90$ between the two dates, and what does the classmate’s method price instead?

Show the answer

The distances are $\ln(95/90) \approx 0.05407$ and $\ln(93/90) \approx 0.03279$ , so

\begin{aligned} &\frac{2 \cdot 0.05407 \cdot 0.03279}{0.04/12} \approx 1.0637, \\[0.4em] &\exp(-1.0637) \approx 0.3452 . \end{aligned}

Although both monthly prices are above the barrier, the continuous path crossed it on that interval with probability about $0.345$ . Recording hits only at the dates prices the monthly monitored contract, which is worth at least as much as the continuously monitored knock-out. In the constructed model it misses enough crossings to lower the hit probability from $0.5983$ to about $0.4917$ . Multiplying each path’s payoff by the bridge survival factors, here $1 - 0.3452 \approx 0.6548$ for this interval, prices the continuously monitored contract without bias.

Barrier options and first passage

Key terms

Barrier option

Hitting time

Reflection principle

Knock-in knock-out parity

Brownian bridge crossing probability

Barrier monitoring

Check yourself

How this lesson connects

Builds on