Barrier Options¶
Introduction¶
Barrier options are path-dependent derivatives that become active (knock-in) or expire worthless (knock-out) when the underlying asset price crosses a predetermined barrier level \(H\) during the life of the option. This conditional activation or deactivation makes barrier options cheaper than their vanilla counterparts, since the holder accepts restricted optionality in exchange for a lower premium.
Barrier options are among the most widely traded exotic derivatives in OTC markets, particularly in foreign exchange and structured equity products.
Prerequisites
- Black–Scholes Formula (vanilla option pricing)
- Exotic Options Overview (classification and motivation)
- Reflection Principle (hitting probabilities for Brownian motion)
Learning Objectives
By the end of this section, you will be able to:
- Classify all eight standard barrier option types
- Write the payoff formula for each barrier type
- State the in–out parity relationship
- Identify analytical pricing formulas under GBM assumptions
- Understand discretization bias in barrier option pricing
Barrier Types and Classification¶
A barrier option is characterized by three features: the option type (call or put), the barrier direction (up or down), and the barrier effect (knock-in or knock-out). This produces eight standard types:
| Barrier Type | Direction | Effect | Condition for Activation/Deactivation |
|---|---|---|---|
| Up-and-in call | Up | Knock-in | Option activates if \(\max_{0 \leq t \leq T} S_t \geq H\) |
| Up-and-out call | Up | Knock-out | Option dies if \(\max_{0 \leq t \leq T} S_t \geq H\) |
| Down-and-in call | Down | Knock-in | Option activates if \(\min_{0 \leq t \leq T} S_t \leq H\) |
| Down-and-out call | Down | Knock-out | Option dies if \(\min_{0 \leq t \leq T} S_t \leq H\) |
| Up-and-in put | Up | Knock-in | Option activates if \(\max_{0 \leq t \leq T} S_t \geq H\) |
| Up-and-out put | Up | Knock-out | Option dies if \(\max_{0 \leq t \leq T} S_t \geq H\) |
| Down-and-in put | Down | Knock-in | Option activates if \(\min_{0 \leq t \leq T} S_t \leq H\) |
| Down-and-out put | Down | Knock-out | Option dies if \(\min_{0 \leq t \leq T} S_t \leq H\) |
Payoff Formulas¶
Knock-Out Options¶
A knock-out option pays like a vanilla option provided the barrier is never breached:
For a down-and-out call with barrier \(H < S_0\):
For an up-and-out call with barrier \(H > S_0\):
Knock-In Options¶
A knock-in option pays only if the barrier has been breached:
For a down-and-in call with barrier \(H < S_0\):
In–Out Parity¶
A fundamental relationship connects knock-in and knock-out options of the same type:
Proof sketch. The barrier is either breached or not breached during \([0, T]\). If breached, the knock-in option pays the vanilla payoff and the knock-out option pays zero. If not breached, the knock-out option pays the vanilla payoff and the knock-in option pays zero. In either case, the combined payoff equals the vanilla payoff exactly.
Numerical Example
For \(S_0 = 100\), \(K = 100\), \(T = 1\), \(r = 0.05\), \(\sigma = 0.20\), \(H = 90\) (down barrier):
| Component | Price |
|---|---|
| Vanilla call (Black–Scholes) | \(\approx 10.45\) |
| Down-and-out call | \(\approx 7.93\) |
| Down-and-in call | \(\approx 2.52\) |
| Sum (KI + KO) | \(\approx 10.45\) ✓ |
Analytical Pricing Under GBM¶
Under geometric Brownian motion with continuous barrier monitoring, closed-form solutions exist. The key mathematical tool is the reflection principle for Brownian motion, which provides the joint distribution of \((S_T, \min_{t} S_t)\) or \((S_T, \max_{t} S_t)\).
Down-and-Out Call (H < K, H < S_0)¶
The Rubinstein–Reiner (1991) formula gives:
where:
The terms involving \((H/S_0)^{2\lambda}\) arise from the reflection principle and encode the probability of the barrier being hit.
Discrete vs. Continuous Monitoring¶
In practice, barriers are typically monitored at discrete intervals (daily closes, for example) rather than continuously. This distinction matters significantly for pricing:
Broadie–Glasserman–Kou Correction¶
For a discrete barrier monitored at \(m\) equally-spaced times, the effective continuous barrier is approximately:
where \(\beta = -\zeta(1/2)/\sqrt{2\pi} \approx 0.5826\), \(\zeta\) is the Riemann zeta function, and the sign depends on whether the barrier is above (\(+\)) or below (\(-\)) the current price.
This correction shifts the barrier outward, reflecting the fact that discrete monitoring makes it harder for the barrier to be breached between observation times.
Pricing Implications¶
| Monitoring | Down-and-Out Price | Explanation |
|---|---|---|
| Continuous | Lower | Barrier more easily breached → higher knockout probability |
| Discrete (daily) | Higher | May "skip over" barrier between observation times |
| Difference | Can be 5–15% | Significant for near-barrier situations |
Rebate Features¶
Many barrier options include a rebate \(R\) paid when the barrier is hit (for knock-out options) or at maturity if the barrier is never hit (for knock-in options):
where \(\tau_H\) is the first hitting time of the barrier. The rebate compensates the holder for the loss of optionality when the barrier is triggered.
Summary¶
| Aspect | Description |
|---|---|
| Definition | Options that activate/deactivate at a barrier level \(H\) |
| Eight types | Up/down × in/out × call/put |
| Key relationship | In–out parity: \(V_{\text{KI}} + V_{\text{KO}} = V_{\text{vanilla}}\) |
| Analytical pricing | Available under GBM with continuous monitoring (reflection principle) |
| Discrete monitoring | Broadie–Glasserman–Kou correction adjusts for discrete observation |
| Cost advantage | Barrier options are cheaper than vanilla (restricted optionality) |
Barrier options introduce path dependency through threshold conditions, creating a family of derivatives that trade optionality for reduced premiums and are priced using the reflection principle of Brownian motion.
Exercises¶
Exercise 1. Prove the in-out parity \(V_{\text{knock-in}} + V_{\text{knock-out}} = V_{\text{vanilla}}\) rigorously by showing that the sum of knock-in and knock-out payoffs equals the vanilla payoff for every possible path \(\{S_t\}_{0 \leq t \leq T}\). State the precise partition of the sample space used in the argument.
Solution to Exercise 1
Let \(\omega\) denote any path \(\{S_t(\omega)\}_{0 \leq t \leq T}\). Define two events:
- \(A = \{\omega : \text{barrier is breached}\}\)
- \(A^c = \{\omega : \text{barrier is not breached}\}\)
These form a partition of the sample space: \(A \cup A^c = \Omega\) and \(A \cap A^c = \emptyset\).
The knock-in payoff is \(\Phi_{\text{KI}}(\omega) = (S_T(\omega) - K)^+ \cdot \mathbf{1}_A(\omega)\) and the knock-out payoff is \(\Phi_{\text{KO}}(\omega) = (S_T(\omega) - K)^+ \cdot \mathbf{1}_{A^c}(\omega)\).
For every path \(\omega\):
since \(\mathbf{1}_A(\omega) + \mathbf{1}_{A^c}(\omega) = 1\) for every \(\omega\). This is the vanilla payoff.
Taking risk-neutral expectations and discounting:
Exercise 2. For a down-and-out call with \(S_0 = 100\), \(K = 95\), \(H = 90\), \(T = 1\), \(r = 5\%\), \(\sigma = 25\%\): (a) Use in-out parity to express the down-and-in call price in terms of the vanilla call and the down-and-out call. (b) If the vanilla call price is \(\$14.23\) and the down-and-out call price is \(\$10.71\), find the down-and-in call price. (c) Explain intuitively why the down-and-out call is cheaper than the vanilla call.
Solution to Exercise 2
(a) By in-out parity: \(V_{\text{KI}} + V_{\text{KO}} = V_{\text{vanilla}}\). Therefore:
(b) Substituting the given values:
(c) The down-and-out call is cheaper than the vanilla call because it provides restricted optionality. If the stock price drops to or below the barrier \(H = 90\) at any time before expiry, the knock-out option expires worthless, even if the stock subsequently recovers above the strike \(K = 95\). The holder accepts this risk of losing the option in exchange for a lower premium. The difference \(\$14.23 - \$10.71 = \$3.52\) represents the value of the "insurance" that the vanilla call provides against the barrier-triggering scenario.
Exercise 3. The Rubinstein-Reiner formula for a down-and-out call contains terms involving \((H/S_0)^{2\lambda}\) where \(\lambda = (r + \frac{1}{2}\sigma^2)/\sigma^2\). Explain the mathematical origin of this exponent in terms of the reflection principle applied to drifted Brownian motion. What happens to this factor when \(r = 0\) and \(\sigma^2 = 2\)?
Solution to Exercise 3
The exponent \(2\lambda\) arises from the reflection principle applied to drifted Brownian motion. Under \(\mathbb{Q}\), the log-price follows \(X_t = \log S_t = x + \mu t + \sigma W_t\) where \(\mu = r - \frac{1}{2}\sigma^2\).
The reflection principle applies directly only to driftless Brownian motion. To remove the drift, we perform a Girsanov-type measure change. Define a new measure \(\hat{\mathbb{Q}}\) via the Radon-Nikodym derivative:
Under \(\hat{\mathbb{Q}}\), the process \(X_t - x = \sigma \hat{W}_t\) is a driftless Brownian motion (scaled by \(\sigma\)).
When reflecting a path that hits barrier \(b = \log H\) about that barrier, the reflected path starts from \(2b - x = \log(H^2/S_0)\). The Radon-Nikodym factor evaluated at the reflected starting point introduces a multiplicative correction:
Since \(\mu/\sigma^2 = (r - \frac{1}{2}\sigma^2)/\sigma^2 = r/\sigma^2 - 1/2 = \lambda - 1\), the factor becomes \((H/S_0)^{2\lambda - 2}\), equivalently \((S_0/H)^{-(2\lambda-2)}\).
When \(r = 0\) and \(\sigma^2 = 2\): \(\lambda = (0 + \frac{1}{2}\cdot 2)/2 = \frac{1}{2}\), so \(2\lambda = 1\) and \((H/S_0)^{2\lambda} = (H/S_0)^1 = H/S_0\). The factor \(2\lambda - 2 = -1\), so the reflection correction becomes \((H/S_0)^{-1} = S_0/H\).
Exercise 4. The Broadie-Glasserman-Kou correction for discrete barrier monitoring is \(H_{\text{eff}} = H \cdot e^{\pm \beta \sigma \sqrt{T/m}}\) where \(\beta \approx 0.5826\). (a) For a down-and-out call with \(H = 90\), \(\sigma = 0.20\), \(T = 1\), and \(m = 252\) (daily monitoring), compute \(H_{\text{eff}}\). (b) Explain why the correction shifts the barrier outward (downward for a down barrier). (c) What happens to \(H_{\text{eff}}\) as \(m \to \infty\)?
Solution to Exercise 4
(a) For a down barrier with \(H = 90\), \(\sigma = 0.20\), \(T = 1\), \(m = 252\):
Computing: \(\sqrt{1/252} \approx 0.06299\), so \(0.5826 \times 0.20 \times 0.06299 \approx 0.007340\).
The effective barrier shifts from \(90\) down to approximately \(89.34\).
(b) The correction shifts the barrier outward (downward for a down barrier, upward for an up barrier). This is because discrete monitoring misses barrier crossings that occur between observation times. A path may dip below \(H\) between two consecutive observation dates without being detected. To replicate the continuous-monitoring behavior with discrete observations, we must lower the barrier to catch some of these paths that would have been knocked out under continuous monitoring. Equivalently, the continuous-monitoring price uses a "tighter" effective barrier.
(c) As \(m \to \infty\), \(\Delta t = T/m \to 0\), so \(\sqrt{T/m} \to 0\), and thus \(H_{\text{eff}} \to H \cdot e^0 = H\). The correction vanishes, confirming that infinitely frequent discrete monitoring converges to continuous monitoring.
Exercise 5. A knock-out barrier option with rebate \(R\) pays \(R \cdot e^{-r(\tau_H - t)}\) at the time the barrier is first hit. Explain why the rebate is discounted from the hitting time \(\tau_H\) rather than from maturity \(T\). Write the total payoff formula for a down-and-out call with rebate, carefully distinguishing the two mutually exclusive events.
Solution to Exercise 5
The rebate is discounted from the hitting time \(\tau_H\) rather than from maturity \(T\) because the rebate is paid at the moment the barrier is hit, not at maturity. Since the holder receives the cash amount \(R\) at time \(\tau_H\), its present value at time \(t < \tau_H\) is \(R \cdot e^{-r(\tau_H - t)}\). If we discounted from \(T\), we would be undervaluing the rebate (since \(\tau_H \leq T\), the holder receives the money sooner).
The total payoff for a down-and-out call with rebate, evaluated at time \(0\), is:
The two events \(\{\min_{0 \leq t \leq T} S_t > H\}\) (barrier not breached) and \(\{\min_{0 \leq t \leq T} S_t \leq H\}\) (barrier breached at time \(\tau_H\)) are mutually exclusive and exhaustive. On the first event, the option survives and pays the standard call payoff at maturity. On the second event, the option is knocked out and the holder receives the rebate \(R\) at the hitting time \(\tau_H\).
Exercise 6. Consider an up-and-out call with \(K = 100\) and \(H = 120\). As the barrier \(H\) approaches the strike \(K\) from above, what happens to the option price? As \(H \to \infty\), what does the option price converge to? Sketch the option price as a function of \(H\) for fixed \(S_0 = 100\) and explain the shape of the curve.
Solution to Exercise 6
As \(H \to K^+\) (barrier approaches strike from above): The up-and-out call price approaches zero. For the call to have a positive payoff, we need \(S_T > K\). But for a call that is in the money at expiry, the stock must have risen above \(K\) at some point. If \(H\) is only slightly above \(K\), then any path where \(S_T > K\) almost certainly has \(\max_{0 \leq t \leq T} S_t \geq H\), which triggers knockout. Therefore, the option is nearly always either out of the money or knocked out, and its value vanishes.
As \(H \to \infty\): The barrier is never reached, so the knockout condition \(\max_{0 \leq t \leq T} S_t \geq H\) has probability approaching zero. The up-and-out call converges to the vanilla European call price \(C_{\text{BS}}(S_0, K, T)\).
Shape of the curve: The option price as a function of \(H\) is a monotonically increasing curve. At \(H = K\) (or slightly above), the price is near zero. As \(H\) increases, the price rises and asymptotically approaches \(C_{\text{BS}}\). The curve is concave, rising steeply near \(H = K\) (since small increases in \(H\) significantly reduce knockout probability) and flattening as \(H\) grows large (since the barrier is already rarely reached). For \(S_0 = 100\) and \(K = 100\), the curve starts near zero at \(H \approx 100\) and converges to approximately \(\$10.45\) as \(H \to \infty\).