Down-and-Out Barrier Call Closed-Form via Brownian Reflection Principle

Scenario

A European down-and-out call option pays $\max(S_T - K, 0)$ at expiration $T$ if and only if the underlying stock price remains strictly above barrier $B$ ($S_t > B$) for all $t \in [0, T]$, where $B < S_0$ and $K \ge B$. Under the risk-neutral measure $\mathbb{Q}$, the asset follows geometric Brownian motion: $dS_t = r S_t dt + \sigma S_t dW_t^{\mathbb{Q}}$

Your Task

1. Transform the stock price process into arithmetic Brownian motion with drift and apply Girsanov's theorem.
2. Apply the Brownian Reflection Principle to derive the joint probability density of the terminal value and running minimum $\min_{0 \le t \le T} S_t$.
3. Prove that the exact closed-form price of the down-and-out call decomposes into a vanilla call and a reflected call: $C_{\text{do}}(S_0, K, B) = C_{\text{BS}}(S_0, K) - \left( \frac{S_0}{B} \right)^{1 - \frac{2r}{\sigma^2}} C_{\text{BS}}\left( \frac{B^2}{S_0}, K \right)$
4. Given $S_0 = 100, K = 100, B = 85, r = 0.05, \sigma = 0.25, T = 1.0$, compute the down-and-out call price and verify that $0 < C_{\text{do}} < C_{\text{BS}}$.

Constraints

• Continuous barrier monitoring.
• $B < S_0$ and $K \ge B$.

Examples For $B = 0$, the reflection term vanishes and $C_{\text{do}} = C_{\text{BS}}$. For $B \to S_0^-$, the knock-out probability approaches 1 and $C_{\text{do}} \to 0$.