The Barrier Claim and Its Pricing Problem
A lower barrier changes a terminal-payoff problem into a path-dependent one: it is not enough that the asset finishes above the strike; it must also avoid the barrier throughout the life of the claim. We combine a drift change for log price with an absorbing-boundary calculation, then verify the resulting price by applying Ito's formula to a stopped discounted process.
Work on a filtered probability space \((\Omega,\mathcal{F},(\mathcal{F}_t)_{t\ge0},\mathbb{P})\) carrying a Brownian motion. Let \(\mathbb{Q}\) be the risk-neutral measure on the finite horizon \([0,T]\), and let \(B_t^{\mathbb{Q}}\) denote the Brownian motion under \(\mathbb{Q}\). Assume constant interest rate \(r\), volatility \(\sigma>0\), and geometric Brownian motion \(dS_t=rS_t\,dt+\sigma S_t\,dB_t^{\mathbb{Q}}\), with \(S_0=x\). This equation has the explicit strong solution \(S_t=x\exp((r-\tfrac12\sigma^2)t+\sigma B_t^{\mathbb{Q}})\). Here “strong” means that the solution is constructed using the given Brownian motion; the formula also makes its law explicit.
Fix a barrier \(L>0\), a strike \(K>L\), and \(x>L\). The claim pays one unit at time \(T\) if the price has remained strictly above \(L\) and finishes strictly above \(K\); otherwise it pays zero. Define \(\tau_L=\inf\{s\ge0:S_s\le L\}\), with \(\inf\varnothing=\infty\), and let \(\tau=T-t\) be the time remaining. Its time-\(t\) value when \(S_t=x>L\) is \(u(t,x)=e^{-r\tau}\mathbb{Q}_x(\tau_L>T-t,\ S_{T-t}>K)\). The strict inequalities do not affect the terminal probability at \(K\), since the lognormal terminal distribution has no atom there. At the barrier the value is zero.
Remove the Log-Price Drift
Under \(\mathbb{Q}\), the log process has drift \(\nu\). On the interval of length \(\tau\), put \(c=\nu/\sigma\), and define a probability measure \(\mathbb{P}^0\) by \(d\mathbb{P}^0/d\mathbb{Q}=\exp(-cB_\tau^{\mathbb{Q}}-\tfrac12c^2\tau)\). This is a genuine density: its expectation under \(\mathbb{Q}\) is one by the normal moment-generating function. It is also the stochastic exponential with deterministic integrand \(-c\). Novikov's condition, \(\mathbb{E}^{\mathbb{Q}}\exp(\tfrac12\int_0^\tau c^2\,ds)<\infty\), holds because the expectation is \(e^{c^2\tau/2}<\infty\). Girsanov's theorem therefore says that \(B_s^0=B_s^{\mathbb{Q}}+cs\) is Brownian motion under \(\mathbb{P}^0\), for \(0\le s\le\tau\). In particular, \(Y_s=y+\sigma B_s^0\) under this auxiliary measure.
The inverse likelihood, expressed in terms of \(B^0\), is \(d\mathbb{Q}/d\mathbb{P}^0=\exp(cB_\tau^0-\tfrac12c^2\tau)\). The stochastic integrals in these measure-change formulas are Ito integrals. In general, on a filtered space under a specified measure, \(\int_0^t H_s\,dB_s\) is defined for progressively measurable, adapted \(H\) with \(\mathbb{E}\int_0^T H_s^2\,ds<\infty\); here the integrands are deterministic constants and meet this condition. The measures \(\mathbb{P}^0\) and \(\mathbb{Q}\) have different roles: \(\mathbb{Q}\) is the pricing measure, while \(\mathbb{P}^0\) is a temporary measure used to compute the probability.
For driftless Brownian motion killed when it reaches zero, the transition density from \(y>0\) to \(z>0\) over time \(\tau>0\) is \(q_\tau(y,z)=\phi_{\sigma^2\tau}(z-y)-\phi_{\sigma^2\tau}(z+y)\), where \(\phi_v(w)=(2\pi v)^{-1/2}\exp(-w^2/(2v))\). The first term is the unrestricted Gaussian density; reflecting paths that hit zero gives the subtracted density. Thus the difference counts only paths that stay above the absorbing boundary. The likelihood for ending at \(z\) is \(\exp(c(z-y)/\sigma-\tfrac12c^2\tau)\), since \(B_\tau^0=(z-y)/\sigma\).
Closed-Form Price and Its Derivation
On the survival event the terminal log distance satisfies \(z>0\); the digital payoff also requires \(z>k\). Integrating the killed density with the likelihood gives the risk-neutral probability. To evaluate the two integrals, use the Gaussian tail identity \(\int_m^\infty e^{\lambda z}\phi_v(z-a)\,dz=e^{\lambda a+\lambda^2v/2}\Phi((a+\lambda v-m)/\sqrt v)\), obtained by completing the square in the exponent. Here \(\lambda=\nu/\sigma^2\), \(v=\sigma^2\tau\), and \(m=k\). Substitution yields the following price.
Proof. Under \(\mathbb{P}^0\), weight the killed density by the inverse likelihood and integrate over \(z>k\). The first Gaussian term has \(a=y\), and the second has \(a=-y\). Completing the square in each term gives \(e^{\lambda y+\lambda^2v/2}\Phi((y+\lambda v-k)/\sqrt v)\) and \(e^{-\lambda y+\lambda^2v/2}\Phi((-y+\lambda v-k)/\sqrt v)\), respectively. The common likelihood factor outside the integrals is \(e^{-\lambda y-\lambda^2v/2}\). Multiplying it through leaves coefficient \(1\) on the first tail and \(e^{-2\lambda y}\) on the second. Since \(\lambda v=\nu\tau\), \(\lambda y=\nu y/\sigma^2\), and \(v=\sigma^2\tau\), the probability is the bracketed expression. Discounting by \(e^{-r\tau}\) proves the formula. At \(y=0\), the two terms agree, giving the absorbing boundary value zero. \(\square\)
Worked Example: Zero Log Drift and Strike at the Starting Price
Take \(\sigma=1\), \(r=\tfrac12\), \(\tau=1\), \(x=eL\), and \(K=eL\). Then \(\nu=r-\tfrac12\sigma^2=0\), \(y=\log(e)=1\), and \(k=\log(e)=1\). The first normal argument is \((1+0-1)/1=0\); the second is \((-1+0-1)/1=-2\). The exponential factor is \(e^0=1\). Therefore $$ u=e^{-1/2}\bigl(\Phi(0)-\Phi(-2)\bigr). $$
Check: Under the driftless measure, integrating the reflected density over \(z>1\) gives \(\int_1^\infty[\phi_1(z-1)-\phi_1(z+1)]\,dz=\Phi(0)-\Phi(-2)\). This recovers the same survival-and-terminal-tail probability before discounting.
Worked Example: Nonzero Drift and a Distinct Strike
Take \(\sigma=1\), \(\nu=1\), \(\tau=1\), \(x=e^2L\), and \(K=eL\). The corresponding interest rate is \(r=\nu+\tfrac12\sigma^2=\tfrac32\). The log distances are \(y=\log(e^2)=2\) and \(k=\log(e)=1\). Thus the first argument is \((2+1-1)/1=2\), the second is \((-2+1-1)/1=-2\), and the correction factor is \(e^{-2(1)(2)/1}=e^{-4}\). The price is $$ u=e^{-3/2}\bigl(\Phi(2)-e^{-4}\Phi(-2)\bigr). $$
Check: The Gaussian tilt shifts the two unrestricted means from \(y=2\) and \(-y=-2\) to \(y+\nu\tau=3\) and \(-y+\nu\tau=-1\). Their tails above \(k=1\) have arguments \((3-1)=2\) and \((-1-1)=-2\). The second tail carries the reflected-path coefficient \(e^{-4}\), as in the formula.
Worked Example: A Larger Initial Distance from the Barrier
Take \(\sigma=1\), \(r=\tfrac12\), \(\tau=1\), \(x=4L\), and \(K=2L\). Then \(\nu=0\), \(y=\log4=2\log2\), and \(k=\log2\). The first argument is \(y-k=\log2\), the second is \(-y-k=-3\log2\), and the correction factor is \(1\). Hence $$ u=e^{-1/2}\bigl(\Phi(\log2)-\Phi(-3\log2)\bigr). $$
Check: The killed-density integral has lower endpoint \(z=k=\log2\); the direct Gaussian tails are \(\int_k^\infty\phi_1(z-y)\,dz=\Phi(y-k)=\Phi(\log2)\) and \(\int_k^\infty\phi_1(z+y)\,dz=\Phi(-y-k)=\Phi(-3\log2)\). Their difference is the undiscounted survival probability.
The Absorbing-Boundary PDE
The price satisfies the Black–Scholes pricing equation on \(t<T\), \(x>L\), with an absorbing boundary at \(L\): $$ u_t(t,x)+rxu_x(t,x)+\tfrac12\sigma^2x^2u_{xx}(t,x)-ru(t,x)=0, \qquad u(t,L)=0. $$ $$ u(T,x)=\mathbf{1}_{\{x>K\}}\quad(x>L). $$ The terminal condition is discontinuous at \(K\); the PDE is asserted classically only for \(t<T\). To see why the formula solves the equation, its probabilistic representation can be written as the killed-density integral just derived. For positive remaining time, the Gaussian density and its time and space derivatives are integrable on \(z>k\); differentiating under the integral is justified on compact subsets with \(\tau>0\). The Gaussian density satisfies the heat equation, so the resulting function satisfies the backward equation for the drifted log process. Equivalently, under \(x=Le^y\), the differential operator in the PDE becomes \(\partial_t+\nu\partial_y+\tfrac12\sigma^2\partial_{yy}-r\). The likelihood-weighted killed-density representation solves this equation, vanishes at \(y=0\), and tends to \(\mathbf{1}_{\{y>k\}}\) as \(\tau\downarrow0\) for \(y\ne k\). At \(y=k\), the first normal CDF tends to \(1/2\) and the second tends to zero: this single point has zero terminal probability and does not change the price.
Verification by Stopped Ito Calculus
Here is a direct verification of the pricing representation. On a space carrying the risk-neutral Brownian motion, set \(X_s=x\exp(\nu s+\sigma B_s^{\mathbb{Q}})\), and let \(\tau_L=\inf\{s\ge0:X_s\le L\}\). Fix \(t<T\), take \(X_0=x>L\), and write \(\tau=T-t\). For integers \(n>x\), stop at the upper exit \(\tau_n=\inf\{s\ge0:X_s\ge n\}\) and at \(T-t-\varepsilon\), where \(0<\varepsilon<T-t\). On the resulting compact time-space region, \(u\) is smooth and its derivatives are bounded.
Apply Ito's formula to \(f(s,z)=e^{-rs}u(t+s,z)\), \(z=X_s\). Its partial derivatives are \(f_s=e^{-rs}(u_t-ru)\), \(f_z=e^{-rs}u_x\), and \(f_{zz}=e^{-rs}u_{xx}\), with the derivatives of \(u\) evaluated at \((t+s,z)\). Since \(dX_s=rX_s\,ds+\sigma X_s\,dB_s^{\mathbb{Q}}\), \(dX_s\,dX_s=\sigma^2X_s^2\,ds\), \(dB_s^{\mathbb{Q}}\,ds=0\), and \(ds\,ds=0\), Ito's formula gives $$ df(s,X_s)=e^{-rs}\left(u_t+ rX_su_x+\tfrac12\sigma^2X_s^2u_{xx}-ru\right)ds +e^{-rs}\sigma X_su_x\,dB_s^{\mathbb{Q}}. $$ The PDE makes the drift term zero. The stopped stochastic integral is a true martingale: on the stopped compact region its integrand is bounded, hence has finite expected squared integral. Its conditional expectation at time zero is its initial value, by the Ito integral martingale property.
At the lower boundary the stopped price is zero. Let \(n\to\infty\) and then \(\varepsilon\downarrow0\). Continuous paths are bounded on the finite interval, so the upper-exit stops eventually do not occur, almost surely. Also, the price and all discounted stopped values are bounded in absolute value by \(e^{|r|T}\), since \(0\le u(s,z)\le e^{-r(T-s)}\le e^{|r|T}\) for \(s\le T\); the additional discount factor is bounded by \(e^{|r|T}\) as well, so a valid common bound for their product is \(e^{2|r|T}\). Dominated convergence therefore applies, including when \(r<0\). As \(\varepsilon\downarrow0\), continuity of paths and the terminal condition give the payoff on \(\{\tau_L>\tau,\ X_\tau>K\}\), and zero otherwise. The event \(X_\tau=K\) has probability zero. We obtain $$ u(t,x)=e^{-r\tau}\mathbb{Q}_x(\tau_L>\tau,\ X_\tau>K), $$ which verifies the claimed risk-neutral price. The stopped discounted process is a martingale on each localized interval; no assertion that an arbitrary unstopped local martingale is a true martingale is needed here. \(\square\)
Common Pitfalls
A common error is to remove the drift from the log process but forget to change the payoff probability by the likelihood ratio. Under the auxiliary measure, the killed density alone gives a driftless probability, not the original risk-neutral probability. The missing factor is \(\exp(cB_\tau^0-\tfrac12c^2\tau)\); omitting it incorrectly removes the drift dependence from the price.
Another error is to treat the barrier as a terminal-only condition. The unrestricted terminal Gaussian density gives the first CDF term, but it also counts paths that crossed the barrier and later returned above it. Subtracting the reflected density is what imposes absorption. Finally, boundedness by one is not available for discounted values when \(r<0\). The bound \(e^{|r|T}\) for \(u\), and \(e^{2|r|T}\) for the discounted stopped quantities used above, is valid for either sign of \(r\).
Check Your Understanding
Use the definitions and hypotheses stated above; in particular, the pricing formula assumes \(x>L\), \(K>L\), \(\sigma>0\), and \(t<T\).
- For \(\sigma=2\) and \(r=2\), calculate \(\nu\). Does the log-price drift vanish?
- Explain why the auxiliary-measure integral has lower endpoint \(k\), rather than zero, when \(K>L\).
- At the barrier \(x=L\), substitute \(y=0\) into the two CDF terms and verify the boundary value.
- For \(r=-1\) and \(T-t=1\), give a finite bound for \(u(t,x)\) that does not assume nonnegative interest rates.
- State which term in the stopped Ito formula is the quadratic-variation correction and identify the PDE that makes the finite-variation term vanish.