Barrier option

A barrier option activates or expires according to whether the underlying price reaches a prescribed level during its lifetime. Its state must include whether the barrier has already been reached; the current price alone generally does not determine its remaining payoff.

An up-and-in claim activates when the stock reaches an upper barrier $b>S_0$ during its lifetime. After activation its terminal payoff is the prescribed function $f(S_T)$; without activation it pays zero. Under continuous stock paths the barrier event is described by the displayed running maximum.

For an up-and-in claim in the dividend-free Black-Scholes model, set $\nu =(\rho -{\sigma }^2/2)/\sigma$, $\kappa =(S_0/b{)}^2$ and assume finite absolute payoff expectations. Its initial price equals the risk-neutral pricing value of the displayed European option. To prove this, express the log-price as $\sigma (W_t+\nu t)$ and apply the joint endpoint and maximum law for drifted Brownian motion. Above the logarithmic barrier $a=\log (b/S_0)/\sigma$, use the unrestricted endpoint density. Below it, the density of paths which hit the barrier is $e^{2a\nu }{\varphi }_T(y-2a-\nu T)$. Substitution $z=y-2a$ gives the scaled payoff, the threshold $\kappa b$, and the factor ${\kappa }^{-\nu /\sigma }$. This is an initial-price identity, not a pathwise equality of terminal payoffs.

An up-and-out power claim pays a power of the terminal stock price provided an upper barrier has never been reached. In the Black-Scholes model, set $A=\log (c/S_0)/\sigma$ and $m_p=(\rho +(p-\frac{1}{2}){\sigma }^2)/\sigma$, where $\sigma >0$, $c>S_0$ and $T>0$. The risk-neutral pricing value isHere $\Phi$ is the standard normal distribution function. Weighting the Brownian motion endpoint by $e^{p\sigma W_T-p^2{\sigma }^{2}T/2}$ changes the logarithmic drift to $m_p$; the finite-horizon maximum of Brownian motion with drift then supplies the survival factor.

A down-and-out claim expires worthless if the stock reaches a lower barrier before maturity. A matching down-and-in claim plus a down-and-out claim pays the ordinary terminal claim on every path; this is in-out parity.

A down-and-in claim with lower barrier $b$ pays a specified terminal payoff only if the stock reaches $b$ before maturity. Before activation its value solves a killed-domain pricing problem; at activation its value equals the ordinary terminal-payoff claim value.

For strike $c>b$, write $C(s,c,\tau )$ for the ordinary European call option value. Before activation the down-and-in claim is $I(s,\tau )=(b/s{)}^{\alpha }C(b^2/s,c,\tau )$, where $\alpha =2\rho /{\sigma }^2-1$. Consequently its left-limit option delta at the barrier is $-\alpha C(b,c,\tau )/b-C_s(b,c,\tau )$, whereas after activation it is $C_s(b,c,\tau )$. Their difference is strictly positive for $\tau >0$: differentiating the killed normal transition density at its boundary gives the integral of a strictly positive function against the positive call payoff. The value is continuous, so changing the bank holding finances the stock rebalance.

For initial spot $s>b$ in the Black-Scholes model, put $\nu =(\rho -{\sigma }^2/2)/\sigma$ and $\kappa =(s/b{)}^2$. A down-and-in claim with terminal payoff $f$ has the same initial price as the terminal claim $g(x)=f(x)1_{x\le b}+{\kappa }^{-\nu /\sigma }f(x/\kappa )1_{x>\kappa b}$. Reflection of the log-price endpoint above the lower barrier changes its normal probability density to $e^{2\nu \ell }{\varphi }_T(y-2\ell -\nu T)$, where $\ell =\log (b/s)/\sigma$. Substitution $z=y-2\ell$ gives the payoff identity. This is price equality, not pathwise equality of payments.