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Let ${\varphi }_T(y)=(2\pi T{)}^{-1/2}{e}^{-y^2/(2T)}$ be the normal distribution density. The Brownian reflection principle gives the killed endpoint densityIndeed, reflection after the first hit of $a$ maps paths ending below $a$ that have hit the barrier to paths ending at $2a-y$; symmetry gives ${\varphi }_T(2a-y)={\varphi }_T(y-2a)$.

By the Girsanov theorem, weighting the path law by $\exp (\nu W_T-{\nu }^{2}T/2)$ makes the coordinate process a Brownian motion with drift $\nu$. Completing the two squares yieldsIntegration up to $x\le a$ proves the joint endpoint and maximum law for drifted Brownian motion:Here $\Phi$ is the standard normal distribution function. Setting $x=a$ also gives a continuous barrier distribution, so equality at the barrier has zero probability.

For the up-and-in option, use the risk-neutral measure of the Black-Scholes model with interest rate $\rho$ and spot volatility $\sigma >0$. ThenAbove $a$, the barrier has certainly been reached. Below $a$, subtracting the killed density from the unrestricted endpoint density leaves $e^{2a\nu }{\varphi }_T(y-2a-\nu T)$. Thus the discounted expected value of its payoff isIn the second integral put $z=y-2a$, and defineThen $S_0{e}^{\sigma y}=S_0{e}^{\sigma z}/\kappa$, $z\le -a$ means $S_0{e}^{\sigma z}\le \kappa b$, and $e^{2a\nu }={\kappa }^{-\nu /\sigma }$. Both integrals now use the unrestricted terminal stock distribution. This proves that the reflected European payoff for an up-and-in option isTheir initial risk-neutral pricing values agree for payoffs with finite absolute expectations in these integrals. The identity concerns prices; the two terminal payoffs need not agree along individual paths.

Let ${\varphi }_t(y)=(2\pi t{)}^{-1/2}{e}^{-y^2/(2t)}$ be the centered normal probability density with variance $t$, and take $t>0$. For a zero-drift Brownian motion $B$, reflect the path after its first hit of $a>0$. The Brownian reflection principle identifies endpoints $y<a$ on paths that have hit $a$ with endpoints $2a-y>a$. Thus the endpoint density on paths staying below $a$ is ${\varphi }_t(y)-{\varphi }_t(2a-y)$ for $y<a$.

To introduce drift $\nu$, use the Girsanov theorem with likelihood $e^{\nu B_t-{\nu }^{2}t/2}$. The coordinate path then has the law of $W_s^{\nu }=B_s+\nu s$. Since the likelihood depends only on the endpoint, the joint event in question has probabilityCompleting the two squares givesConsequently the joint endpoint and maximum law for drifted Brownian motion isThe endpoint and the maximum have no relevant atom at the boundary, so the same integral includes $x=a$.

For the down-and-in claim, use non-dividend-paying Black-Scholes model dynamics under the risk-neutral measure. Put $T=t_0$ andLet ${\tau }_b$ be the first time the stock reaches $b$. If $Y_T\le \ell$, continuity ensures ${\tau }_b\le T$, so these terminal values contribute the payoff $f(S_T)1_{S_T\le b}$. For $y>\ell$, lower-barrier reflection followed by the same exponential tilt gives the endpoint density on paths that have hit the barrier:Indeed at zero drift the reflected density is ${\varphi }_T(y-2\ell )$, and multiplying by $e^{\nu y-{\nu }^{2}T/2}$ gives the displayed expression.

Thus the remaining, undiscounted payoff expectation isSet $z=y-2\ell$ andThen $e^{2\nu \ell }={\kappa }^{-\nu /\sigma }$, $S_0{e}^{\sigma y}=S_0{e}^{\sigma z}/\kappa$, and $z>-\ell$ is equivalent to $S_0{e}^{\sigma z}>\kappa b$. The integration variable $z$ has the ordinary terminal log-price density. Adding the first contribution therefore proves the static terminal-payoff representation of a down-and-in claim:This holds for nonnegative measurable payoffs, and for signed payoffs whenever the expectations are absolutely integrable. It is an equality of prices, not equality of pathwise terminal payments.

For the call, write $C(s,c,\tau )$ for the ordinary European call option value, with remaining lifetime $\tau >0$ and strike $c>b$. Before activation take current spot $s>b$. The first term in $g$ is zero because the call pays nothing below $b$. The second term reduces to ${\kappa }^{-\nu /\sigma -1}(x-\kappa c{)}^{+}$, since $\kappa c>\kappa b$. The homogeneity $C(s,\kappa c,\tau )=\kappa C(s/\kappa ,c,\tau )$ therefore gives the inactive down-and-in claim valueAt $s=b$ its value is $C(b,c,\tau )$, as required by activation. Its pre-activation option delta, however, has boundary limitwhereas the active contract's delta hedge holds $C_s(b,c,\tau )$ shares. The stock holding consequently jumps byThis jump is strictly positive, even for negative interest rates. To verify that there is no exceptional cancellation, put $z=\log (s/b)/\sigma$ and $y=\log (S_T/b)/\sigma$. The density for log-price paths killed at the lower barrier isThe down-and-out claim is the ordinary call minus $I$. Differentiate its value at $z=0+$: the bracket vanishes there and its derivative is $2y{\varphi }_{\tau }(y)/\tau$. Consequently its derivative with respect to the spot at $b$ isThis derivative is exactly the displayed delta jump. Thus on reaching the barrier before maturity the number of shares changes discontinuously. The portfolio value remains continuous; the bank position changes to finance the share adjustment. Hitting precisely at maturity has probability zero and does not alter this conclusion. This is the delta jump at activation of a down-and-in call.

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.