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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.

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.