= Solution
Let $\phi_t(y)=(2\pi t)^{-1/2}e^{-y^2/(2t)}$ 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 $\phi_t(y)-\phi_t(2a-y)$ for $y<a$.
To introduce drift $\nu$, use the <Girsanov theorem> with likelihood $e^{\nu B_t-\nu^2t/2}$. 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 probability
$$
\int_{-\infty}^x e^{\nu y-\nu^2t/2}\bigl[\phi_t(y)-\phi_t(2a-y)\bigr]dy.
$$
Completing the two squares gives
$$
e^{\nu y-\nu^2t/2}\phi_t(y)=\phi_t(y-\nu t),\qquad e^{\nu y-\nu^2t/2}\phi_t(2a-y)=e^{2a\nu}\phi_t(y-2a-\nu t).
$$
Consequently the <joint endpoint and maximum law for drifted Brownian motion> is
$$
\boxed{\mathbb P(W_t^\nu\le x,M_t^\nu<a)=\Phi\left(\frac{x-\nu t}{\sqrt t}\right)-e^{2a\nu}\Phi\left(\frac{x-2a-\nu t}{\sqrt t}\right),\quad x\le a.}
$$
The 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 $T=t_0$ and
$$
Y_t=\frac1\sigma\log(S_t/S_0)=W_t^Q+\nu t,\qquad\nu=\frac{\rho-\sigma^2/2}{\sigma},\qquad\ell=\frac1\sigma\log(b/S_0)<0.
$$
Let $\tau_b$ 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 $f(S_T)\mathbf1_{S_T\le b}$. For $y>\ell$, lower-barrier reflection followed by the same exponential tilt gives the endpoint density on paths that have hit the barrier:
$$
\mathbb P_Q(Y_T\in dy,\tau_b\le T)/dy=e^{2\nu\ell}\phi_T(y-2\ell-\nu T).
$$
Indeed at zero drift the reflected density is $\phi_T(y-2\ell)$, and multiplying by $e^{\nu y-\nu^2T/2}$ gives the displayed expression.
Thus the remaining, undiscounted payoff <expectation> is
$$
e^{2\nu\ell}\int_\ell^\infty f(S_0e^{\sigma y})\phi_T(y-2\ell-\nu T)dy.
$$
Set $z=y-2\ell$ and
$$
\boxed{\kappa=(S_0/b)^2>1,\qquad\nu=(\rho-\sigma^2/2)/\sigma.}
$$
Then $e^{2\nu\ell}=\kappa^{-\nu/\sigma}$, $S_0e^{\sigma y}=S_0e^{\sigma z}/\kappa$, and $z>-\ell$ is equivalent to $S_0e^{\sigma z}>\kappa b$. 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>:
$$
\boxed{V_0=e^{-\rho T}\mathbb E_Q[g(S_T)],\qquad g(x)=f(x)\mathbf1_{x\le b}+\kappa^{-\nu/\sigma}f(x/\kappa)\mathbf1_{x>\kappa b}.}
$$
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> value
$$
I(s,\tau)=\left(\frac bs\right)^\alpha C(b^2/s,c,\tau),\qquad\alpha=\frac{2\nu}{\sigma}=\frac{2\rho}{\sigma^2}-1.
$$
At $s=b$ its value is $C(b,c,\tau)$, as required by activation. Its pre-activation <option delta>, however, has boundary limit
$$
I_s(b+,\tau)=-\frac\alpha bC(b,c,\tau)-C_s(b,c,\tau),
$$
whereas the active contract's <delta hedge> holds $C_s(b,c,\tau)$ shares. The <stock> holding consequently jumps by
$$
\boxed{\Delta_{\rm after}-\Delta_{\rm before}=2C_s(b,c,\tau)+\frac\alpha bC(b,c,\tau).}
$$
This 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 is
$$
e^{\nu(y-z)-\nu^2\tau/2}\bigl[\phi_\tau(y-z)-\phi_\tau(y+z)\bigr],\qquad z,y>0.
$$
The <down-and-out claim> is the ordinary call minus $I$. Differentiate its value at $z=0+$: the bracket vanishes there and its derivative is $2y\phi_\tau(y)/\tau$. Consequently its derivative with respect to the spot at $b$ is
$$
\frac{e^{-\rho\tau-\nu^2\tau/2}}{\sigma b}\int_{\log(c/b)/\sigma}^{\infty}(be^{\sigma y}-c)e^{\nu y}\frac{2y}{\tau}\phi_\tau(y)dy>0.
$$
This 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>.
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