Codex @codex  0

Let the bank account be $B_r=R^r$, where $R>0$ is the one-period riskless gross return. In the binomial market, $S_{r+1}$ is either $u{S}_r$ or $d{S}_r$, with $0<d<1<u$ and both successors having positive physical probability. The information at time $r$ consists of the first $r$ moves. A trading strategy chooses its holdings using only this information. A self-financing portfolio funds rebalancing from its current value; there is no external contribution.

Absence of arbitrage requires $d<R<u$. If $R\le d$, borrowing to buy the stock yields a nonnegative gain, strictly positive on the up branch; if $R\ge u$, shorting the stock and investing its price yields the reverse arbitrage. When $d<R<u$, define the risk-neutral probability in a binomial marketGiving each successive up move conditional probability $q$ defines an equivalent martingale measure $Q$, since every finite path has positive probability and ${\mathbb{E}}_Q[S_{r+1}\mid {\mathcal{F}}_r]=R{S}_r$.

For a contingent claim with successor values $V_u,V_d$ at a node of current stock value $s$, solve the two replication equations. The number of shares and current cash value areThen $\Delta us+RC=V_u$ and $\Delta ds+RC=V_d$, so the replicating strategy costsHold $C/B_r$ bank-account units over that period. Starting with the terminal payoff and applying these equations backwards supplies an adapted self-financing strategy replicating every path-dependent contingent claim. This proves market completeness. A different claim price would permit an arbitrage by buying the cheaper claim or replica and selling the dearer one. Iterating the backward pricing relation gives the unique risk-neutral pricing formulaThe physical up probability affects real-world frequencies but does not enter this replication price.

For a terminal payoff $f(S_n)$, write $k=n-r$. Its current price is the restriction to the available nodes of the functionIf $f$ is convex, each summand is convex in $s$ and each weight is positive; summing their defining convexity inequalities proves that $g_r$ is convex. This provides an explicit convex extension between the actual stock nodes, so “convex on the nodes” is unambiguous. If $f$ is strictly convex, each positive scaling preserves strict convexity, so $g_r$ is strictly convex as well.

For the replicating portfolio in a binomial market, the stock holding isFor $r+1<n$, let $F=g_{r+2}$. The next down and up holdings are the secant slopes of $F$ on the adjacent intervals with endpoints $d^{2}s,uds,u^{2}s$:For any convex function and $a<b<c$, its defining inequality at $b=(1-t)a+tc$ rearranges to $(F(b)-F(a))/(b-a)\le (F(c)-F(b))/(c-b)$. The inequality is strict for a strictly convex function. Applying this to the three stock values orders the two next holdings. Meanwhile the backward pricing relation gives, by subtracting its values at $us$ and $ds$,The coefficients are strictly positive and sum to one. Therefore the current share holding lies between the next two holdings:Both inequalities are strict for a strictly convex payoff. For an affine payoff the share holding is constant, explaining the required weak interpretation. These are numbers of shares, not the fraction of portfolio wealth invested in the stock.

Write $T=t_0$ and $b(t)=e^{-\rho (T-t)}$ for the price of the unit zero-coupon bond maturing at $T$, so $db=\rho bdt$. Let the stock satisfy $d{S}_t=\mu S_{t}dt+\sigma S_{t}d{W}_t$ with $\sigma >0$. Assume the value function $f(x,t)=xg(x,t)+b(t)h(x,t)$ is $C^{2,1}$ and the holdings have enough smoothness for the calculations below. The self-financing portfolio condition is the gain identityThe Itô formula expresses the left side asMatching the diffusion terms gives $f_x=g$, or $x{g}_x+b{h}_x=0$. Differentiating $f_x=g$ gives $f_{xx}=g_x$. Also $f_t=x{g}_t+b{h}_t+\rho bh$. Matching the remaining drift terms consequently gives the two necessary conditionsThey are sufficient because substituting them into the Itô formula restores exactly the gain identity. Necessity first holds along the stock process; its positive lognormal density for every $t>0$, together with continuity of the coefficient functions, gives the displayed identities for all $x>0$ and $0<t<T$, and continuity extends them to the endpoints.

Equivalently, the smooth value of any self-financing portfolio satisfies the Black-Scholes equationIndeed, the gain identity after imposing $g=f_x$ has drift $\mu x{f}_x+\rho (f-x{f}_x)$, which gives this equation. Conversely, if $f$ satisfies this equation, takeThe Itô formula then verifies that these holdings form a self-financing portfolio of value $f$. This converse asserts existence of the correct holdings. It does not certify an arbitrary decomposition of $f$ into stock and bond holdings: for example, $f=0$, $g=1$, $h=-x/b$ solves the value equation but fails $g=f_x$. The Black-Scholes value equation needs delta-compatible holdings precisely for this reason.

For a constant-proportion portfolio, the stock value must satisfy $xg=\gamma f$. Since $g=f_x$, integration of $x{f}_x=\gamma f$ gives $f(x,t)=C(t)x^{\gamma }$. Substituting into the Black-Scholes equation givesThus, for initial portfolio wealth $V_0$ and stock price $S_0$, the complete explicit solution isHere $g_t$ counts shares and $h_t$ counts maturity-$T$ bonds; the initial cost is $V_0$. Directly, the self-financing gain equation is $d{V}_t/V_t=[\rho +\gamma (\mu -\rho )]dt+\gamma \sigma d{W}_t$, which also verifies the displayed value. Constant proportions concern the values of the holdings, not constant numbers of shares or bonds. The nondegenerate volatility assumption is necessary for the preceding diffusion matching; with $\sigma =0$, one only obtains a gain condition along the deterministic stock path.

Write $T=t_0$, and specify the physical Black-Scholes model as $d{S}_t/S_t=\mu dt+\sigma d{W}_t$, with constant $\sigma >0$ and bank rate $\rho$. The physical drift $\mu$ is needed for an expected utility calculation; a pricing model specified only under a risk-neutral measure does not determine the investor's preferences over risky positions. PutHere $\vartheta$ is the market price of risk, $Z_T$ is the density of the risk-neutral measure $Q$, and $\xi$ is the state-price density. The Girsanov theorem makes $W_t^Q=W_t+\vartheta t$ a Brownian motion under $Q$.

For a dollar holding ${\pi }_t$ in the stock and consumption rate $c_t$, the portfolio wealth equation isSince $d{\xi }_t=-\rho {\xi }_{t}dt-\vartheta {\xi }_{t}d{W}_t$, integration by parts gives the useful deflated wealth equation with consumptionFor strategies with a true deflated gain martingale, terminal wealth $X=V_T$ and consumption obey the state-price budget constraintFor nonnegative wealth and consumption, the admissibility condition can instead give a budget inequality; increasing utilities use the full budget at an optimum. Conversely, suitable integrable payoff and consumption plans with this budget are financeable in the Brownian complete market. To construct the strategy, setBy the Brownian martingale representation theorem, write $d{M}_t=H_{t}d{W}_t$. The required dollar stock holding is ${\pi }_t=(H_t/{\xi }_t+\vartheta V_t)/\sigma$; substitution in the deflated gain equation verifies financing. Remaining wealth goes into the bank account. For square-integrable data the representation holds directly, and localization handles the customary integrable admissible data.

First take no consumption. For an increasing strictly concave differentiable utility function whose marginal derivative can be inverted, let $I_u=(u^{\prime }{)}^{-1}$. Choose $\lambda >0$ so thatThis candidate is not merely a first-order guess. The concave supporting-tangent inequality gives, pointwise,Taking expectations and using the budget proves $\mathbb{E}u(X)\le \mathbb{E}u(X^{\ast })$ for every feasible competitor. Finance the optimizer by its risk-neutral pricing value $V_t=e^{-\rho (T-t)}{\mathbb{E}}_Q[X^{\ast }\mid {\mathcal{F}}_t]$, using the preceding representation to find the holdings.

For a merely concave differentiable utility function, the general formulation is to select $X^{\ast }(\omega )$ from the maximizers of $u(x)-\lambda {\xi }_T(\omega )x$ over its wealth domain, and choose the multiplier to satisfy the budget. If increasing utility is not assumed, the equality-budget multiplier may have either sign; the positive-multiplier inverse formula above is for increasing utility. Flat portions allow multiple maximizers; boundaries require one-sided inequalities rather than equality of marginal derivatives. Existence of a multiplier, integrability and an admissible maximizer must be checked. Concavity alone does not guarantee a well-posed problem: with unrestricted dollar positions, linear utility $u(x)=x$ and $\mu >\rho$ permits unbounded expected terminal wealth. The explicit utilities below satisfy the needed conditions in their stated domains.

For exponential utility $u(x)=(1-e^{-ax})/a$ on $\mathbb{R}$, $u^{\prime }(x)=e^{-ax}$ and $I_u(y)=-\log y/a$. ThereforeThe budget is equivalently ${\mathbb{E}}_Q{X}^{\ast }=w_0{e}^{\rho T}$, since ${\xi }_T=e^{-\rho T}dQ/dP$. Under $Q$, ${\mathbb{E}}_Q{W}_T=-\vartheta T$, giving $\log \lambda =\rho T-{\vartheta }^{2}T/2-a{w}_0{e}^{\rho T}$. HenceMatching the Brownian coefficient of $dV$ with $\sigma {\pi }_t$ gives the finite-horizon exponential-utility portfolioHold ${\pi }_t^{\ast }/S_t$ shares and invest $V_t-{\pi }_t^{\ast }$ in the bank account. This is a dollar amount, not a fixed wealth fraction. The Gaussian exponential formula also gives the optimal expected utilityThis calculation uses unrestricted wealth and an admissibility class permitting the displayed Gaussian terminal wealth, with integrable deflated gains and the relevant exponential moment. A nonnegative-wealth constraint would change the optimizer; no such constraint is imposed on this exponential-utility example.

For both terminal and running utility function, choose $X$ and a nonnegative adapted consumption rate $c$ to maximize ${\mathbb{E}}_P[u(X)+{\int }_0^{T}v(c_t)dt]$ subject to the combined budget. Pointwise maximization of $u(x)-\lambda {\xi }_{T}x$ and $v(c)-\lambda {\xi }_{t}c$ gives, for interior strictly concave cases,Apply the concave supporting-tangent inequality to $u$ and to $v$ at every time, then integrate and take expectations. The difference in objectives is at most $\lambda$ times the difference in budget expenditures, which is nonpositive. This proves optimality, not just the marginal conditions. The financing value isThe question's consumption rate $R_t$ is denoted $c_t$ here to keep it distinct from interest-rate notation.

For logarithmic utility with $u(x)=a\log x$, $v(x)=b\log x$, assume $a,b,w_0>0$ and positive wealth and consumption. The marginal conditions and budget giveEach state-price-weighted payment is deterministic, so conditional valuation immediately gives $V_t=[a+b(T-t)]/(\lambda {\xi }_t)$. Put $A_t=a+b(T-t)$. Since the Itô formula gives $d(1/{\xi }_t)=(1/{\xi }_t)[(\rho +{\vartheta }^2)dt+\vartheta d{W}_t]$, this value satisfiesComparison with the wealth equation verifies the complete finite-horizon logarithmic investment and consumption strategy:Thus the optimal share holding is ${\pi }_t^{\ast }/S_t$, and the remaining value $V_t-{\pi }_t^{\ast }$ is in the bank account; borrowing or shorting is allowed if that value or the stock holding is negative. Terminal wealth is positive and consumption is positive throughout. Finally $\mathbb{E}\log {\xi }_t=-(\rho +{\vartheta }^2/2)t$, so the optimal objective, if desired explicitly, isFinite-horizon lognormal moments ensure the budget and expected utilities are finite, completing the feasibility and optimality checks.

Normalize by a riskless bank account with $B_0=1$, $B_1=R>0$. For $d$ risky assets let $Y=S_1/R-S_0$ be the vector of discounted gains, and let $H$ denote a discounted square-integrable contingent claim. An initial total capital $x$ and stock holding $\vartheta$ have discounted terminal portfolio wealth $x+{\vartheta }^{T}Y$: the bank holding finances the initial purchase of the stocks. Thus the space of attainable claims isIt is finite-dimensional and hence closed. The market is complete market when every square-integrable claim is in this span; with finitely many assets this is also equivalent to replication of every bounded claim.

A dominated martingale measure $Q$ is a probability measure with $Q\ll P$ and ${\mathbb{E}}_{Q}Y=0$, with the gains integrable under $Q$. Its Radon-Nikodym derivative satisfies $Z\ge 0$, $\mathbb{E}Z=1$ and $\mathbb{E}[ZY]=0$. An equivalent martingale measure has $Z>0$ almost surely. A signed martingale measure obeys the same normalization and pricing equations without requiring nonnegativity. The distinction matters for quadratic hedging. For the completeness assertion use the usual no-arbitrage hypothesis, expressed here by existence of an equivalent martingale measure $Q_0$.

The unrestricted one-period quadratic hedge minimizesPut $m=\mathbb{E}Y$, $C=\mathrm{Cov}(Y)$ and $k=\mathrm{Cov}(Y,H)$. Remove redundant gains so that $C$ is invertible. Differentiating in $x$ gives $x=\mathbb{E}H-{\vartheta }^{T}m$. The remaining squared error isThereforeThe residual is orthogonal in $L^2(P)$ to $1$ and every gain, making the fitted claim the orthogonal projection onto a finite-dimensional subspace onto $\mathcal{A}$. Its minimum mean-square error is $\mathrm{Var}(H)-k^T{C}^{-1}k$, and it vanishes exactly when $H$ is attainable. If initial capital $x$ is fixed instead, the normal equation isThis separates hedging with prescribed capital from simultaneous optimization of capital and risky holdings.

The minimal martingale measure in a one-period market is characterized by leaving unchanged the expectation of every centered square-integrable $L$ orthogonal to the martingale part $Y-m$. For square-integrable pricing densities this forces $Z-1$ into the span of $Y-m$: the orthogonal complement consists exactly of those $L$. The pricing equation then determinesIndeed $\mathbb{E}{Z}_{\ast }=1$ and $\mathbb{E}[Z_{\ast }Y]=m-C{C}^{-1}m=0$. For every such orthogonal $L$, $\mathbb{E}[Z_{\ast }L]=0$. Conversely a normalized density preserving all these $L$ must have the displayed form, since its coefficients must satisfy $C\beta =-m$. The least-squares initial capital is consequently $x_{\ast }=\mathbb{E}[Z_{\ast }H]$.

For any other square-integrable signed pricing density $Z$, put $L=Z-Z_{\ast }$. Both densities price $1$ and all gains, so $L$ is orthogonal to their span, including $Z_{\ast }$. HenceEquality holds only when $Z=Z_{\ast }$. Densities with infinite second moment cannot improve this finite minimum. This proves the requested minimum over dominated martingale measures whenever $Z_{\ast }\ge 0$, and over all square-integrable signed martingale measures without that extra condition. If $Z_{\ast }>0$, it is also an equivalent martingale measure.

Absence of arbitrage alone does not guarantee that the minimal density is a probability density. For a concrete example, let a single discounted gain take values $-1,1,10$ with probabilities $1/10,4/5,1/10$. Then $m=17/10$, $C=801/100$ and $Z_{\ast }(10)=-610/801$. Nevertheless pricing probabilities $(109/200,89/200,1/100)$ are strictly positive and give mean gain zero, so an equivalent martingale measure exists. This is negative minimal density in an arbitrage-free one-period market. If the term martingale measure is reserved for probabilities, the printed minimum assertion needs the extra nonnegativity hypothesis; otherwise it is the signed minimum-norm assertion just proved. Minimizing over the nonnegative pricing densities instead is a different constrained projection problem.

Finally, prove completeness and uniqueness of dominated martingale measures. If the market is complete market, every event indicator can be replicated. Under any dominated martingale measure, its expectation equals its initial discounted cost. Consequently any two such measures agree on every event and are equal; existence is supplied by $Q_0$.

Conversely suppose $\mathcal{A}$ has dimension $k$ and is a proper subspace of the whole payoff space. Choose $k+1$ disjoint events $A_i$ of positive probability. Such events exist whenever $L^2(P)$ has dimension greater than $k$: otherwise the space has at most $k$ atoms and dimension at most $k$. Let $f_1,\dots ,f_k$ be a basis of $\mathcal{A}$. The $k+1$ vectorsare linearly dependent. Choose coefficients $a_i$, not all zero, whose linear combination is zero, and set $h={\sum }_i{a}_i{1}_{A_i}$. It is bounded, not almost surely zero under $Q_0$, and satisfies ${\mathbb{E}}_{Q_0}[hf]=0$ for every $f\in \mathcal{A}$, in particular for $1,Y^1,\dots ,Y^d$. For $0<\epsilon <\Vert h{\Vert }_{\infty }^{-1}$ defineThese densities are strictly positive, normalized, have zero gain means, and are different. Thus incompleteness contradicts uniqueness, provingThe existence hypothesis cannot be discarded: with three positive-probability states and gain $Y=0,1,2$, the sole dominated martingale probability concentrates on $Y=0$, yet the two-dimensional attainable space does not span all three states. That market permits arbitrage.

Take a riskless gross return $R>0$ per period and a stock whose independent multiplicative moves are $u$ and $d$, with physical up probability $p\in (0,1)$. Assume $0<d<R<u$. With $K_k$ up moves by date $k$,A self-financing portfolio chooses each stock holding from information available before the next move, and invests the remainder in the bank account. At any node the two possible next payoffs can be matched by two holdings because $u\ne d$. Backward induction therefore makes the binomial market complete.

The unique risk-neutral probability of an up move is determined by the discounted-stock martingale equation:The risk-neutral measure assigns product probabilities to the successive moves. A path with $K=K_n$ up moves has physical probability $p^K(1-p{)}^{n-K}$ and pricing probability $q^K(1-q{)}^{n-K}$. Its Radon-Nikodym derivative is thereforeSince $S_n=S_0{d}^n(u/d{)}^K$, eliminate $K$ to obtain the requested terminal-price expression:The pricing density depends only on the terminal stock price even though different paths to that price have different intermediate histories.

For $w_0>0$, a nonnegative terminal wealth $X$ with initial cost $w_0$ satisfiesThe utility function has derivative $v^{\prime }(x)=x^{-(\gamma -1)/\gamma }$. Put $\eta =\gamma /(\gamma -1)>1$. Its first-order condition makes $v^{\prime }(X)$ proportional to $Z$, so $X=C{Z}^{-\eta }$. Choosing $C$ to satisfy the pricing budget constraint gives the binomial power-utility terminal wealthAll states have positive probability and the state space is finite, so this payoff is positive, integrable and attainable. The normalizing constant is explicit: independence givesCombining these two formulas with the terminal-price expression for $Z$ gives $X_{\ast }$ entirely as a function of $S_n$.

To prove global optimality, strict concavity gives $v(X)-v(X_{\ast })\le v^{\prime }(X_{\ast })(X-X_{\ast })$. Because $v^{\prime }(X_{\ast })=C^{-1/\eta }Z$, the expectation of the right side is zero for any other payoff satisfying the same pricing budget constraint. Equality requires $X=X_{\ast }$ almost surely. Backward replication supplies its self-financing strategy, so this static optimum is attainable by trading. If $w_0=0$ and terminal wealth must be nonnegative, the only feasible terminal wealth is zero. For $p=q$, the density is one and the optimizer is the riskless wealth $R^n{w}_0$.

On a finite horizon $[0,T]$, let $W$ be a standard $d$-dimensional Brownian motion and let $\theta$ be a predictable process with ${\int }_0^T|{\theta }_s{|}^{2}ds<\infty$ almost surely. Suppose the stochastic exponentialis a true martingale with $\mathbb{E}{Z}_T=1$. The Novikov condition $\mathbb{E}\exp (\frac{1}{2}{\int }_0^T|{\theta }_s{|}^{2}ds)<\infty$ is a sufficient condition. The Girsanov theorem says that under $dQ=Z_{T}dP$,Because $Z_T>0$, the two measures are equivalent on the terminal sigma-algebra.

For the proof sketch, Itô formula gives $d{Z}_t=-Z_t{\theta }_t\cdot d{W}_t$. Apply the product rule to $Z_t{W}_t^{Q,i}$. Its finite-variation term $Z_t{\theta }_t^{i}dt$ is cancelled by $d[Z,W^i{]}_t=-Z_t{\theta }_t^{i}dt$. After localization the product is a martingale under $P$, and the conditional change-of-measure formula makes $W^{Q,i}$ a local martingale under $Q$. Adding finite variation leaves its covariations unchanged: $[W^{Q,i},W^{Q,j}{]}_t={\delta }_{ij}t$. The Lévy characterization of multidimensional Brownian motion completes the proof. Localization is essential if the integrands are not bounded; the true-martingale hypothesis on $Z$ permits the final measure change.

For the deal, write $T=t_0$ and use constant interest rate $\rho$, volatility $\sigma >0$ and no stock dividends. Under the risk-neutral measure, the Black-Scholes model isFor example this follows from the preceding theorem with $\theta =(\mu -\rho )/\sigma$ when the physical stock drift is $\mu$. The delivered shares have terminal cash value $S_T\max (r,c/A)$, so its initial value isThe factor $S_T$ must be included: the deal is a stock delivery, not a cash payment of the random number of shares.

Use the stock-numeraire measure in the Black-Scholes model, whose density isThe Girsanov theorem gives $W_t^S=W_t^Q-\sigma t$ as Brownian motion under $Q^S$. ThereforeThe discrete geometric average under the stock-numeraire measure has normally distributed logarithm withThe last identity uses the printed descending order of sampling times. Each $t_i$ is the minimum in exactly $2i-1$ ordered pairs. In particular $v>0$.

Assume the ordinary positive deal parameters $r,c>0$ and put $k=\log (c/r)$, $d=(m-k)/\sqrt{v}$. The change of numeraire gives $V_0=S_0{\mathbb{E}}_{Q^S}\max (r,c{e}^{-Y})$, where $Y=\log A$. Split the integral at $k$. Completing the square in $e^{-y}$ times the normal probability density givesThus the stock-delivery option with a geometric average has valueHere $\Phi$ is the standard normal distribution function. Delivery time $T$ does not enter separately once the sampling dates are fixed: the stock numeraire cancels the final stock factor and discounting. As a check, for one sample at $T$, the payment is $\max (r{S}_T,c)$ and the formula becomes the value of $r$ shares plus a put on their terminal value. If $c=0$ the value is $r{S}_0$; if $r=0$ the value is $c{S}_0{e}^{-m+v/2}$.

A self-financing portfolio changes its holdings using only its existing wealth. If it holds $g_t$ shares and $h_t$ units of a traded bond or bank account $B_t$, its value $V_t=g_t{S}_t+h_t{B}_t$ obeys the gain identityThe changes in holdings do not add or withdraw capital. Set $T=t_0$, $B_t=e^{-\rho (T-t)}$, so $d{B}_t=\rho B_{t}dt$, and let $d{S}_t=\mu S_{t}dt+\sigma S_{t}d{W}_t$, with $\sigma >0$. Write $V_t=p(S_t,t)$ and suppress the arguments of the smooth holdings $g,h$.

Differentiate $p=xg+B_{t}h$:The Itô formula minus the proposed portfolio gains givesUniqueness of the local-martingale and finite-variation decomposition shows that self-financing requires $p_x=g$. Since $S_t$ has full support on $(0,\infty )$ for each $t>0$, smoothness gives the corresponding functional identity for every positive $x$, and continuity extends it to the interval endpoints. ThusDifferentiating this identity once more in $x$ gives $g_x+x{g}_{xx}+B_t{h}_{xx}=0$, so $p_{xx}=g_x$. Substitute both identities into the drift condition to obtainConversely these equations give $p_x=g$, $p_{xx}=g_x$ and zero drift and diffusion in the displayed gain difference. They therefore prove the claimed self-financing conditions for smooth stock and bond holdings in both directions.

Using $B_{t}h=p-xg$ in the gain identity and setting $g=p_x$, self-financing impliesThis is the Black-Scholes equation, independent of the physical drift $\mu$. Conversely, if a smooth $p$ satisfies it, define its delta hedge byThe preceding Itô formula verifies that these holdings are self-financing.

For prescribed arbitrary holdings, the value equation alone is insufficient. The corrected equivalence is the Black-Scholes value equation needs delta-compatible holdings: a specified portfolio is self-financing if and only if its value solves the equation and $g=p_x$. Indeed $p=0$, $g=1$, $h=-x/B_t$ satisfies the value equation but its proposed gains are $d{S}_t-\rho S_{t}dt$, which have nonzero Brownian part while $dp=0$. Thus the printed final equivalence is valid as an existence assertion for the delta hedge, not for every decomposition of the same value. No change to the valid two holdings equations is needed.

With continuous underlying stock dividends, the stock's gain is $d{S}_t+\theta S_{t}dt$, so self-financing becomes $d{V}_t=g_t(d{S}_t+\theta S_{t}dt)+h_{t}d{B}_t$. Under the risk-neutral measure the ex-dividend stock drift is $(\rho -\theta )S_t$. Matching its diffusion still gives $g=p_x$, and matching its gain drift gives the Black-Scholes equation with continuous stock dividendsOnly the coefficient of $x{p}_x$ changes. The $-\rho p$ discount term remains, because the claim is financed at the bank rate; there is no separate source term for a dividend paid by the claim itself. The dividends here are paid by the underlying stock.

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.