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Put $X_t=g_t(x)-W_t$, $Y_t=g_t(y)-W_t$ and $D_t=Y_t-X_t$. Before $T(x)$, order preservation gives $0<X_t<Y_t$ and $D_t>0$. The common noise cancels from their difference:Since $D$ has finite variation, applying the Itô formula to $Z=Y/D$ gives the SLE two-boundary-point ratio diffusionWith the strictly increasing clock $u(t)={\int }_0^t{D}_s^{-2}ds$, the Dambis-Dubins-Schwarz theorem yields a Brownian motion $B$ for which the time-changed process obeysThe sign of the new Brownian motion has absorbed the minus sign above.

The domain Markov property of a chordal Loewner chain says that, conditionally on the past before $T(x)$, the mapped future is a fresh SLE. Its boundary marked points are $X_t,Y_t$. HenceLocalize, for example, where $D_t\ge 1/n$, $Z_t\in [1+1/n,n]$ and $t\le n$. On each such stopped interval this is a bounded conditional-expectation martingale, and therefore a genuine martingale. No smoothness assumption on $f$ is needed for this argument. The exit calculation below identifies it explicitly.

Let $D$ be the unit disc and normalize its positive Poisson kernel at the boundary point $1$ byIt is harmonic in $D$. Let $p_D(t,z,w)$ be the killed Brownian transition density, for planar Brownian motion with generator $\frac{1}{2}\Delta$. The Doob h-transform has sub-Markov transition densityIts missing mass represents paths already absorbed at the distinguished boundary point. Equivalently, for a stopping time $\sigma$ before exit from a compact subdomain, its law is weighted relative to ordinary killed Brownian motion by $h(B_{\sigma })/h(B_0)$. These stopped laws are consistent and define a diffusion up to its lifetime. Its generator isThis is a rigorous Brownian motion conditioned to exit at a boundary point, not conditioning on an event of positive probability. Exhaustion of $D$ shows that its terminal boundary limit is $1$; its lifetime is finite. For example its expected lifetime from $0$ is ${\int }_D{G}_D^{\mathrm{BM}}(0,w)h(w)dw$, which is finite: the Green function vanishes linearly near the smooth boundary and cancels the Poisson-kernel singularity there, while its logarithmic singularity at $0$ is integrable.

One can also condition ordinary Brownian motion on exiting through an arc $I_{ϵ}$ about $1$, and then let its length decrease to zero. The conditional harmonic functions ${\omega }_D(z,I_{ϵ})/{\omega }_D(0,I_{ϵ})$ converge locally uniformly to $h(z)$. Thus their stopped conditional laws converge to this same diffusion. Both constructions specify the conditioning unambiguously.

Here the map direction is opposite to Question 2: write $\varphi :U\to D$, with $\varphi (0)=0$ and $\varphi (1)=1$. The separation condition makes $U$ agree with $D$ near $1$. Schwarz reflection across that analytic boundary arc therefore defines the boundary derivative, whose modulus is positive.

For ordinary Brownian motion $W$ started at $0$, let ${\sigma }_U$ be its first exit from $U$. Take sufficiently short arcs $I_{ϵ}$ around $1$. They are also arcs of $\partial U$. The event that $W$ stays in $U$ until its disc exit and exits in $I_{ϵ}$ is exactly the event $W_{{\sigma }_U}\in I_{ϵ}$. Therefore the conditional avoidance probability isConformal invariance of planar Brownian motion gives ${\omega }_U(0,I_{ϵ})={\omega }_D(0,\varphi (I_{ϵ}))$. At the disc center harmonic measure is normalized arc length, so the ratio tends to $|{\varphi }^{\prime }(1)|$.

To justify that this limit is the avoidance probability for the point-conditioned diffusion, one may use the stopped transform directly. Its probability of reaching a smooth boundary arc at $1$ before any other boundary of $U$ is the mass at that pole in the transformed harmonic measure. All other exit points have weights $h(w)$ and represent failure. Equivalently integrate the Poisson-kernel density at $1$ for $U$ and divide by that for $D$. This gives the boundary-density ratio above, without assuming that a full-path avoidance event is a continuity set for arbitrary weak convergence. The kernels transform by the boundary Jacobian:ConsequentlyThere is no extra factor of $|{\Phi }_U^{\prime }(0)|$. The starting point is fixed and the conditioning concerns boundary harmonic-measure density. In this normalization the boundary derivative is a positive real, so it may also be written ${\Phi }_U^{\prime }(1)$. It is at most one by the probability interpretation. Given avoidance, the mapped path is the same conditioned Brownian motion up to conformal time change; this is radial restriction for a conditioned Brownian path.

One precise relation is to radial SLE with parameter $2$. Its orientation is from the boundary point $1$ to the interior target $0$. With conformal-radius time it is defined byThere is a coupling of the conditioned Brownian path $B$ with this simple radial curve $\eta$ such that, for each initial curve segment, the first point at which $B$ encounters it is its tip. More precisely, if $s_t=\inf \{s:B_s\in \eta [0,t]\}$, then $B_{s_t}=\eta (t)$. Reversing the Brownian path turns these first-hit times into the last-visit times that characterize loop-erasure of planar Brownian motion. ThusThis is an existence statement for a coupling, not a pointwise equality of the two processes or a claim that a naive finite-loop deletion algorithm applies to Brownian motion.

Here is the martingale mechanism behind the coupling. Put $a_t=e^{i{\theta }_t}$ andThis is the positive slit-domain Poisson kernel at the tip, normalized by $Q_t(0)=1$. Writing $q_t=g_t(z)/a_t$, the radial equation givesThe drift of $(1+q_t)/(1-q_t)$ is $(2-\kappa )q_t(1+q_t)/(1-q_t{)}^3$, so $Q_t(z)$ is a local martingale at $\kappa =2$.

Start independent stopped $B$ and radial SLE2, and weight their joint laws, before meeting, byFor fixed $s$ this is an SLE local martingale. For fixed $t$ it is a Brownian-transform local martingale, because ${\mathcal{L}}^h(Q_t/h)=(2h{)}^{-1}\Delta Q_t=0$. Since $L(0,t)=L(s,0)=1$, localized two-parameter weighting preserves both original marginals. For a fixed slit, the changed Brownian law is the transform by $Q_t$, so it hits that slit at the pole, its tip. Compatible stopped couplings, followed by exhaustion, give the first-hit-at-tip property. This completion is the substantive continuum coupling step. Discrete loop-erased random walks and their radial-SLE2 limit give a parallel approximation picture.

For example, in this coupling avoidance by $B$ implies avoidance by $\eta$, since the radial curve is contained in the Brownian trace. The probability from part (b) therefore supplies a lower bound $|{\varphi }^{\prime }(1)|\le \mathbb{P}(\eta \subset U)$. The two avoidance laws are not identical: unweighted simple radial ${\mathrm{SLE}}_2$ does not have the Brownian radial-restriction property. This is a way to transfer path-containment information while retaining the distinct geometry and orientation of the two random curves.

For the quadratic Ornstein-Uhlenbeck state-price density, apply the Itô's formula under the reference probability measure. Since $f^{\prime }(x)=x$ and $f^{\prime \prime }(x)=1$,The stochastic integral is a true martingale on finite horizons by the finite moments of the Ornstein-Uhlenbeck process. A uniform sufficient condition is $\boxed{\alpha \ge {\sigma }^2/(2a)}$, making the finite-variation drift nonpositive. This is also the condition for nonpositive drift at every state.

Use the positive state-price density $\zeta$ to price a terminal payoff $H$ by ${\Pi }_t(H)=\mathbb{E}[{\zeta }_{T}H\mid {\mathcal{F}}_t]/{\zeta }_t$. The riskless account must cancel the drift of $\zeta$, giving the short rateThis pricing system is consistent with a money-market risk-neutral measure: with $M_t=e^{{\int }_0^t{r}_{u}du}$, the process ${\zeta }_t{M}_t/{\zeta }_0$ is a stochastic exponential whose diffusion coefficient $\sigma X_t/f(X_t)$ is bounded. The Novikov condition makes it a true density martingale. The original Brownian motion is for the reference measure, not automatically for this new pricing measure.

For $X_0=x_0$, the OU second moment is $\mathbb{E}{X}_T^2=x_0^2{e}^{-2\lambda T}+{\sigma }^2(1-e^{-2\lambda T})/(2\lambda )$. Thus the zero-coupon bond price isConditionally at time $t$, replace $T$ by $T-t$ and $x_0$ by $X_t$.

The short rate is nonnegative under the condition above, increases with $X_t^2$, and lies between $\alpha -{\sigma }^2/(2a)$ and $\alpha +2\lambda$. The model is tractable but imposes fixed rate bounds, identical rates for opposite values of $X$, and a restricted one-factor term structure. Its long-maturity yield is $\alpha$. These structural restrictions, including exclusion of negative rates under the supermartingale condition, may be unsuitable for a general interest-rate fit.

Use the independent increments of Brownian motion, rather than merely checking that an Itô formula drift vanishes. For $0\le s<t$, put $h=t-s$ and write $B_t=B_s+Z$, where $Z$ is independent of ${\mathcal{F}}_s$ and has normal distribution $N(0,h)$. Its first four moments are $0,h,0,3h^2$. ThereforeFor the cubic expression, the coefficient of $B_s$ after conditioning is $3(t-s)+\alpha (t)$, so $\alpha (t)=-3t$ makes it $\alpha (s)$. For the quartic expression, choose $\beta (t)=-6t$. Its conditioned coefficient of $B_s^2$ is then $6(t-s)-6t=-6s$. The constant term becomeswhich equals $3s^2$ when $\gamma (t)=3t^2$. Thus a standard choice isThe resulting stochastic processes are Hermite polynomial martingales $H_3(B_t,t)$ and $H_4(B_t,t)$. They are genuine integrable martingales, since Gaussian moments are finite at every finite time and the displayed conditional identities establish the martingale property directly.

The choice is not unique. Constants $c,d,e\in \mathbb{R}$ give the valid family $\alpha (t)=c-3t$, $\beta (t)=d-6t$, and $\gamma (t)=3t^2-dt+e$: these add $c{B}_t$ to the cubic martingale and $d(B_t^2-t)+e$ to the quartic one. The boxed choice sets these harmless additions to zero.

The Skorokhod embedding of a centered random walk states that a random walk with independent identically distributed centered steps of finite variance ${\sigma }^2$ can be realized on an appropriate probability space aswhere $B$ is a standard Brownian motion and the $T_n$ are finite stopping times. More precisely, the stopped positions have the same joint law as the given random walk, and the pairsmay be chosen independent and identically distributed. Their spatial component has the step law, and $\mathbb{E}(T_n-T_{n-1})={\sigma }^2$. Repeating the one-step Skorokhod embedding theorem with the Strong Markov property gives this formulation. In the present normalization, the mean time increment is one, and the strong law of large numbers gives $T_n/n\to 1$ almost surely.

The Donsker invariance principle states that the linearly interpolated diffusively rescaled random walkconverges weakly as a random element of $C[0,1]$, equipped with the uniform norm, to standard Brownian motion restricted to $[0,1]$. At $t=1$ the fractional term is zero. The only step assumptions needed here are zero mean, unit variance, and independent identical distributions; a higher moment or bounded support is not required. This is a functional central limit theorem, concerning the entire interpolated path rather than only its endpoint.

Sample Brownian motion at integer times and put $Z_n=B_n/\sqrt{n}$. Every $Z_n$ has normal distribution $N(0,1)$, although the $Z_n$ are correlated. LetFor every fixed integer $m$,because $B_m/\sqrt{n}\to 0$ almost surely. The right-hand expression depends only on the future independent increments $B_j-B_{j-1}$ with $j>m$. Thus $\{L\ge a\}$ is, up to a null set, a tail event of those independent increments, and its probability is zero or one by the Kolmogorov zero-one law.

For any finite real $a$, every $Z_n$ exceeds $a$ with the same positive probability $p_a=\mathbb{P}(N(0,1)\ge a)$. For each $N$,Taking the decreasing intersection over $N$ shows that $Z_n\ge a$ infinitely often with probability at least $p_a$. This event implies $L\ge a$, so $\mathbb{P}(L\ge a)>0$. The zero-one law makes it one. Intersecting over positive integers $a$ gives $L=+\infty$ almost surely. The continuous-time limit superior is at least the one along integers, henceThis proves that Brownian fluctuations exceed the square-root scale without needing the lower bound in the law of the iterated logarithm.

Let ${\Delta }_j=W_{t_j}-W_{t_{j-1}}$ for $1\le j\le n$. These increments are jointly normal, because each is obtained by evaluating the isonormal Gaussian process on an interval indicator. Indicators of distinct intervals are orthogonal in $L^2([0,\infty ))$, soBy uncorrelated jointly Gaussian variables are independent, the increments on all these disjoint intervals are independent. Reversing the sign of the first increment, as in the printed list, preserves this independence.

The three requested properties have now been obtained without an existence theorem for Brownian motion. One can also obtain continuous paths: the Gaussian fourth moment gives $\mathbb{E}|W_t-W_s{|}^4=3|t-s{|}^2$. The Kolmogorov continuity theorem therefore supplies a continuous modification on every finite time interval, which can be chosen consistently on the half-line. Modification preserves every finite-dimensional distribution and hence the independent Gaussian increments. This yields Brownian motion itself.

For $0\le t\le T$, set $Z_t=\cosh x{Y}_t$. Part (b) gives $Z_0=1$, $\mathbb{E}{Z}_T=1$, and $Z_t>0$. Its stochastic differential isThus $Z$ is the stochastic exponential of $\int \eta dW$. The Novikov condition also holds, since $\exp (\frac{1}{2}{\int }_0^T{\eta }_t^{2}dt)\le e^{T/2}$.

The Girsanov theorem says that under the measure with Radon-Nikodym derivative $Z_T$, the process $W_t-{\int }_0^t{\eta }_{s}ds$ is a Brownian motion up to $T$. Substituting the sign of $\eta$ and the original stochastic differential equation givesThe density is strictly positive, so $\mathbb{Q}$ and $\mathbb{P}$ are equivalent probability measures.

Fix $T>0$. The given normal distribution and part (d) implyOne can deduce determinism without any moment assumption on the bracket. Put $Z=e^{-⟨X{⟩}_T}$. Taking $\lambda =1,2$ gives $\mathbb{E}Z=e^{-T}$ and $\mathbb{E}{Z}^2=e^{-2T}$, so $\mathrm{Var}(Z)=0$. Therefore $⟨X{⟩}_T=T$ almost surely. Applying this at every rational time and using continuity of quadratic variation gives $⟨X{⟩}_t=t$ simultaneously for all $t\ge 0$ outside a single null set.

The Lévy characterization of Brownian motion states that a continuous local martingale starting at zero with this bracket is Brownian motion in its filtration. To see the independent-increment conclusion directly, the Itô formula shows that $e^{i\theta X_t+{\theta }^{2}t/2}$ is a martingale on any fixed bounded time interval: it is a local martingale with a deterministic bound on its modulus. ThusThe deterministic conditional characteristic function identifies an $N(0,t-s)$ increment independent of ${\mathcal{F}}_s$. Together with the given path continuity and $X_0=0$, this proves $X$ is Brownian motion.

For the drifted Brownian motion $Y_t=y+B_t+at$ on $(0,\infty )$, the diffusion generator is $\mathcal{L}=\frac{1}{2}\frac{d^2}{d{z}^2}+a\frac{d}{dz}$. Seek a bounded solution of $\mathcal{L}u=\lambda u$ with $u(0)=1$. The exponential ansatz $u(z)=e^{rz}$ givesBecause $\lambda >0$, the plus root is positive and the minus root is negative. Boundedness on $[0,\infty )$ therefore selectsIt satisfies the boundary condition and all hypotheses of part (d). HenceThis is the first-passage Laplace transform for Brownian motion with drift, with $e^{-\lambda \infty }=0$. At $a=0$ it reduces to $e^{-y\sqrt{2\lambda }}$. As $\lambda \downarrow 0$, it gives $\mathbb{P}(S<\infty )=e^{-2y\max (a,0)}$, agreeing with certainty of hitting when the drift points towards zero and a possible escape when it points away.

The multidimensional Itô formula includes a mixed second derivative multiplied by the quadratic covariation, with no extra factor $1/2$. HereConsequently the Itô formula for $M_t=U(t,{\sigma }_t,X_t)$ has drift equal to the left side of the stated backward partial differential equation. That drift vanishes, leavingA stochastic integral against Brownian motion with locally square-integrable predictable integrand is a continuous local martingale. The smoothness of $U$ and localization of the diffusion and its coefficients give this integrability on the model's lifetime. Thus $M$ is a local martingale, as required. The partial differential equation cancellation alone does not establish a true martingale or justify replacing $U$ by a terminal-payoff expectation without an additional integrability argument.

The joint generator. With $\theta$ again measured in dollars, portfolio wealth obeysIts noise and the factor noise are driven by the same Brownian motion. Their quadratic covariation is $\sigma (x)\theta \alpha (x)dt$, so the diffusion generator has a cross derivative. Dynamic programming and the Itô formula giveAll coefficient functions in this formula are evaluated at $x$. The cross derivative is essential: it gives intertemporal hedging demand. Normalizing CRRA utility as $U(c)=c^p/p$, where $p=1-R$, the two optimizations give, at nonzero volatility,and henceHere $p(-1/R)=1-1/R$.

Homogeneity and the reduced equation. Scaling initial portfolio wealth and both controls preserves the portfolio wealth constraint and multiplies the objective by the positive number $b^p$. ThusThis expression is valid for both signs of $p$: $V$ is negative when $R>1$, but its portfolio wealth derivative is positive. Its derivatives areSubstitution yieldsThe resulting feedback isThe first portfolio term is myopic, and the second is intertemporal hedging demand.

Constant market price of risk. If $\mu -r=\sigma \kappa$ and volatility is nonzero, the portfolio term becomes $p(\kappa f+\alpha f^{\prime }{)}^2/(2Rf)$, so the magnitude of stock volatility disappears. Applying the power transformation of a complete-market investment equation $f=g^R$ cancels the squared-gradient terms and gives the further reductionFor ${\gamma }_M>0$ its economic solution is $g=1/{\gamma }_M$. ThusTo see why this solves the investment-consumption problem, optimize directly over Brownian portfolio exposures, writing $y=\sigma (x)\theta$. The portfolio wealth equation becomes $dw=ydW+(rw+\kappa y-c)dt$, which no longer contains $X$. With nonzero volatility the same admissible exposure processes are available for every factor state, so the attainable wealth-consumption pairs, and therefore the value, are exactly those of the Merton consumption-investment problem. This also excludes extraneous solutions of the linear equation without imposing artificial factor boundary data.

The printed boundedness assumptions do not ensure nonzero volatility or a finite value. The unsimplified HJB equation remains the correct control equation at a zero of $\sigma$. There the hedge term vanishes; if $\mu \ne r$, the riskless excess return gives arbitrage with unrestricted holdings. Under $\mu -r=\sigma \kappa$, a zero-volatility state offers only the bank exposure at that instant. For example $\sigma \equiv 0$, $\mu \equiv r$ satisfies this relation for any chosen $\kappa$, but its value uses ${\gamma }_0=[\rho -pr]/R$, not a fictitious nonzero risk premium. The constant-value formula using $\kappa$ presupposes access to the Brownian exposure, with sufficient integrability for the corresponding holdings. Additive utility constants again only shift $V$ by a constant divided by $\rho$.

Available portfolio wealth and the ruin boundary. Put $h=\bar{r}D$, the constant interest payment on the fixed loan. The loan principal is already included in available portfolio wealth; it is not a growing portfolio holding. ThereforeIn particular the interest outflow is $\bar{r}D$, not $(\bar{r}-r)D$. Writing $w=x+D$ would instead give net portfolio wealth drift $rx+(\mu -r)\theta -(\bar{r}-r)D-c$, which explains the distinction.

Let ${\tau }_0$ denote the ruin time, to avoid confusing it with a fixed terminal horizon. The objective stops at ${\tau }_0$; consequently the absorbing boundary is $V(0)=0$, without an obligation to keep financing the loan after ruin. Dynamic programming gives, for $w>0$,For increasing strictly concave value, put $z=V^{\prime }(w)>0$ and use inverse marginal utility $I$. The optimal controls and the optimized HJB equation arewhere $\tilde{U}(z)={\sup }_{c\ge 0}[U(c)-zc]$. For CRRA utility with $0<R<1$, write $p=1-R$ and $q=1-1/R<0$. Then

Dualization and the printed constant. Use the convex wealth-variable Legendre dualAt an interior maximizing portfolio wealth, $J^{\prime }=-w$, $J^{\prime \prime }=-1/V^{\prime \prime }>0$ and $V=J-z{J}^{\prime }$. The dual HJB equation is the linear Euler differential equationA trial power $z^t$ givesDirect substitution givesThe PDF prints an additional factor $1/2$ before ${\kappa }^2/(2R)$ in its definition of ${\gamma }_M$. That printed definition is inconsistent with its own identity for $Q(q)$. The expression above is the one used here; assume this corrected ${\gamma }_M>0$.

Solution when $\kappa \ne 0$. Assume positive discount $\rho$, nonzero $\sigma$, and $h>0$. Let $m>0$ and $n<0$ be the two roots of $Q$:Since $Q(q)<0$, $n<q<0<m$. Put $C=R/(p{\gamma }_M)$. For $r\ne 0$, the general interior solution isThe appropriate large-wealth condition is the Merton consumption-investment problem boundIndeed any original control consumes $c+h$ in the debt-free comparison model until ruin, and $U(c+h)\ge U(c)$. Dualizing this bound gives $0\le J(z)\le C{z}^q$. Because $n<q$, convexity and this upper bound force $B=0$ as $z\downarrow 0$.

At the other endpoint portfolio wealth reaches zero. If $z_{\ast }=V^{\prime }(0+)$, the dual ruin boundary with debt service requiresSolving these two equations givesFor $r>0$, $m>1$, and for $r<0$, $0<m<1$, so the quantity defining $z_{\ast }$ is positive in either case. These formulas determine the entire value. For each $w>0$, choose the unique $z\in (0,z_{\ast })$ satisfyingThenBoth terms in the bracket are positive, even when $r<0$ and $A<0$. Thus $J^{\prime \prime }>0$, $-J^{\prime }$ decreases from infinity to zero as $z$ increases, and the portfolio wealth inversion really is unique. The extended dual is continuously differentiable and convex.

There is no additional condition $J^{\prime \prime }(z_{\ast }^{-})=0$. In factAvailable portfolio wealth is killed at zero; the portfolio can have a nonzero limiting volatility immediately before ruin. Imposing a reflecting-boundary or zero-curvature condition would solve a different problem.

For $r=0$, the linear forcing resonates with the root $m=1$. Put $b=\rho +{\kappa }^2/2=Q^{\prime }(1)$. The dual and boundary constants instead areExtend by zero for $z\ge z_{\ast }$. Hereand the controls remain $c=z^{-1/R}$ and $\theta =\kappa z{J}^{\prime \prime }/\sigma$, with $J^{\prime \prime }=h/(bz)+Cq(q-1)z^{q-2}>0$.

Verification and transversality. The candidate is nonnegative, increasing, strictly concave and zero at ruin. Its HJB equation makes the discounted value plus accrued utility a local supermartingale for every admissible control, and a local martingale for the stated feedback. Localization at positive lower and finite upper portfolio wealth levels gives the finite-horizon comparison. The investment value transversality condition follows from the same debt-free bound: applying the Itô formula to $w^p$ and maximizing its risky term givesThe nonnegative consumption and debt-service drifts only decrease this bound. Thus the expected terminal candidate tends to zero. The feedback has at most linear growth, including a finite limit as portfolio wealth decreases to zero; stopping it at ruin gives an admissible policy. Letting localization levels and then the horizon tend to their limits proves that the candidate is the value, rather than just a formal dual solution. If $D=0$, the absorbing-debt boundary disappears and the ordinary Merton consumption-investment problem formula is recovered.

Zero market price of risk. With $\sigma \ne 0$ and $\kappa =0$, the dual equation is first order. If $\rho >r$, the preceding formulas remain valid with $m=\rho /(\rho -r)$ for $r\ne 0$, and the logarithmic formula with $b=\rho$ for $r=0$; there is no $B$ term and ${\theta }^{\ast }=0$. The same boundary and transversality argument verifies this deterministic consumption policy.

The remaining finite-value case has $r>0$ and $pr<\rho \le r$. Put ${\gamma }_0=(\rho -pr)/R$, $w_c=h/(rR)$ andThe correct convex dual and its corresponding value areThe two value branches have the same value and derivative at $w_c$. Above $w_c$, hold no stock and consume ${\gamma }_0(w-h/r)$; the surplus over $h/r$ grows at rate $r-{\gamma }_0\ge 0$. If $\rho =r$, the lower branch is attained by zero stock holding and constant consumption $ph/R$: portfolio wealth solves $\dot{w}=r(w-w_c)$ until ruin, and direct integration gives $z_{c}w$.

If $\rho <r$ and $0<w<w_c$, the lower branch is a supremum attained in a limit of increasingly rapid fair stock lotteries between zero and $w_c$, followed by the upper-branch policy on success. The success probability tends to $w/w_c$ and the fixed service cost during the lottery tends to zero. This is possible because unrestricted dollar holdings in the nonzero-volatility stock produce a fair Brownian motion exposure even when its excess drift is zero. The supporting linear branch has optimized waiting residual $(r-\rho )z_c(w-w_c)<0$; the fast lotteries, rather than a finite feedback optimizer, supply the missing control limit. The piecewise candidate is concave, has nonpositive waiting residual everywhere, and the preceding moment bound still supplies an upper-bound verification. This degenerate case can have a supremum without an ordinary maximizing strategy.

Pricing kernel and replication. In the nondegenerate Black-Scholes model, put $\kappa =(\mu -r)/\sigma$ and normalize the state-price density by ${\zeta }_0=1$. The process isThe density $e^{rt}{\zeta }_t$ changes probability to the risk-neutral measure. Under that measure $W_t+\kappa t$ is a Brownian motion and the stock drift is $r$. Thus an integrable contingent claim $H$ has time-$t$ priceIn the usual augmented natural Brownian filtration, the Brownian martingale representation theorem supplies a replicating strategy; this is the complete market assumption. For a nonnegative admissible trading strategy without intermediate consumption, the state-price budget constraint is $\mathbb{E}[{\zeta }_T{w}_T]\le w_0$, with equality for a fully invested replicated claim. In particular