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Backward induction makes the Snell envelope integrable and adapted: $|U_t|\le |Y_t|+\mathbb{E}(|U_{t+1}|\mid {\mathcal{F}}_t)$. Its definition gives $U_t\ge Y_t$ and $U_t\ge \mathbb{E}(U_{t+1}\mid {\mathcal{F}}_t)$, so it is a supermartingale dominating the reward.

For a stopping time $\tau$ taking values in $\{0,\dots ,T\}$, expand its stopped value asThe indicators are ${\mathcal{F}}_t$-measurable. Taking conditional expectations in each summand makes its expectation nonpositive by the supermartingale property. Since ${\mathcal{F}}_0$ is trivial, $U_0$ is deterministic andThis proves the finite-horizon optional sampling theorem directly in the instance needed here, without assuming nonnegative rewards.

A unit zero-coupon bond pays one at its maturity, so $P_t(t)=1$. Monotonicity in maturity gives $P_t(t+1)\le 1$. Using the state-price density representation with $T=t+1$,The process is positive and integrable, as noted in part (a), and adapted. Thus it is a supermartingale. Strictly decreasing maturity prices yield a strict one-step conditional inequality; weak decrease is already enough for the conclusion.

Let $h_t$ and $k_t$ be predictable holdings in the stock and account, with the stochastic integrability needed for a self-financing portfolio. Then $V=hS+kB$ and $dV=hdS+kdB$. The Itô product rule and self-financing identity giveSince $V=hS+kB$ and $\sigma \lambda =\mu -r$, the finite-variation term is zero. Thereforeand the deflated wealth is a local martingale.

For zero-capital nonnegative wealth under a local deflator, a nonnegative local martingale is a supermartingale, by localization and the Conditional Fatou lemma. Under the required nonnegative-wealth condition, $ZV\ge 0$ and starts at zero, so $0\le \mathbb{E}(Z_t{V}_t)\le 0$. Thus $Z_t{V}_t=0$ almost surely at every fixed $t$. Strict positivity of $Z$ gives $V_t=0$ almost surely. Taking a countable intersection over rational times and then using continuous wealth paths strengthens this toThis argument only needs the local deflator, so it remains valid without promoting $ZB$ to a true martingale.

The stopped processes are nonnegative martingales. Their expectations equal the finite deterministic value $X_0$. The Fatou lemma gives, for each fixed integer $t$,Thus every $X_t$ is integrable. Apply the preceding discrete-time criterion to conclude $X$ is a martingale, not merely a supermartingale. This is the nonnegative discrete-time local martingale is a martingale result; its conclusion does not extend to arbitrary continuous-time nonnegative local martingales.

Choose the market price of risk ${\lambda }_t=(\mu -r{S}_t)/\sigma$. The Girsanov theorem, with the hypotheses allowed in the question, gives an equivalent martingale measure $\mathbb{Q}$ under whichThus the stock under the risk-neutral measure is a linear Gaussian diffusion. In particular, for $\tau =T-t$ its conditional mean and standard deviation arePut $\varphi (x)=(2\pi {)}^{-1/2}{e}^{-x^2/2}$ and let $\Phi$ be the standard normal distribution function. For $N\sim N(0,1)$,since ${\int }_{-d}^{\infty }x\varphi (x)dx=\varphi (d)$. Discounting this expectation gives the call price in an arithmetic stock model with interest. A particularly convenient expression isAt maturity define $C(T,s)=(s-K{)}^{+}$. This value is nonnegative because it is a discounted expectation of a nonnegative payoff.

For $t<T$, hold ${\pi }_t=C_s(t,S_t)$ shares and hold ${\beta }_t=[C(t,S_t)-{\pi }_t{S}_t]/B_t$ units of the continuous-time bank account. The pricing function solvesThe Itô formula under the physical measure therefore givesThis proves self-financing and terminal replication, with wealth always $C(t,S_t)\ge 0$. The coefficients are locally smooth before maturity, and the strategy extends to maturity through its continuous wealth limit and the square-integrable discounted payoff representation.

To see minimality, any other nonnegative self-financing portfolio replicating the payoff has discounted wealth a nonnegative local martingale under $\mathbb{Q}$, hence a supermartingale. Its initial capital $x$ must satisfy $x\ge {\mathbb{E}}^{\mathbb{Q}}[e^{-rT}(S_T-K{)}^{+}]=C(0,S_0)$. The strategy constructed above attains equality. ThusThe additive physical diffusion may take negative stock values; the formula and nonnegative replicating wealth remain valid. Replacing it by a multiplicative Black–Scholes diffusion would give the wrong price and hedge.

The integral against the local-martingale vector $YP$ is a local martingale, provided the predictable holdings are stochastically integrable. Integrating part (b) givesBoth $Y_t{X}_t$ and the cumulative deflated consumption are nonnegative. Hence $M_t\ge -Y_0{X}_0$. With the usual finite deterministic initial capital, the shifted processis a nonnegative local martingale, and is therefore a supermartingale. For completeness, a localizing sequence turns it into true martingales; conditional Fatou lemma for their nonnegative stopped values gives the supermartingale inequality and ordinary Fatou gives integrability at each time. Subtracting the initial constant provesThis is supermartingale control of deflated consumption gains. It uses $c\ge 0$ as well as $X\ge 0$; an unrestricted stochastic integral is not necessarily a true supermartingale.

Write $V_t=⟨X{⟩}_t$ for the quadratic variation and setThe Hölder factorization of stochastic exponentials follows by adding exponents:ThusThe subtraction inside the numerator is $\sqrt{pq}-1$, outside the square root.

The stochastic exponential $\mathcal{E}(\sqrt{pq}X)$ starts at one and is a nonnegative local martingale, hence a supermartingale. The optional sampling theorem for a supermartingale gives expectation at most one at bounded stopping times. For a finite, possibly unbounded, stopping time $T$, apply this to $T\wedge n$, then use continuity and the Fatou lemma:The Holder inequality with exponents $q$ and $q/(q-1)$ now givesThe inequality also holds when the right side is infinite.

Use deterministic times tending to infinity and the Fatou lemma:Apply the terminal scaling inequality for stochastic exponentials to $X=aM$ and $r=1/a$. Part (c) gives $\mathbb{E}[\mathcal{E}(aM{)}_{\infty }]=1$, soLetting $a\uparrow 1$ shows that the terminal expectation is at least one. The nonnegative local martingale $\mathcal{E}(M)$ starts at one and is a supermartingale, so the Fatou lemma gives the opposite inequality. Consequently

The terminal expectation criterion for a nonnegative local martingale now closes the argument. Conditional Fatou lemma applied to the supermartingale at times tending to infinity givesThe left side has expectation one and the right side at most one, so equality holds almost surely. Thusand uniform integrability of conditional expectations proves that $\mathcal{E}(M)$ is a uniformly integrable martingale. This establishes the required stopped-moment form of the Kazamaki criterion.

Write $\alpha =2r/{\sigma }^2\in (0,1)$. In a continuation interval, the Euler differential equation has power solutions $s$ and $s^{-\alpha }$: substitution of $s^{\beta }$ gives $(\beta -1)(\beta +\alpha )=0$. A decreasing bounded value uses the second solution.

Let $b$ be the exercise boundary. Value matching and smooth fit require $A{b}^{-\alpha }=(1+b{)}^{-1}$ and $-\alpha A{b}^{-\alpha -1}=-(1+b{)}^{-2}$. Dividing gives $b/(1+b)=\alpha$. ConsequentlyThis is a perpetual reciprocal-payoff American option.

To verify the obstacle inequality, observe that $s^{\alpha }/(1+s)$ increases up to $b$ and decreases afterwards, because its logarithmic derivative is $\alpha /s-1/(1+s)$. Therefore $V(s)\ge (1+s{)}^{-1}$ for $s>b$. In the continuation region $({\mathcal{L}}_r-r)V=0$. In the exercise region,The quadratic is convex. At the two endpoints of $[0,b]$ it equals $-\alpha /2$ and $-\alpha /[2(1-\alpha )]$, respectively, so it is negative throughout that interval. Thus the obstacle problem is satisfied on both regions, with value matching and smooth fit at $b$. The second derivative has a jump at $b$; the equation is understood piecewise and in the generalized Itô sense described in part (a).

For optimality, use the Risk-neutral measure for the Black-Scholes model, under which $d{S}_t=r{S}_{t}dt+\sigma S_{t}d{W}_t^{\mathbb{Q}}$. The Itô formula makes $e^{-rt}V(S_t)$ a nonnegative supermartingale, so every exercise time $\tau$ satisfiesBefore hitting the exercise region, its drift vanishes. Since $0\le V\le 1$, the stopped process is a bounded martingale. Apply the optional sampling theorem at ${\tau }_{\ast }\wedge T$ and let $T\to \infty$. The contribution from $\{{\tau }_{\ast }>T\}$ is at most $e^{-rT}$, while the boundary value equals the payoff. Hence equality holds forIf $S_0\le b$, exercise immediately; otherwise wait for the first down-crossing. If that time is infinite, the discounted payout is zero. This proves both the value and the optimal stopping policy, and part (a) supplies its superhedge.

For $S_0>0$, the stochastic exponential solution stays strictly positive. Put $U_t=\sqrt{S_t}$. The Itô formula givesThus $U$ is a nonnegative local supermartingale. To justify the true supermartingale property, stop where $U$ or its stochastic integral exceeds successive bounds. The stopped Itô formula gives $\mathbb{E}[U_{T\wedge {\tau }_m}\mid {\mathcal{F}}_t]\le U_{t\wedge {\tau }_m}$ for $T\ge t$. The conditional Fatou lemma and ${\tau }_m\uparrow \infty$ yieldIn particular $\mathbb{E}{U}_T\le U_0<\infty$. The case $S_0=0$ is the identically zero process. This is the square-root stock supermartingale.

Normalize $B_0=1$ and put ${\tilde{S}}_t=e^{-rt}{S}_t$. Its dynamics are $d{\tilde{S}}_t={\tilde{S}}_t\sigma (t,S_t)d{W}_t$. Bounded volatility makes this stochastic exponential a true martingale on each finite horizon, by the Novikov condition. It also gives a finite second moment: stopping the Itô formula for $S^2$ and applying the Gronwall inequality yields $\mathbb{E}{S}_T^2\le S_0^2{e}^{(2r+L^2)T}$ when $\sigma \le L$.

The discounted payoff $\xi =e^{-rT}(S_T-K{)}^{+}$ is thus square-integrable. Define the nonnegative martingaleThe Brownian martingale representation theorem says that every square-integrable martingale in the Brownian filtration has a representation $M_t=M_0+{\int }_0^t{h}_{u}d{W}_u$ with predictable $h$ and $\mathbb{E}{\int }_0^T{h}_u^{2}du<\infty$. Since $\tilde{S}>0$ and $\sigma >0$, chooseThen discounted gains satisfy $d{M}_t={\pi }_{t}d{\tilde{S}}_t$. Consequently $X_t=B_t{M}_t={\varphi }_t{B}_t+{\pi }_t{S}_t$ is self-financing, nonnegative, and hence an admissible trading strategy. At maturity $X_T=(S_T-K{)}^{+}$, proving claim replication at cost $C(T,K)$.

For minimality, discounted wealth of any admissible self-financing portfolio is a local martingale bounded below, hence a supermartingale by localization and the conditional Fatou lemma. Thus any such replication with initial wealth $x$ obeysTogether with the constructed portfolio, this provesThis is Brownian representation replication in a local volatility market. The given drift $r$ means that the original probability measure already serves as the risk-neutral measure.

Assume $\sigma \ne 0$ and the usual Brownian filtration, so the single-risky-asset market is a complete market. With market price of risk $\kappa =(\mu -r)/\sigma$, the normalized state-price density isThe Itô formula shows that ${\zeta }_t{w}_t$ is a local martingale for a self-financing portfolio. For nonnegative admissible wealth it is a supermartingale, yielding the state-price budget constraintEvery integrable nonnegative terminal claim with equality is attainable in the complete market, by the Brownian martingale representation theorem. Its wealth process is $w_t={\zeta }_t^{-1}\mathbb{E}[{\zeta }_{T}X\mid {\mathcal{F}}_t]\ge 0$.

The Inada conditions $u^{\prime }(0+)=\infty$ and $u^{\prime }(\infty )=0$, together with strict concavity, make $I=(u^{\prime }{)}^{-1}$ a decreasing map from $(0,\infty )$ onto $(0,\infty )$. Pointwise optimization of $u(x)-y{\zeta }_{T}x$ gives the optimal terminal wealthwith scalar multiplier $y>0$. The question assumes a multiplier giving the required budget; utility expectations must also be well defined.

For any admissible terminal wealth $X$, the supporting-line inequality for a concave function givesTaking expectations and using the state-price budget constraint proves optimality:Strict concavity makes the optimal terminal wealth unique up to almost-sure equality.

The Novikov condition holds on every horizon, sinceTherefore $Z=\mathcal{E}(M)$ is a true martingale and the Girsanov theorem applies through each finite $T$.

The deterministic bound also gives a genuine infinite-horizon density. For any $p>1$, the stochastic exponential identity isA nonnegative local martingale is a supermartingale, soThis $L^p$ bound gives uniform integrability. Also $M$ is a L2-bounded continuous martingale, since localization and the Itô isometry give $\mathbb{E}{M}_t^2=\mathbb{E}[M{]}_t\le C$. The L2 martingale convergence theorem gives a finite limit $M_{\infty }$; hence $Z_{\infty }=\exp (M_{\infty }-[M{]}_{\infty }/2)>0$ and $\mathbb{E}{Z}_{\infty }=1$. Weighting by $Z_{\infty }$ defines an equivalent probability measure on ${\mathcal{F}}_{\infty }$. This is the bounded-bracket criterion for a stochastic exponential.

Backward induction makes every $U_t$ adapted and integrable: the conditional expectation of an integrable random variable is integrable, and $|\max (a,b)|\le |a|+|b|$. The Snell envelope recursion directly gives $U_t\ge \mathbb{E}[U_{t+1}\mid {\mathcal{F}}_t]$, so $U$ is a supermartingale.

If $Z$ is a submartingale, iterating its defining inequality gives $\mathbb{E}[Z_T\mid {\mathcal{F}}_t]\ge Z_t$. Starting at $U_T=Z_T$ and applying the tower property of conditional expectation backwards showsThis last conditional expectation is a true martingale. The conclusions areIn the second case waiting until $T$ attains the optimal stopping value. It is not generally true that $U_t=Z_t$: that further equality holds when $Z$ itself is a martingale.

Use the Chordal Loewner equation ${\partial }_t{g}_t=2/(g_t-U_t)$ with $U_t=\sqrt{\kappa }{B}_t$. For a fixed $z\in \mathbb{H}$, let $T_z$ be its Loewner swallowing time, and write $Z_t=X_t+i{Y}_t=g_t(z)-U_t$ and $J_t=|g_t^{\prime }(z)|$ before that time. The Loewner conformal radius is ${{\rm Y}}_t=Y_t/J_t$, half of the conformal radius of $D_t=\mathbb{H}\setminus K_t$ at $z$. The Koebe quarter theorem bounds the conformal radius above by four times the distance to the boundary. For the reverse comparison, if $\varphi :\mathbb{D}\to D_t$ maps zero to $z$, apply the Schwarz lemma to ${\varphi }^{-1}(z+dw)$ on the unit disc, where $d=\mathrm{dist}(z,\partial D_t)$; it gives $d\le |{\varphi }^{\prime }(0)|$. ThusWe use the basic trace theorems that, for $\kappa >8$, the Loewner trace is continuous and Transience of chordal SLE gives $|\gamma (t)|\to \infty$. These facts make its image relatively closed in $\mathbb{H}$; they do not assume the space-filling conclusion. The continuity and transience statements are available in Rohde and Schramm's basic trace theorems.

Take $\rho =\kappa -8>0$ in the supplied SLE interior-point martingale. The derivative exponent vanishes, leavingA nonnegative local martingale is a supermartingale. Applying the optional stopping theorem after localization at its first hit of a level $R$ gives the maximal inequality for a nonnegative supermartingaleThus $C={\sup }_{t<T_z}{M}_t<\infty$ almost surely.

Suppose the Loewner trace avoids some open ball $B(z,r)$ whose closure is inside $\mathbb{H}$. Before $T_z$, the whole open ball is in $D_t$: a connected open ball disjoint from the trace cannot be partly in the unbounded connected component. Hence ${{\rm Y}}_t\ge r/2$. The bound on $M$ forces $S_t\ge c>0$ throughout this interval, for a positive random constant $c$.

The Chordal Loewner equation and its derivative yieldThe first identity forces $T_z<\infty$, since otherwise $Y_t^2\le Y_0^2-4c^{2}t$ becomes negative. At a finite maximal lifetime, $Y_t\to 0$; otherwise the continuous Loewner driving function and the ordinary differential equation continue past that time. Integrating the last identity, using $S_t\ge c$, givescontradicting ${{\rm Y}}_t\ge r/2$. Thus every fixed rational ball inside $\mathbb{H}$ is hit almost surely. A countable intersection makes the trace a dense subset almost surely, and its relatively closed image, established by continuity and transience, then contains all of $\mathbb{H}$:This proves space-filling SLE above parameter eight. It rules out unvisited open regions, rather than inferring visits merely from membership in the filled compact H-hulls. The choice $\rho =\kappa -8$ does not address the critical value eight.

Use the standard definition of a local martingale, including an integrable initial value, and let $T_n\uparrow \infty$ be a localizing sequence. If $X\ge 0$, the Fatou lemma gives $\mathbb{E}{X}_t\le {\mathrm{liminf}}_n\mathbb{E}{X}_{t\wedge T_n}=\mathbb{E}{X}_0<\infty$. For $s\le t$, the Conditional Fatou lemma givesFor clarity, the Conditional Fatou lemma itself follows by setting $U_m={\inf }_{n\ge m}{U}_n$ for nonnegative $U_n$, applying the conditional version of the monotone convergence theorem to $U_m\uparrow \mathrm{liminf}{U}_n$, and observing $\mathbb{E}(U_m\mid {\mathcal{F}}_s)\le {\inf }_{n\ge m}\mathbb{E}(U_n\mid {\mathcal{F}}_s)$. Hence a nonnegative local martingale is a supermartingale.

If $|X_t|\le C$ simultaneously for all $t$, with a deterministic $C<\infty$, the dominated convergence theorem applied to the stopped martingale identity gives, for each $A\in {\mathcal{F}}_s$,This is exactly the conditional expectation identity, so a bounded local martingale is a true martingale. The same proof works for deterministic bounds on every finite time interval; merely having a finite random pathwise bound supplies no dominating integrable random variable and is a different convention.