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This part's conditional-expectation representation is its own hypothesis; it does not need the incorrect unrestricted claim in part (b). Fix $T$, and writeThe assumed representation makes $M(T)$ a martingale. The stochastic Fubini theorem givesApply the Itô formula to $e^{-I}$ and use the equation for $U$ from part (a), including the cross-variation. The result isUniqueness of the continuous semimartingale decomposition makes the drift vanish. Initially this is a $dtd\mathbb{P}$ statement; the assumed continuity in time and maturity extends it to the continuous versions simultaneously. Let $T\downarrow t$ to obtainSubstitute back and differentiate the maturity integrals using their continuous integrands:This is the forward drift restriction for square-root stock claims. The $-{\sigma }_t/2$ term comes from the product cross-variation and must be retained. For the uninformative zero-stock case, the representation does not identify $f$; as usual a positive initial stock price is understood.

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.

Differentiating the European call option price in strike gives, for $T>0$,The strike derivative uses dominated convergence; the second uses the continuous density. Also,Differentiate the supplied time-integral identity and the discount factor. Continuity of the density supplies the diffusion-term derivative. For the tail first moment, continuity in time follows from continuous stock paths, locally uniformly bounded second moments, and the absence of an atom at $K$. ThereforeHence the Dupire equation isIts initial condition is $C(0,K)=(S_0-K{)}^{+}$; natural strike boundaries are $C(T,0)=S_0$ and $C(T,K)\to 0$ as $K\to \infty$. These are consistent with the discounted stock martingale and integrable tails. Where $C_{KK}>0$, the same identity gives local volatility recovery from call prices:The equation evolves in maturity and strike, unlike the backward option-value equation in calendar time and spot.

The Brownian martingale representation theorem argument in part (a) also replicates the bounded put payoff. The discounted stock is a true martingale, so the elementary terminal payoff identity yields put-call paritySet $H(T,K)=-S_0+K{e}^{-rT}$. Then $H_T=-rK{e}^{-rT}$, $H_K=e^{-rT}$, and $H_{KK}=0$. ThusLinearity of the Dupire equation and $P=C+H$ giveThe initial payoff is $P(0,K)=(K-S_0{)}^{+}$, and $P(T,0)=0$. Therefore calls and puts obey the same maturity-strike differential equation, with their respective initial and boundary data.

Use the natural urn filtration. At stage $n$, put $s=X_n+Y_n=n+2$ and $d=X_n-Y_n$. The difference becomes $d-1$ with probability $X_n/s$ and $d+1$ with probability $Y_n/s$. Hence its conditional next mean is $d+(Y_n-X_n)/s=d(1-1/s)$. Since the next total is $s+1$,Adaptedness is immediate, and every $M_n$ is bounded by a deterministic finite number, so integrable. This proves the martingale property.

Nevertheless it does not converge to a finite limit on any sample path. The increments are $d+s$ or $d-s$, because $M_{n+1}=s(d\pm 1)$ and $M_n=(s-1)d$. Both colours remain present, so $|d|\le s-2$ and $|M_{n+1}-M_n|\ge 2$. Convergence of a real sequence would force its increments to tend to zero. There is therefore no almost-sure finite convergence, illustrating why the boundedness condition in the preceding part matters.

Conditional expectation makes $Y_n$ measurable with respect to ${\mathcal{F}}_n$, and conditional Jensen gives $E|Y_n|\le E|Y|<\infty$. The tower property of conditional expectation for nested sigma-algebras givesThus this is a martingale. Its finite limit exists almost surely for every integrable $Y$ under the stated increasing filtration: the martingale convergence theorem applies to the uniform bound on $E|Y_n|$. In fact the family is uniformly integrable. One way to see this is to approximate $Y$ in $L^1$ by a bounded variable, whose conditional expectations remain bounded, while the conditional expectations of the error have uniformly small $L^1$ norm. Consequently convergence is also in $L^1$, with limit $E[Y\mid \sigma ({\bigcup }_n{\mathcal{F}}_n)]$. Mere martingale status without a suitable bound would not imply convergence.

A martingale relative to a filtration $({\mathcal{F}}_n)$ is an adapted sequence of integrable random variables with $E[X_{n+1}\mid {\mathcal{F}}_n]=X_n$ almost surely for every $n$. Adapted means $X_n$ is ${\mathcal{F}}_n$-measurable. Iterating gives $E[X_m\mid {\mathcal{F}}_n]=X_n$ for all $m\ge n$.

The almost sure submartingale convergence theorem says that if a submartingale $(Y_n)$ satisfiesthen there is an integrable random variable $Y_{\infty }$ such thatIndeed the positive-part bound also bounds the negative parts, since $\mathbb{E}{Y}_n\ge \mathbb{E}{Y}_0$; the theorem gives a finite limit. In particular, the Martingale convergence theorem applies to any martingale with ${\sup }_n\mathbb{E}|M_n|<\infty$.

For a nonnegative martingale, $\mathbb{E}{M}_n=\mathbb{E}{M}_0$ supplies this bound automatically. Its limit obeys $\mathbb{E}{M}_{\infty }\le \mathbb{E}{M}_0$ by the Fatou lemma. Almost sure convergence alone need not preserve the expectation or give convergence in L1. Uniform integrability is the additional condition that supplies convergence in L1 and the identity $M_n=\mathbb{E}[M_{\infty }\mid {\mathcal{F}}_n]$.

Take independent and identically distributed random variables ${\xi }_k$ with the Bernoulli distribution of parameter $1/2$, and their natural filtration. SetThe event at $n=0$ is the whole sample space. Independence givesso this is a nonnegative martingale; it is integrable with $\mathbb{E}{M}_n=1$. The first zero among the ${\xi }_k$ is almost surely finite, since the probability of having only ones through time $n$ is $2^{-n}\to 0$. After that first zero, every $M_n$ is zero. Hence the almost sure limit is zero:This fair-coin doubling martingale also exhibits failure of uniform integrability: convergence in L1 would force the expectations to converge to zero.

Let $({\mathcal{F}}_n)$ be the natural filtration of the random walk on a graph. For a bounded harmonic function on a graph, the transition rule and the Markov property giveThus $h(X_n)$ is a bounded martingale. The Martingale convergence theorem gives a finite almost sure limit $L$.

On the probability-one recurrence event, every vertex $v$ is visited along a sequence $n_k\to \infty$. Along that sequence $h(X_{n_k})=h(v)$, while convergence of the whole sequence forces $h(v)=L$. Therefore every pair of vertices has the same value. This is the bounded harmonic function theorem on a recurrent graph:Recurrence makes each vertex value a subsequential limit of the same convergent martingale. A connected locally finite graph is countable, since it is the union of its finite distance balls, so these almost sure assertions can also be combined by a countable intersection. The one-vertex case is immediate.

Write $D_k=M_k-M_{k-1}$ for the martingale difference sequence. Bounded increments mean that one deterministic $C<\infty$ satisfies $|D_k|\le C$ for every $k$, almost surely. Then $\mathbb{E}{D}_k=0$ and, for $j<k$,The increments are uncorrelated random variables, uniformly bounded in $L^2$. Applying the supplied strong law for uniformly L2-bounded uncorrelated random variables gives $n^{-1}{\sum }_{k=1}^n{D}_k\to 0$ almost surely. Also $M_0/n\to 0$, because $M_0$ is finite almost surely. Therefore a martingale with bounded increments has zero linear growth:This is the strong law for martingales with bounded increments.

For the submartingale, let $\Delta Y_k=Y_k-Y_{k-1}$ and define its predictable drift and compensator byThis is the Doob decomposition in discrete time. The process $N_n$ is a martingale. Since $|\Delta Y_k|\le C$, we have $0\le a_k\le C$, so $|N_k-N_{k-1}|\le 2C$. The result just proved gives $N_n/n\to 0$ almost surely, and $A_n\ge 0$. Consequently the strong law for submartingales with bounded increments isThe compensator may grow, so the conclusion is a lower bound rather than a claim of convergence to zero.

For each deterministic $n$, use the nonnegativity and the martingale property to obtainA nonnegative random variable with zero expectation is zero almost surely. Hence $M_{n+1}=0$ almost surely on $\{M_n=0\}$. Intersecting these probability-one statements over all $n$ makes them hold simultaneously, and induction givesZero is absorbing for a nonnegative martingale. This is the absorption at zero of a nonnegative martingale. On $\{T=\infty \}$ the assertion has no finite-time requirement. No unbounded-time use of the optional stopping theorem is needed.

The overshoot hypothesis as printed cannot hold for every $R>0$. For $R\le 1$, $T_R=0$ and $M_{T_R}=1$; choosing $R<1/\tau$ contradicts $M_{T_R}\le \tau R$. The intended and sufficient assumption is the overshoot bound for $R\ge 1$, on the event $\{T_R<\infty \}$. We use those thresholds below; when $R\le 1$, the requested stopped-time bound is trivial because $T\wedge T_R=0$.

Fix $R\ge 1$ and set $S=T\wedge T_R$. By part (a), the stopped martingale satisfies $0\le M_{S\wedge n}\le \tau R$: before stopping it is below $R$, at $T$ it is zero, and at $T_R$ the assumed overshoot bound applies. Thus the square of this stopped process is integrable. For $\Delta M_n=M_{n+1}-M_n$,The event $\{S>n\}$ is ${\mathcal{F}}_n$-measurable and contained in $\{T>n\}$. Conditional expectation of the linear term is zero, while the variance assumption makes the remaining conditional expectation nonnegative. The stopped increments here are bounded by the stopping and overshoot bounds, so all conditional expectations are legitimate. ConsequentlySince $X_0=1$, we have ${\sigma }^2\mathbb{E}(S\wedge n)\le \mathbb{E}{M}_{S\wedge n}^2-1$. Bounded-time optional stopping theorem also gives $\mathbb{E}{M}_{S\wedge n}=1$, soApply the monotone convergence theorem to $S\wedge n$. This proves the stronger estimate $(\tau R-1)/{\sigma }^2$ and hence the requested absorption-time bound from conditional variance and overshoot:In particular this stopping time is finite almost surely. For $0<R\le 1$, $S=0$ and $X_n=1$ is a constant submartingale, while the stopping-time expectation is zero.

First take finite endpoints $\ell <r$ and a starting point $z\in (\ell ,r)$. For the Brownian exit time $H=\inf \{t:B_t\notin (\ell ,r)\}$, the stopped path stays in $[\ell ,r]$. The process $(B_t-z{)}^2-t$ is a martingale. Bounded-time optional stopping theorem givesBy the monotone convergence theorem, $\mathbb{E}H<\infty$, so exit from every bounded interval is almost sure:For completeness, a proper unbounded interval also has almost sure exit. Applying optional stopping theorem to $B_{H\wedge t}$ and then bounded convergence theorem gives the endpoint probabilitiesFor a fixed lower level $\ell$, let $r\to \infty$. Hitting $\ell$ before $r$ is a subset of ever hitting $\ell$, and its probability tends to one. Similarly every fixed higher level is hit almost surely. After a level is reached, the same argument from that level shows that a level strictly outside a given proper interval is eventually reached. This proves exit even when endpoints are included. A starting point already outside needs no argument. The whole real line has no exit, so the phrase “every interval” is understood to exclude that trivial exception. Countably many rational levels make the exit conclusion simultaneous for all proper intervals. The bounded-interval assertion is the one used below.

The domain is the union of the horizontal and vertical strips. The four connected components of its boundary carry the constants $a,b,c,d$ in the northeast, northwest, southwest and southeast, respectively.

First justify finiteness of $T$. A small open ball around $(2,2)$ lies outside the domain. By part (b), planar Brownian motion visits that ball almost surely in finite time. A continuous path from $z\in D$ to that ball must meet $\partial D$ first. Hence $T<\infty$ almost surely, and $B_T\in \partial D$ by continuity.

The harmonic function $u$ is smooth inside $D$. The Itô formula, localized on relatively compact subdomains, makes $u(B_{t\wedge T})$ a local martingale. Its boundary definition uses the continuous extension of $u$. Since $u$ is bounded, this stopped process is a true martingale with uniform integrability: pass through the compact exhaustion using bounded convergence theorem. In particular ${\mathbb{E}}_{z}u(B_{t\wedge T})=u(z)$.

Now $B_{t\wedge T}\to B_T$ almost surely as $t\to \infty$. The dominated convergence theorem, boundedness of $u$, and its continuous boundary values give the bounded harmonic representation with almost sure Brownian exit:Almost sure exit and boundedness of the harmonic function are both used. The domain's unboundedness causes no loss of expectation here because the stopped function values are uniformly bounded.