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For integer paths with increments in $\{-1,0,1\}$, the positive part obeys a telescoping identity with slope $f(x)=1_{\{x>0\}}+\frac{1}{2}{1}_{\{x=0\}}$ and a correction $\frac{1}{2}{1}_{\{S_{t-1}=K\}}(\Delta S_t{)}^2$ at the kink. For a martingale, the slope term is a martingale transform.

Put $t_j=j{2}^{-n}$ and ${\Delta }_j=X_{t_{j+1}}-X_{t_j}$. Telescoping $X_{t_{j+1}}^2-X_{t_j}^2=2X_{t_j}{\Delta }_j+{\Delta }_j^2$ givesThe martingale transform summands are orthogonal in $L^2$: for an earlier summand, conditioning on the sigma-algebra at the start of the later increment makes the cross expectation zero. HenceThe martingale increments themselves are also orthogonal. Since $X_0=0$, their variance sum equals $\mathbb{E}{X}_1^2\le C^2$. ThereforeOnly discrete martingale orthogonality is used here; no pre-existing quadratic variation calculation is needed.

For $n>m$, define $r_j=\lfloor j{2}^{m-n}\rfloor 2^{-m}$ and $h_j=X_{j{2}^{-n}}-X_{r_j}$. Each $h_j$ is measurable at time $j{2}^{-n}$, so the summands in the given difference formula are orthogonal martingale transforms. ThereforeLetPath continuity on the compact interval gives ${\omega }_m\to 0$ almost surely, and ${\omega }_m\le 2C$. Since $0\le j{2}^{-n}-r_j<2^{-m}$, $|h_j|\le {\omega }_m$. Thus, by the Cauchy-Schwarz inequality and part (b),The last step is the dominated convergence theorem. The bound is uniform in $n>m$, and the other ordering follows by symmetry. Hence the terminal martingale transforms are Cauchy in $L^2$.

Predictability means $K_s$ is ${\mathcal{F}}_{s-1}$-measurable. If $|K_s|\le L$, each product $K_s(M_s-M_{s-1})$ is integrable, and the finite sum defining $Y_t$ is integrable. Pulling the bounded predictable factor out of the conditional expectation givesTherefore the bounded predictable martingale transform $Y$ is a martingale starting at zero.

Stop just before a large predictable coefficient would be used. Setwith the infimum of the empty set equal to infinity. Because $K_{t+1}$ is ${\mathcal{F}}_t$-measurable, ${\sigma }_j$ is a stopping time. The increment of the stopped martingale transform isIts coefficient is predictable and bounded by $j$: the first coefficient exceeding $j$ occurs at the step after stopping, and is never included. Part (c) makes $Y^{{\sigma }_j}$ a martingale. Since the finitely many $K_s$ on any fixed finite horizon are finite almost surely, ${\sigma }_j\uparrow \infty$ almost surely. ThusThis is predictable-coefficient localization of a martingale transform. Stopping after taking the large increment would not give the required bound.

On a finite sample space, all real-valued holdings over the finite interval $0,\dots ,T$ are bounded after null states are discarded. The gains identity is therefore a bounded predictable martingale transform of the vector martingale $P$, summed over its coordinates. It follows that $X$ is a martingale, soHere the replication cost is a prescribed deterministic initial capital, as in the definition of attainability. The stronger intermediate identity is $X_t=\mathbb{E}[\xi \mid {\mathcal{F}}_t]$. If initial capital is instead allowed to be ${\mathcal{F}}_0$-measurable and random, the corresponding statement is $X_0=\mathbb{E}[\xi \mid {\mathcal{F}}_0]$; its unconditional expectation still equals $\mathbb{E}\xi$.

Write $z^{-}=\max (-z,0)$. For $a<b$, let $U_N[a,b]$ be the upcrossing count obtained by successively buying at an observation at or below $a$ and selling at an observation at or above $b$, with the first purchase allowed at time zero. The Doob upcrossings lemma for an integrable discrete-time martingale isThe weaker last bound is also the usual Doob upcrossing inequality.

To prove it, hold one unit between each purchase and sale, and zero units otherwise. If $H_k$ records the holding over $(k,k+1]$, the decision is ${\mathcal{F}}_k$-measurable, with $H_k\in \{0,1\}$. The resulting martingale transform has gainEach completed upcrossing earns at least $b-a$. An unfinished holding loses at most $(X_N-a{)}^{-}$. If a purchase occurs at zero, its price is $X_0\le a$, giving the additional discount $(X_0-a{)}^{-}$, whether that holding has been sold or remains open. If no purchase occurs at zero, this extra term is zero. Therefore, path by path,Taking expectations proves the bound.

The Martingale convergence theorem, stated without proof, says that a discrete-time martingale satisfying ${\sup }_n\mathbb{E}|X_n|<\infty$ converges almost surely to a finite integrable random variable $X_{\infty }$. In particular every nonnegative martingale converges almost surely to a finite limit. Under uniform integrability, the uniformly integrable martingale convergence theorem additionally gives convergence in L1 and $X_n=\mathbb{E}[X_{\infty }\mid {\mathcal{F}}_n]$.

For a discrete-time martingale $(M_n)$ and real $a<b$, let $U_N[a,b]$ be the upcrossing count: buy on the first observation at or below $a$, sell on the next observation at or above $b$, and repeat up to time $N$, counting only completed pairs. Writing $z^{-}=max(-z,0)$, the Doob upcrossing inequality isThis convention permits the first purchase at time zero. Equivalently, the right side is $\mathbb{E}(M_N-a{)}^{+}-\mathbb{E}(M_0-a)$, because the martingale has constant expected value.

For completeness, let $H_k\in \{0,1\}$ indicate whether the trading rule holds one unit during $(k-1,k]$. This is a predictable process, so the martingale transform $G_N={\sum }_{k=1}^N{H}_k(M_k-M_{k-1})$ has zero expected value. Every completed trade gains at least $b-a$, and an unfinished trade loses at most $(M_N-a{)}^{-}$. Thus $G_N\ge (b-a)U_N[a,b]-(M_N-a{)}^{-}$, giving the inequality.

Take an increasing localizing sequence ${\tau }_n$ for $X$ and define the stopping timesThe predictable process condition $K_{u+1}\in {\mathcal{F}}_u$ makes $R_n$ a stopping time. For each fixed finite time range, the finitely many coefficients $K_1,\dots ,K_t$ are finite almost surely; hence $T_n\uparrow \infty$ almost surely.

The stopped martingale transform can be writtenIts coefficient is ${\mathcal{F}}_{s-1}$-measurable and bounded in absolute value by $n$. Each summand is consequently integrable with zero conditional expectation given ${\mathcal{F}}_{s-1}$, since $X^{{\tau }_n}$ is a true martingale. Thus $M^{T_n}$ is a true martingale andThe original PDF fixes $M_0=0$; the TeX extraction omits this equality.

For an integer $m$ and $d\in \{-1,0,1\}$, verify the single-step identityIf $m\ge 1$, the positive part is linear across the step and $f(m)=1$; if $m\le -1$, it remains zero and $f(m)=0$. At $m=0$, $d^{+}=(d+d^2)/2$, which is checked for the three permitted values of $d$.

Apply this with $m=S_{t-1}-K$ and $d=S_t-S_{t-1}$, and telescope over $t$. The result is the discrete Tanaka formulaThe first sum is a martingale transform; the second records the correction at the kink of the positive part. The algebraic identity itself needs only the integer values and permitted increments, not the martingale property.

Taking expectations in the single-step discrete Tanaka formula, the martingale transform term vanishes because $f$ is bounded and ${\mathcal{F}}_T$-measurable. HenceAt a state with $p_K=\mathbb{P}(S_T=K)>0$, the tower property of conditional expectation gives $\mathbb{E}[S_{T+1}\mid S_T=K]=K$, so the conditional second moment of the increment is ${\sigma }^2(T,K)$. Part (a) gives $p_K=C(T,K+1)-2C(T,K)+C(T,K-1)$. Combining these identities yields the discrete Dupire equationThe conditional variance needs a positive-probability conditioning state. The bound $|K-S_0|\le T$ does not guarantee $p_K>0$. If $p_K=0$, both the call-price curvature and the time increment above are zero; the equation can be extended by assigning an arbitrary finite value to ${\sigma }^2(T,K)$ at such a state, but its conditional variance is not determined there.

For a finite horizon $N$, let $U_N[a,b]$ be the number of completed upcrossings by time $N$. Use the predictable strategy that holds one unit of the process after a visit below $a$ until the next visit above $b$. For a supermartingale, the expected gain of this nonnegative predictable martingale transform is nonpositive. Pathwise, the completed trades earn at least $(b-a)U_N[a,b]$, while an unfinished final trade can lose at most $(X_N-a{)}^{-}$. HenceTaking expectations and using $X_N\ge 0$ gives the Doob upcrossing inequalityAs $N\to \infty$, monotone convergence yields

Because $H_k$ is previsible, $H_k$ is ${\mathcal{F}}_{k-1}$-measurable. The finite sum ${\hat{X}}_n$ is therefore ${\mathcal{F}}_n$-measurable. Boundedness of $H$ and integrability of the martingale increments implyFinally,Thusso $\hat{X}$ is a martingale. It is the martingale transform of $X$ by $H$.

Localize when the predictable integrand first exceeds level $n$, just before that increment is taken. The resulting stopped integrand is bounded, so its martingale transform is a martingale by the stated result. The localization times increase to infinity because each finite collection of $H_t$ is finite almost surely. Hence $X$ is a local martingale.

The process is adapted because $X_0$ is ${\mathcal{F}}_0$-measurable, $A_k$ is ${\mathcal{F}}_{k-1}$-measurable, and $M_k-M_{k-1}$ is ${\mathcal{F}}_k$-measurable. Using the predictability of $A_n$ and the martingale increment property,The assumed integrability therefore proves that $X$ is the martingale transform of $M$ by $A$ and is a martingale.

A martingale deflator is a strictly positive adapted process $Y$ such that every deflated cum-dividend asset gain has zero conditional drift:Using the definitions of ${\pi }^H$ and ${\xi }^H$,The holdings $H_t$ are ${\mathcal{F}}_{t-1}$-measurable, so the right side is a martingale transform of the deflated asset-gain local martingale. Hence $Z$ is a local martingale.

For a finite-valued predictable process $K$, this is a stopping time and increases to infinity. The stopped martingale transform uses coefficients $K_s{1}_{\{s\le {\sigma }_j\}}$, which are predictable and bounded by $j$. Thus a transform of a discrete-time martingale is a local martingale without a global boundedness assumption. The stop is before, not after, the first large coefficient is used.