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Let a simple symmetric random walk start from a deterministic integer $S_0=i$, and put $L_n={\sum }_{k=0}^{n-1}{1}_{\{S_k=0\}}$. Then $M_n=|S_n|-L_n$ is a martingale with $M_0=|i|$. The visit count includes time $0$ if $i=0$.

For every integer $s$ and step $\xi \in \{-1,1\}$, the pathwise identityholds with $\mathrm{sgn}(0)=0$. Subtract the indicator function and sum over time to obtainThe sign function of the current state is a bounded predictable process for the next increment, so the sum is a martingale transform. Alternatively, averaging the two possible next absolute values shows directly that the conditional drift is $1_{\{S_k=0\}}$. Bounds $|S_n|\le |i|+n$ and $L_n\le n$ verify integrability at each finite time.

This is an absolute-value version of the discrete Tanaka formula. The correction records the occasions when the random walk is at the corner of the absolute-value function; away from zero that function is locally linear along a single step.

In the completed natural Brownian filtration, a bounded continuously differentiable $\varphi$ with bounded derivative satisfiesThe integrand is unique up to $d\mathbb{P}dt$-almost everywhere equality. Gaussian integration by parts first gives the duality between the payoff and every square-integrable Itô integral. The Brownian martingale representation theorem then identifies the integrand as the projection of ${\varphi }^{\prime }(W_T)1_{\{t\le T\}}$ onto the predictable processes in the product $L^2$ space. This explains the conditional expectation in the formula, rather than a nonadapted terminal derivative.

For independent return innovations ${\xi }_n$ and a predictable process of holdings ${\theta }_n$, wealth obeysThe investor chooses the holdings to maximize expected utility of terminal wealth.

An elementary predictable process may use finitely many ordered stopping times ${\tau }_j$, with bounded ${\mathcal{F}}_{{\tau }_j}$-measurable coefficients $h_j$ on $({\tau }_j,{\tau }_{j+1}]$. Its elementary stochastic integral is ${\sum }_j{h}_j(X_{t\wedge {\tau }_{j+1}}-X_{t\wedge {\tau }_j})$. The stopped indicators are left-continuous and adapted, so the process is predictable; common refinement shows the sum is independent of its representation. Deterministic endpoints recover the usual simple predictable process. These integrands give the precise good-integrator test in the Bichteler-Dellacherie theorem.

For a filtration $({\mathcal{F}}_t)$, the information strictly before time $t$ isA predictable process evaluated at a deterministic positive time $t$ is ${\mathcal{F}}_{t-}$-measurable.

For a continuous local martingale and a locally bounded predictable process, stop at a localizing time, an integrand bound, a time bound, and level bounds for the centered martingale and its bracket. The stopped martingale is bounded and its bracket and integrand are deterministically bounded. The Itô isometry, applied to $1_F{1}_{(s,t]}H$ for $F\in {\mathcal{F}}_s$, gives the conditional second-moment identity. Expanding the square proves that $(H\cdot M{)}^2-H^2\cdot [M]$ is a martingale after stopping. Uniqueness of an increasing square compensator identifies the bracket, and increasing stopping times patch the identities into the displayed local result.

Let $f(y)=|y|$ away from zero. Its first and second partial derivatives areThus its Laplace operator is $\Delta f(y)=(n-1)/|y|$. The quadratic covariations of the independent Brownian coordinates satisfy $d[B_j,B_k{]}_s={\delta }_{jk}ds$. The stopped Itô formula therefore givesFor $t<\tau$, this is the required integral representation.

To identify the driving Brownian motion, the stopped stochastic term alone is insufficient: its quadratic variation is $t\wedge \tau$. Extend its unit direction after exit. Define a vector predictable process $u(s)$ byPredictability follows from the continuous adapted direction before stopping and the left-continuous adapted indicator $1_{\{s\le \tau \}}$. This $W$ is a zero-starting continuous martingale, andThe Lévy characterization of Brownian motion identifies $W$ as a one-dimensional Brownian motion. Before $\tau$, the preceding representation becomes the stochastic differential equation of the dimension-$n$ Bessel process:The fully stopped equation has both integrals stopped at $\tau$.

For $n=2$, apply the one-dimensional Itô formula to the logarithm. Before $\tau$, $d[X{]}_t=dt$, so the two drift terms cancel:Consequently,Indeed $X_s\ge r$ before stopping, so the Itô isometry bounds the stochastic term's second moment by $t/r^2$, and $\log X_{t\wedge \tau }\in [\log r,\log R]$. This is the martingale assertion under the paper's standing stopping assumption. Without stopping, the same calculation establishes only a local martingale; the stopped qualification is essential for the stated argument.

For a submartingale $(Z_n)$ and $a<b$, let $U_N(a,b)$ count the completed upcrossings of $[a,b]$ by time $N$. The Doob upcrossing inequality givesFor a martingale $(X_n)$, a useful sharper form isHere $x^{+}=\max (x,0)$ and $x^{-}=\max (-x,0)$. To see the martingale form, use a bounded predictable process $H_k\in \{0,1\}$ that holds one unit after an observation at or below $a$ and releases it on the next observation at or above $b$. Its gain $G_N={\sum }_{k=1}^N{H}_k(X_k-X_{k-1})$ is at least $(b-a)U_N-(X_N-a{)}^{-}$: completed trades earn at least $b-a$, and any unfinished trade loses at most $(X_N-a{)}^{-}$. Since $\mathbb{E}{G}_N=0$, the bound follows. For the submartingale version, apply the same strategy to $(Z_n-a{)}^{+}$, itself a submartingale. Each purchase occurs at value zero, so its gain is at least $(b-a)U_N$. The complementary predictable strategy has nonnegative expected gain, bounding this strategy's expected gain by $\mathbb{E}(Z_N-a{)}^{+}-\mathbb{E}(Z_0-a{)}^{+}$, and in particular by the stated right side.

The almost-sure martingale convergence theorem says that if $C={\sup }_n\mathbb{E}|X_n|<\infty$, then $X_n$ converges almost surely to a finite, integrable random variable $Y$, with $\mathbb{E}|Y|\le C$. Indeed, the Doob upcrossing inequality bounds the expected total number of upcrossings of every rational interval by $(C+|a|)/(b-a)$. By monotone convergence, each such total count is finite almost surely; intersect these probability-one events over all rational $a<b$. If the lower and upper limits of a sample path differed, some rational interval strictly between them would be crossed infinitely often. Thus each remaining path has a limit in the extended real line. The Fatou lemma givesso the limit is finite almost surely and integrable. Almost-sure convergence by itself neither preserves expectations nor supplies this uniform first-moment bound.

For the loss of expectation, let $({\eta }_k)$ be independent variables with the fair Bernoulli distribution and setConditional on the past, survival doubles the value with probability $1/2$ and otherwise sets it to zero, so this is a nonnegative martingale. An infinite string of successes has probability zero. The process therefore eventually vanishes almost surely, giving yes: $X_0=1$ and $Y=0$ almost surely are possible. Its expectations remain one at every finite time; the process is not uniformly integrable.

A nonintegrable limit is also possible for a signed martingale. Take independent ${\xi }_k$ withand let $M_0=1$, $M_n=1+{\sum }_{k=1}^n{\xi }_k$. Every increment has mean zero and a finite absolute expectation, so each $M_n$ is integrable and $(M_n)$ is a martingale. Since ${\sum }_k\mathbb{P}({\xi }_k\ne 0)<\infty$, the Borel-Cantelli lemma makes only finitely many increments nonzero almost surely; hence the finite limit $Y=1+{\sum }_k{\xi }_k$ exists almost surely. To compute its absolute expectation, let $A_k$ be the event that only the $k$th increment is nonzero. Independence givesThe infinite product is positive because its tail logarithms are bounded in absolute value by a constant times the summable series ${\sum }_j{2}^{-j}$. The events $A_k$ are disjoint, and on $A_k$, $|Y|\ge 4^k-1$. ThereforeThus yes: a finite almost-sure martingale limit can have infinite absolute expectation, even with deterministic $M_0=1$. This is almost-sure martingale convergence to a nonintegrable limit; the missing hypothesis is the uniform first-moment bound. For a nonnegative martingale, in contrast, the Fatou lemma would force the limit to be integrable.

For a simple predictable process, define its stochastic integral byCommon refinement of the deterministic partitions makes this definition independent of the representation. Each summand is a martingale increment multiplied by information already available at its left endpoint, so the resulting process is a continuous square-integrable martingale starting at zero. Its terminal value is the same sum with $t$ replaced by $t_n$.

Write ${\Delta }_{k}M=M_{t_{k+1}}-M_{t_k}$. If $j<k$, then $Z_{t_j}{\Delta }_{j}M{Z}_{t_k}$ is ${\mathcal{F}}_{t_k}$-measurable. By the conditional expectation identity for increments of a martingale,The conditional bracket isometry for stopped martingale increments givesFor completeness, this identity follows by applying the optional stopping theorem to $M$ and $M^2-[M]$ and expanding the square. The latter process is a true uniformly integrable martingale here: the Doob L2 maximal inequality bounds $\mathbb{E}{\sup }_t|M_t{|}^2$, stopped square identities and Fatou's lemma give $\mathbb{E}[M{]}_{\infty }<\infty$, and its absolute supremum is bounded by ${\sup }_t|M_t{|}^2+[M{]}_{\infty }$. Consequently all the stopped identities pass to the limit.

Multiplying the conditional identity by $Z_{t_k}^2$, taking expectations, and using the vanished cross terms proves the Itô isometry:

To extend the stochastic integral, use the finite quadratic-variation measure ${\nu }_M(C)=\mathbb{E}\int 1_{C}d[M]$ on the predictable sigma-algebra. The density of simple predictable processes for finite measures gives approximations to every $H\in L^2({\nu }_M)$. The Itô isometry makes the corresponding terminal integrals Cauchy in $L^2$, and the Doob L2 maximal inequality makes the whole processes Cauchy in expected squared supremum. Their limit defines a continuous martingale, and the isometry persists. For a locally bounded predictable process, choose increasing stopping times tending to infinity on which it is bounded, also restrict the time horizon, and perform this construction after stopping. The isometry shows that the stopped definitions agree on overlaps, so they patch into a uniquely defined continuous local martingale $H\cdot M$.

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 $c{S}_0{e}^{-m+v/2}$.

For an elementary previsible process $H={\sum }_{j=0}^{r-1}{H}_j{1}_{(t_j,t_{j+1}]}$, with bounded ${\mathcal{F}}_{t_j}$-measurable coefficients, defineThe Itô isometry isMore generally the squared $L^2$ distance between the integrals of $H$ and $K$ is the integral of $(H-K{)}^2$ on the right. The integral is a zero-starting continuous square-integrable martingale.

Elementary predictable processes are dense in $L^2(M)$. Given $H$ in this space, choose elementary $H_n$ converging in its norm and define the Itô integral by the $L^2$ limit of $H_n\cdot M$. The isometry proves that this is independent of the approximating sequence and remains valid for the limit. The Doob L2 maximal inequality bounds the expected squared path supremum of the difference by four times its terminal second moment; hence the convergence also gives a continuous martingale version, with uniform convergence on compacts in probability.

For a continuous local martingale $M$ and locally bounded process $H$ that is previsible, choose increasing stopping times ${\tau }_n\uparrow \infty$ such that $M^{{\tau }_n}$ is L2-bounded, $|H|\le K_n$ up to ${\tau }_n$, and ${\tau }_n\le n$. This can be done by combining localizers with first exits of $M$ from bounded intervals and the local bounds of $H$. Then $H{1}_{[0,{\tau }_n]}$ is in the integrand space for $M^{{\tau }_n}$. Define the integral on each stopped interval by the preceding construction. The stopping property of elementary integrals, extended by the isometry, makes these definitions agree on overlaps. Pasting givesThis is a localization construction; global square integrability of the resulting process is not asserted.

A left-continuous adapted process is a predictable process. Define the elementary predictable approximationAt every $s>0$, its sample time approaches $s$ from the left, including when $s$ is itself a grid point. Thus $H_s^{(n)}\to H_s$ pathwise by left continuity.

Use the continuous semimartingale decomposition $X=X_0+M+A$, with $M$ a continuous local martingale and $A$ a continuous finite-variation process. Localize so that on a fixed horizon $T$, the integrands are bounded by a deterministic constant, $[M{]}_T$ is bounded, and the total-variation process of $A$ at $T$ is bounded. This is possible using local boundedness and the finite pathwise clock and variation.

For the finite-variation integral, the dominated convergence theorem givesalmost surely. For the martingale integral, the Itô isometry and the Doob L2 maximal inequality giveHere the localized bounds justify dominated convergence also in expectation. Remove localization to conclude that the full elementary integrals converge in uniform convergence on compacts in probability to $\int HdX$.

The sum of completed intervals differs from the full elementary integral at time $t$ byIts maximum on $[0,T]$ tends to zero almost surely, by the local bound on $H$ and uniform continuity of $X$. ThereforeThis proves left-endpoint approximation of a continuous semimartingale integral, including the unfinished interval convention.

Let $\tau =T-t>0$ and $I_{t,T}={\int }_t^T{\sigma }_s^{2}ds$. The stochastic exponential solution givesBoundedness of the predictable process $\sigma$ implies the Novikov condition on every finite horizon, so this exponential has conditional expectation $\mathbb{E}[M_{t,T}\mid {\mathcal{F}}_t]=1$. The pathwise bounds $a^2\tau \le I_{t,T}\le b^2\tau$ and positivity of $M_{t,T}$ giveThis proves the required bound in the independent-volatility case, and the exponential argument in fact does not need independence. In the usual filtration generated by Brownian motion and an independent volatility history, one can alternatively condition on the whole volatility path: the Brownian integral is then conditionally Gaussian with variance $I_{t,T}$. The result isthe conditional square-root price under independent volatility. The outer conditional expectation cannot in general be removed.

On $\{S_t>0\}$, these positive price bounds justify the unique square-root stock implied volatilityThere is a genuine zero-price qualification: if $S_t=0$, the nonnegative stock is absorbed at zero and both claim prices are zero for every volatility parameter. The price equation then does not determine a unique $\Sigma$. One may set $\Sigma =a$ on that event to retain the bounds; identification by inversion requires $S_t>0$.

First take a bounded elementary predictable process $\alpha ={\sum }_j{K}_j{1}_{(s_j,t_j]}$, with each $K_j$ bounded and ${\mathcal{F}}_{s_j}$-measurable and with finite time support. The Itô integral is the corresponding finite sum ${\sum }_j{K}_j(W_{t_j}-W_{s_j})$. Applying part (a) term by term givesSuch elementary predictable processes are dense among predictable processes in $L^2(d\mathbb{P}du)$. The Itô isometry makes the left functional continuous, with boundThe Cauchy-Schwarz inequality makes the right functional continuous, with bound $\Vert {\varphi }^{\prime }{\Vert }_{\infty }\sqrt{T}{\left(\mathbb{E}\int {\alpha }_u^{2}du\right)}^{1/2}$. Approximation therefore proves the same identity for every allowed predictable $\alpha$. The integral over the infinite time interval is the $L^2$ limit of its finite-horizon Itô integrals.

Multiply the representation in part (c) by ${\int }_0^{\infty }{\alpha }_{u}d{W}_u$ and take expectations. This Itô integral has mean zero, and the bilinear form of the Itô isometry givesEquating this with part (b), and writing both ordinary integrals with the same time variable, yieldsAll terms are integrable by the Cauchy-Schwarz inequality, the assumed square integrability of $\alpha$, and boundedness of ${\varphi }^{\prime }$. This is an orthogonality statement against predictable processes; its second term need not itself be predictable.

Set $Z_t=\mathbb{E}[{\varphi }^{\prime }(W_T)\mid {\mathcal{F}}_t]$. The Brownian martingale representation theorem applied to the bounded terminal variable ${\varphi }^{\prime }(W_T)$ supplies a continuous adapted version of $Z$, so it is predictable. Henceis predictable and satisfies $\mathbb{E}\int {\gamma }_t^{2}dt\le T\Vert {\varphi }^{\prime }{\Vert }_{\infty }^2$. For every square-integrable predictable process $\alpha$, conditioning at each deterministic time and using Fubini theorem givesThus part (d) says $\mathbb{E}\int ({\beta }_t-{\gamma }_t){\alpha }_{t}dt=0$. Taking $\alpha =\beta -\gamma$, which is an allowed predictable square-integrable process, makes its squared norm zero. We concludeThis is the Clark-Ocone formula for a smooth Brownian terminal payoff, with exactly the uniqueness established in part (c).

We prove the stronger finite branching bound in a complete market: modulo null sets, ${\mathcal{F}}_t$ is generated by at most $n^t$ atoms. Here an atom is a positive-probability event which cannot be split into two positive-probability measurable pieces.

At time zero there is one atom because ${\mathcal{F}}_0$ is trivial. Suppose ${\mathcal{F}}_{t-1}$ has at most $n^{t-1}$ atoms, and fix one such atom $A$. Any predictable process holdings over $(t-1,t]$ are constant on $A$, so the restriction to $A$ of every time-$t$ replicable payoff lies ina vector space of dimension of a vector space at most $n$. If $A$ contained $n+1$ disjoint positive-probability ${\mathcal{F}}_t$ events, their indicators would have linear independence on $A$. Market completeness would replicate each indicator, contradicting that dimension bound.

For clarity, the absence of $n+1$ disjoint positive pieces implies that $A$ is a union of at most $n$ atoms: start with $A$ and repeatedly split any non-atom into two positive pieces. Each split increases the count by one. The process must stop before the count exceeds $n$, and at termination every piece is an atom. Thus each time-$(t-1)$ atom has at most $n$ successors. Induction gives at most $n^t$ atoms at time $t$.

Each disjoint positive-probability ${\mathcal{F}}_t$ event contains at least one distinct atom. ThereforeThe argument includes $t=0$ and uses the ability to replicate the indicator claims at every intermediate maturity.

For the simple predictable process from part (a), define its stochastic integral byThe time-zero value contributes nothing. Write $D_i=M_{t\wedge t_{i+1}}-M_{t\wedge t_i}$. If $i<j$ and $t>t_j$, then $h_i{D}_i{h}_j$ is ${\mathcal{F}}_{t_j}$-measurable and $\mathbb{E}[D_j\mid {\mathcal{F}}_{t_j}]=0$. If $t\le t_j$, then $D_j=0$. Boundedness ensures integrability, so all off-diagonal terms vanish by conditional expectation. ConsequentlyThe quadratic variation theorem says $M^2-[M]$ is a local martingale. A bounded continuous martingale is square-integrable, and localization gives the conditional second-moment identityMultiplication by $h_i^2$ and summation therefore give the Itô isometryFor Brownian motion, $[M{]}_s=s$. More generally the Itô isometry holds for every predictable process with finite right-hand side, against a continuous square-integrable martingale. Indeed, on a fixed horizon the quadratic-variation measure ${\mu }_M(E)=\mathbb{E}{\int }_0^t{1}_{E}d[M]$ is finite. Part (c) approximates the integrand by simple predictable processes; the displayed equality makes the elementary integrals Cauchy in $L^2$ and defines their unique limit. The finite sum is exactly the elementary case of the Itô isometry.

The reference Brownian motion must start at $x$, and both laws must use the same stopped-path rule. Under the reference probability measure let $R_s=x+W_s$, let $\tau$ be its exit time from $(a,b)$, and put $T=t\wedge \tau$ and $c=(\delta -1)/2$. The Girsanov density for a stopped Bessel process, as a functional of the stopped reference path, isUse the predictable process ${\theta }_s=c{R}_s^{-1}{1}_{\{s\le \tau \}}$, with any harmless choice at the endpoint. On $[0,t]$, $\int {\theta }_s^{2}ds\le c^{2}t/a^2$. The Novikov condition holds; under the density $D$, the Girsanov theorem makes ${\tilde{W}}_s=W_s-{\int }_0^s{\theta }_{u}du$ a Brownian motion. Before exit, $d{R}_s=c{R}_s^{-1}ds+d{\tilde{W}}_s$, which is the required stochastic differential equation. The drift is a locally Lipschitz function on $(0,\infty )$, so pathwise uniqueness and uniqueness in law hold up to this exit time.

For a version involving just the endpoint and an ordinary integral, the Itô formula for $\log R$ givesThusThe stopped density is measurable from the stopped path, so it is also the Radon-Nikodym derivative of the stopped-path laws. The stopping times in this formula are evaluated on the reference path, rather than taken from a different realization of $X$. A zero-start reference Brownian motion would have a different deterministic initial value and could not dominate this law.

A simple predictable process has the form $H_s={\sum }_{j=0}^{m-1}{h}_j{1}_{(t_j,t_{j+1}]}(s)$, where $0=t_0<\cdots <t_m<\infty$ are deterministic and $h_j$ is bounded and ${\mathcal{F}}_{t_j}$-measurable. A possible ${\mathcal{F}}_0$-measurable value at zero is irrelevant to this Itô integral. DefineEach coefficient is known before the corresponding increment. The conditional expectation of each subsequent increment is zero, so the result is a continuous martingale starting at zero. It is an L2-bounded continuous martingale because it is a finite sum of bounded coefficients times stopped increments of an L2-bounded continuous martingale. This definition is independent of the chosen subdivision: splitting an interval simply splits its increment into a telescoping sum. General admissible predictable processes are then integrated by completion using the Itô isometry; unbounded step coefficients are allowed when the weighted $L^2$ condition in the quadratic-variation measure in the next part holds.