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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.

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.

Fix a L2-bounded continuous martingale $M$. Its quadratic-variation measure on the predictable sigma-algebra isIt is finite, since ${\nu }_M(\Omega \times [0,\infty ))=\mathbb{E}[M{]}_{\infty }=\mathbb{E}{M}_{\infty }^2-\mathbb{E}{M}_0^2$. The source Hilbert space is $L^2(M)=L^2(\Omega \times [0,\infty ),\mathcal{P},{\nu }_M)$: predictable processes with finite $\mathbb{E}\int H^{2}d[M]$, identified when equal ${\nu }_M$-almost everywhere. Its norm is $\Vert H{\Vert }_{L^2(M)}^2=\mathbb{E}\int H^{2}d[M]$.

The target ${\mathcal{M}}_0^2$ consists of L2-bounded continuous martingales starting at zero, identified up to indistinguishability of stochastic processes, with norm $\Vert K{\Vert }_{{\mathcal{M}}_0^2}=\Vert K_{\infty }{\Vert }_2$. It is a Hilbert space: terminal values belong to the closed linear subspace of $L^2({\mathcal{F}}_{\infty })$ whose conditional expectation given ${\mathcal{F}}_0$ is zero and whose associated martingales have continuous versions. Closure of that continuous-version subspace follows from the Doob L2 maximal inequality and an almost surely uniformly convergent subsequence.

The Itô isometry says that integration extends uniquely from simple predictable processes to a linear isometry $L^2(M)⟶{\mathcal{M}}_0^2$, withFor every finite $t$, the same identity holds with the integral restricted to $[0,t]$ and the left side $\mathbb{E}|(H\cdot M{)}_t{|}^2$. No claim of surjectivity onto all of ${\mathcal{M}}_0^2$ is needed: that would require an additional Martingale representation theorem.

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.