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.

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).

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.

The terminal variable $F=\varphi (W_T)$ is bounded and hence square-integrable. In the completed natural Brownian filtration, the Brownian martingale representation theorem says that any square-integrable ${\mathcal{F}}_T$-measurable variable admits a representationwith $\beta$ predictable and $\mathbb{E}{\int }_0^T{\beta }_u^{2}du<\infty$. Extend $\beta$ by zero after $T$. Thus the requested constant and integrability areExpectation determines $c$ uniquely. If two integrands give the same representation, the Itô isometry gives $\mathbb{E}{\int }_0^{\infty }({\beta }_u-{\tilde{\beta }}_u{)}^{2}du=0$. Consequently $\beta$ is unique up to $d\mathbb{P}du$-almost everywhere equality, rather than pointwise equality at every time. The corresponding integral martingales are indistinguishable.

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.

If $T\le s$, then $\varphi (W_T)K$ is ${\mathcal{F}}_s$-measurable. Independence and centring of the future Brownian increment make the desired left side zero; the time multiplier on the right is zero as well.

Suppose $T>s$. Conditional on ${\mathcal{F}}_s$, write $W_T=W_s+U$ and $W_t-W_s=V$. The pair $(U,V)$ is jointly normal and independent of ${\mathcal{F}}_s$, withApply the supplied Gaussian integration by parts formula to $z\mapsto \varphi (W_s+z)$, treating the known $W_s$ as its parameter. This givesMultiply by the bounded ${\mathcal{F}}_s$-measurable $K$ and use the defining property of conditional expectation. Thus the required expectation identity holds for every $s<t$, including intervals crossing or lying after $T$.

The density $Z=\exp (X(k)-\Vert k{\Vert }^2/2)$ is strictly positive. The Gaussian moment-generating function gives $\mathbb{E}Z=1$, so it defines an equivalent probability measure.

The joint normal distribution of $(X(h),X(k))$, with covariance $⟨h,k⟩$, gives the mixed exponential formulaMultiplying by the normalizing and centring factors therefore yieldsThe characteristic function identifies the answer:This is exponential tilting of an isonormal Gaussian process: the mean shifts by the inner product while its covariance remains unchanged.

For any $r_1,\dots ,r_n\in \mathbb{R}$, the linearity from part (a) givesalmost surely, and the right side has a normal distribution. This is the defining linear-combination criterion for a multivariate normal distribution; singular covariance matrices are allowed.

Passing to the limit in the inner products of the partial sums givesThe passage to the limit is justified by Cauchy-Schwarz inequality and $L^2$ convergence. Equivalently, the vector's characteristic function isThus both joint normality and the complete covariance matrix follow from the Hilbert-space inner product.

For $t>e$, put $\Phi (t)=\sqrt{2t\log \log t}$. Fix $a>1$ and $\epsilon >0$, and setfor $n$ large enough that $a^n>e$. The Brownian reflection principle and the Gaussian tail estimate giveSince $(1+\epsilon {)}^2>1$, the sum of these probabilities is finite. The Borel-Cantelli first lemma implies that, almost surely, for all sufficiently large $n$,Now take $t\in [a^{n-1},a^n]$. The function $\Phi$ is increasing for $t>e$, soHere the positive upper bound for $B_t$ can first be divided by $\Phi (t)$, and the denominator can then be bounded below by $\Phi (a^{n-1})$; this remains valid even when $B_t<0$. Moreover,Thus, for each fixed pair $(a,\epsilon )$,Use the countable choices $a_m=1+1/m$ and ${\epsilon }_m=1/m$, intersect their probability-one events, and let $m\to \infty$. This proves the Brownian upper law of the iterated logarithm:Only large times are involved. The hint's related monotonicity assertion for $\Phi (t)/t$ is also valid eventually: the derivative of $2\log \log t/t$ is $2(1/\log t-\log \log t)/t^2$, which is negative for $t\ge e^e$. No monotonicity at the small-time edge of the logarithmic expression is required.

Sample Brownian motion at integer times and put $Z_n=B_n/\sqrt{n}$. Every $Z_n$ has normal distribution $N(0,1)$, although the $Z_n$ are correlated. LetFor every fixed integer $m$,because $B_m/\sqrt{n}\to 0$ almost surely. The right-hand expression depends only on the future independent increments $B_j-B_{j-1}$ with $j>m$. Thus $\{L\ge a\}$ is, up to a null set, a tail event of those independent increments, and its probability is zero or one by the Kolmogorov zero-one law.

For any finite real $a$, every $Z_n$ exceeds $a$ with the same positive probability $p_a=\mathbb{P}(N(0,1)\ge a)$. For each $N$,Taking the decreasing intersection over $N$ shows that $Z_n\ge a$ infinitely often with probability at least $p_a$. This event implies $L\ge a$, so $\mathbb{P}(L\ge a)>0$. The zero-one law makes it one. Intersecting over positive integers $a$ gives $L=+\infty$ almost surely. The continuous-time limit superior is at least the one along integers, henceThis proves that Brownian fluctuations exceed the square-root scale without needing the lower bound in the law of the iterated logarithm.

Let $W_n$ be the interpolation in part (a). The operationis continuous, since $|I(f)-I(g)|\le \Vert f-g{\Vert }_{\infty }$. By the Donsker invariance principle and the continuous mapping theorem,The exact trapezoidal integral of the linear interpolation isTherefore the statistic in question differs from $I(W_n)$ by $S_n/(2n^{3/2})$. Using independence, zero means, and unit variances givesThis error tends to zero in $L^2$, hence in probability. The Slutsky theorem now proves the integrated random-walk limit:The limiting law can also be made explicit. The time integral is a Gaussian random variable, as a mean-square limit of linear combinations of a Gaussian process. It is centered, and the covariance identity $\mathbb{E}(B_s{B}_t)=\min (s,t)$ givesThus the terminal value of integrated Brownian motion here has law $N(0,1/3)$.

The Skorokhod embedding of a centered random walk states that a random walk with independent identically distributed centered steps of finite variance ${\sigma }^2$ can be realized on an appropriate probability space aswhere $B$ is a standard Brownian motion and the $T_n$ are finite stopping times. More precisely, the stopped positions have the same joint law as the given random walk, and the pairsmay be chosen independent and identically distributed. Their spatial component has the step law, and $\mathbb{E}(T_n-T_{n-1})={\sigma }^2$. Repeating the one-step Skorokhod embedding theorem with the Strong Markov property gives this formulation. In the present normalization, the mean time increment is one, and the strong law of large numbers gives $T_n/n\to 1$ almost surely.

The Donsker invariance principle states that the linearly interpolated diffusively rescaled random walkconverges weakly as a random element of $C[0,1]$, equipped with the uniform norm, to standard Brownian motion restricted to $[0,1]$. At $t=1$ the fractional term is zero. The only step assumptions needed here are zero mean, unit variance, and independent identical distributions; a higher moment or bounded support is not required. This is a functional central limit theorem, concerning the entire interpolated path rather than only its endpoint.

Fix $t$ and put $\sigma =t\wedge \tau$. The stopping time property implies that $\sigma$ is an ${\mathcal{F}}_t$-measurable random variable with values in $[0,t]$: for $s<t$,and for $s\ge t$ the event is the whole space. By part (c), the restriction of $X$ to $[0,t]\times \Omega$ is $\mathcal{B}([0,t])\otimes {\mathcal{F}}_t$-measurable.

The evaluation map $\omega \mapsto (\sigma (\omega ),\omega )$ is measurable from $(\Omega ,{\mathcal{F}}_t)$ to this product space: the inverse image of a measurable rectangle $A\times C$ is $\{\sigma \in A\}\cap C$. Composing it with the jointly measurable stochastic process gives an ${\mathcal{F}}_t$-measurable random variableSince this holds for every fixed $t$, the stopped process is adapted. If path regularity holds only almost surely, first apply the proof to its pathwise regular representative; with a completed filtration, the original stopped variable differs only on a null event and is also ${\mathcal{F}}_t$-measurable. The adaptedness of a stopped right-continuous process requires neither boundedness of $\tau$ nor a martingale assumption; $\tau =\infty$ is harmless because $t\wedge \tau \le t$.

Relative to a filtration $({\mathcal{F}}_t)$, progressive measurability means that, for every $T<\infty$, the mapis measurable for the product sigma-algebra $\mathcal{B}([0,T])\otimes {\mathcal{F}}_T$ and the Borel sigma-algebra on $\mathbb{R}$.

Fix $T>0$ and divide $[0,T]$ into $2^m$ equal subintervals with mesh $h_m=T/2^m$. DefineSince $X$ is adapted, every random variable $X_{k{h}_m}$ is ${\mathcal{F}}_{k{h}_m}$-measurable, hence ${\mathcal{F}}_T$-measurable. Each approximation is consequently $\mathcal{B}([0,T])\otimes {\mathcal{F}}_T$-measurable. For $t<T$, its sampling time lies strictly to the right of $t$, tends to $t$, and never exceeds $T$. Right continuity implies $X_t^{(m)}\to X_t$; at $T$ equality is exact. Thus $X$ is the pointwise limit of measurable functions on this product space. As $T$ was arbitrary, $X$ is progressively measurable. This is the theorem that right-continuous adapted processes are progressively measurable.

The right-endpoint approximations need not themselves be adapted at their intermediate times. What the proof requires is their joint measurability with respect to the single terminal sigma-algebra ${\mathcal{F}}_T$. The proof uses the pathwise càdlàg convention. If path regularity is assumed only almost surely, under a completed filtration setting the stochastic process to zero on its common exceptional null event gives an indistinguishable progressively measurable version. Arbitrary values on that null event need not make the original stochastic process progressively measurable: even with a complete filtration, a null sample point may be assigned a non-Borel time function. This is why almost sure path regularity does not ensure progressive measurability.

For each nonnegative rational $q$, equality of the two versions gives $\mathbb{P}(X_q=Y_q)=1$. Intersect these countably many full-probability events with the full-probability event on which both paths are càdlàg. Call the resulting event $A$; then $\mathbb{P}(A)=1$.

Fix $\omega \in A$ and any real $t\ge 0$. Choose rational numbers $q_n>t$ decreasing to $t$. Right continuity givesThe same event $A$ works for every $t$, because the argument is pathwise after $\omega$ is fixed. The stochastic processes are therefore indistinguishable. This proves that càdlàg versions are indistinguishable; the left limits are not needed for this implication, since right continuity alone suffices.