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.

A version of a stochastic process means a stochastic process on the same probability space such that, for every fixed $t$,The exceptional null set may depend on $t$. Indistinguishability of stochastic processes means that there is one null set outside which $X_t=Y_t$ for all $t\ge 0$ simultaneously.

For an example separating the definitions, let $U$ have uniform distribution on $(0,1)$, and setFor every fixed $t$, $\mathbb{P}(U=t)=0$, so $Y$ is a version of a stochastic process with original stochastic process $X$. But for every sample outcome the stochastic processes differ at its time $t=U$. HenceThe spike path of $Y$ is not right-continuous at $U$, which explains why the next part's regularity assumption rules out this example.

Write $\tau ={\tau }_x$ and $T_t=t\wedge \tau$. Path continuity gives $|B_{T_t}|\le x$. Stop the martingale $B_s^2-s$ at the bounded stopping time $T_t$ and apply the optional stopping theorem:By the monotone convergence theorem, $\mathbb{E}\tau \le x^2$, so $\tau <\infty$ almost surely. Path continuity then gives $B_{\tau }\in \{-x,x\}$. The dominated convergence theorem for the bounded variables $B_{T_t}^2$ shows

Next stop the quartic Hermite polynomial martingale from part (a), again only at $T_t$. Its expectation is zero, soMonotone convergence proves $\mathbb{E}{\tau }^2\le 2x^4$, establishing the needed second-moment integrability before the final passage to the limit. Since $T_t{B}_{T_t}^2\le x^2\tau$ and $B_{T_t}^4\le x^4$, dominated convergence givesConsequently the Brownian symmetric interval-exit moments are

To obtain the Laplace transform of symmetric Brownian interval-exit time, put $a=\sqrt{2\lambda }$. The Exponential martingale for Brownian motion shows thatis a martingale with $H_0=1$. Bounded-time stopping gives $\mathbb{E}{H}_{T_t}=1$. Its stopped values are bounded by $\cosh (ax)$, so dominated convergence applies as $t\to \infty$. Since $\cosh (a{B}_{\tau })=\cosh (ax)$, it yieldsEvery use of stopping at $\tau$ has thus been justified through bounded stopping and an explicit integrability or domination argument.

Use the independent increments of Brownian motion, rather than merely checking that an Itô formula drift vanishes. For $0\le s<t$, put $h=t-s$ and write $B_t=B_s+Z$, where $Z$ is independent of ${\mathcal{F}}_s$ and has normal distribution $N(0,h)$. Its first four moments are $0,h,0,3h^2$. ThereforeFor the cubic expression, the coefficient of $B_s$ after conditioning is $3(t-s)+\alpha (t)$, so $\alpha (t)=-3t$ makes it $\alpha (s)$. For the quartic expression, choose $\beta (t)=-6t$. Its conditioned coefficient of $B_s^2$ is then $6(t-s)-6t=-6s$. The constant term becomeswhich equals $3s^2$ when $\gamma (t)=3t^2$. Thus a standard choice isThe resulting stochastic processes are Hermite polynomial martingales $H_3(B_t,t)$ and $H_4(B_t,t)$. They are genuine integrable martingales, since Gaussian moments are finite at every finite time and the displayed conditional identities establish the martingale property directly.

The choice is not unique. Constants $c,d,e\in \mathbb{R}$ give the valid family $\alpha (t)=c-3t$, $\beta (t)=d-6t$, and $\gamma (t)=3t^2-dt+e$: these add $c{B}_t$ to the cubic martingale and $d(B_t^2-t)+e$ to the quartic one. The boxed choice sets these harmless additions to zero.

Take two independent rate-one Poisson processes $N^{+}$ and $N^{-}$, and letThe difference of independent Poisson processes starts at zero and has stationary increments and independent increments. Its paths are càdlàg. For an interval of length $h$, the probability of any jump is $1-e^{-2h}\to 0$, proving stochastic continuity. Thus $X$ is a Lévy process. The characteristic function of a Poisson distribution with mean $t$ is $\exp (t(e^{iu}-1))$, so independence givesThe sample paths are integer-valued step functions with jumps $+1$ or $-1$. On every bounded interval there are only finitely many jumps, and independent Poisson arrival times coincide with probability zero. The combined arrival rate is two: holding times are independent exponentials of rate two, and each jump direction has probability $1/2$, independently of the holding times. This is equivalently a Compound Poisson process of rate two with Rademacher distribution jump sizes. Its paths have finite variation on compact time intervals, although there are infinitely many jumps over the whole half-line almost surely.