Lemma 1.1.6: {{\cite[Equation 8.1]{stroock}}} | Brownian Motion and Stochastic Calculus

Lemma 1.1.6 ([Equation 8.1, Str24]).label Let $\wien: \cb(\sps) \to [0, 1]$ be the classical Wiener measure on $(\sps, \cb(\real^{d}))$, then for every $\lambda \in \sps^{*}$,

(1)
$\lambda$ viewed as a random variable on $\sps$ is a centred Gaussian.
(2)
$\ev^{\wien}[\abs{\lambda}^{2}] = \iint_{[0, \infty)^2}s \wedge t d\lambda(s)d\lambda(t)$.
(3)
$\wh \wien(\lambda) = \exp\braks{-\frac{1}{2}\iint_{[0, \infty)^2}s \wedge t d\lambda(s)d\lambda(t)}$.

Proof. $(1)$, Finite: First suppose that there exists $0 = t_{0} \le t_{1} < \cdots < t_{n} < \infty$ and $\seqf{\xi_j}\subset \real^{d}$ such that

\[\angles{\theta, \lambda}_{\sps}= \sum_{j = 1}^{n} \angles{\theta(t_j), \xi_j}_{\real^d}\]

for all $\theta \in \sps$. This allows decomposing

\begin{align*}\angles{\theta, \lambda}_{\sps}&= \sum_{j = 1}^{n} \sum_{k = 1}^{j}\angles{\theta(t_k) - \theta(t_{k - 1}), \xi_j}_{\real^d}\\&= \sum_{k = 1}^{n} \angles{\theta(t_k) - \theta(t_{k - 1}), \sum_{j = k}^n\xi_j}_{\real^d}\end{align*}

as a sum of independent centred Gaussian random variables. Thus $\theta \mapsto \angles{\theta, \lambda}_{\real^d}$ is a centred Gaussian random variable.

$(1)$, with Riemann-Stieltjes: Let $\mu: \cb((0, \infty)) \to \real$ be a finite signed measure, and $F(t) = \mu((0, t])$ be the distribution function of $\mu$, then $\norm{F}_{\text{var}}= \norm{\mu}_{\text{var}}< \infty$.

For any indicator function of the form $\one_{(a, b]}$ with $0 \le a < b < \infty$,

\[\int_{0}^{\infty} \one_{(a, b]}(t)dF(t) = F(b) - F(a) = \mu((0, t]) = \int_{(0, \infty)}\one_{(a, b]}(t)d\mu(t)\]

where the integral on the left is a Riemann-Stieltjes integral. As simple functions that vanish at $0$ are dense in $C_{0}((0, \infty); \real)$, and

\[\abs{\int_0^\infty \phi(t)dF(t)}\le \norm{\phi}_{u}\norm{F}_{\text{var}}\quad \abs{\int_{(0, \infty)}\phi(t)d\mu(t)}\le \norm{\phi}_{u}\norm{\mu}_{\text{var}}\]

for any simple function $\phi: (0, \infty) \to \real$,

\[\int_{0}^{\infty} \phi(t)dF(t) = \int_{(0, \infty)}\phi(t)d\mu(t)\]

for all $\phi \in C_{0}((0, \infty); \real)$.

As seen in the proof of Lemma 1.1.2, there exists a $\real^{d}$-valued finite Borel measure $\mu: \cb((0, \infty)) \to \real^{d}$ such that for any $\theta \in \sps$,

\[\angles{\theta, \lambda}_{\sps}= \int_{(0, \infty)}\Lambda \theta d\mu = \int_{(0, \infty)}\frac{\theta(t)}{1 + t}d\mu(t)\]

Let $F(t) = \mu((0, t])$, then since $\Lambda \theta \in C_{0}((0, \infty); \real)$,

\[\angles{\theta, \lambda}_{\sps}= \int_{0}^{\infty} \frac{\theta(t)}{1 + t}dF(t)\]

can be expressed as a Riemann-Stieltjes integral. Thus

\[\angles{\theta, \lambda}_{\sps}= \lim_{t \to \infty}\limv{n}\sum_{j = 1}^{2^n}\theta(2^{-n}jt)[F(2^{-n}jt) - F((2^{-n}- 1)jt)]\]

As the above holds for all $\theta \in \sps$, $\lambda$ viewed as a random variable is a^[1] sequential limit of Gaussian random variables, so it is also Gaussian.

$(1)$, with Approximations: As seen in Lemma 1.1.2, there exists a $\real^{d}$-valued Borel measure $\mu: \cb((0, \infty)) \to \real^{d}$ such that

\[\angles{\theta, \lambda}_{\sps}= \int_{(0, \infty)}\frac{\theta(t)}{1 + t}d\mu(t) \quad \forall \theta \in \sps\]

Since there exists $\real^{d}$-valued Borel measures $\seq{\mu_k}$ with finite support such that $\mu_{k} \Rightarrow \mu$, each $\theta \mapsto \int \frac{\theta(t)}{1 + t}d\mu_{k}(t)$ is Gaussian random variable and $\theta \mapsto \angles{\theta, \lambda}_{\sps}$ is a pointwise limit of Gaussian random variables, thus it is also Gaussian.

$(2)$: Firstly, for any $0 \le s, t < \infty$ and $\xi, \eta \in \real^{d}$,

\[\ev^{\wien}(\angles{\theta(s), \xi}\cdot \angles{\theta(t), \eta}) = (s \wedge t)\angles{\xi, \eta}\]

Let $F: [0, \infty) \to \real^{d}$ be measurable such that $d\lambda = Fd\abs{\lambda}$, then

\begin{align*}\angles{\theta, \lambda}_{\sps}^{2}&= \braks{\int_{[0, \infty)} \theta(t) d\lambda(t)}^{2} = \int_{[0, \infty)}\theta(t)\int_{[0, \infty)}\theta(s)d\lambda(s)d\lambda(t) \\&= \int_{[0, \infty)^2}\angles{\theta(s), F(s)}_{\real^d}\angles{\theta(t), F(t)}_{\real^d}d\abs{\lambda}(s)d\abs{\lambda}(t)\end{align*}

By Fubini’s theorem, the expectation can be pushed inside the integral:

\begin{align*}\ev^{\wien}[\abs{\lambda}^{2}]&= \iint_{[0, \infty)^2}\ev^{\wien}\braks{\angles{\theta(s), F(s)}_{\real^d}\angles{\theta(t), F(t)}_{\real^d}}d\abs{\lambda}(s)d\abs{\lambda}(t) \\&= \iint_{[0, \infty)^2}(s \wedge t)\angles{F(s), F(t)}_{\real^d}d\abs{\lambda}(s)d\abs{\lambda}(t) \\&= \iint_{[0, \infty)^2}s \wedge t d\lambda(s)d\lambda(t)\end{align*}

$(3)$: $\lambda$ is a Gaussian random variable with variance $\iint_{[0, \infty)^2}s \wedge t d\lambda(s)d\lambda(t)$.$\square$

I’m pretty sure I can just use one single limit here, but just to be safe I nested it.keyboard_return

Direct References

circleLemma 1.1.2: [Lemma 8.1.1, Str24]