How to show the following definition gives Wiener measure

Question

On the first page of Ustunel's lecture notes, he defines the Wiener measure in the following way:

Let $W = C_0([0,1]), \omega \in W, t\in [0,1]$, define $W_t(\omega) = \omega(t)$. If we denote by $\mathcal{B}_t = \sigma\{W_s; s\leq t\}$, then there is one and only one measure $\mu$ on $W$ such that

1) $\mu \{W_0(\omega) = 0\} = 1$

2) $\forall f \in C_b^{\infty}$, the stochastic process $$(t, \omega ) \mapsto f(W_t(\omega)) - \dfrac{1}{2}\int_0^tf''(W_s(\omega))ds$$

is a $(\mathcal{B}_t, \mu)$-martingale. $\mu$ is called the Wiener measure

I am more familiar with the definition which supposes that we have already a Brownian motion $B_t$ available and then define $$\nu\left(\{\omega: \omega_{t_1} \in A_1, \cdots, \omega_{t_n} \in A_ n\}\right) = P(B_{t_1} \in A_1, \cdots, B_{t_n} \in A_ n)$$

My question is why $\mu$ and $\nu$ are the same? Of couree if we begin with $\nu$ and use Its's formula, we can see the two conditions defning $\mu$ are verified. But if we begin with the definition of $\mu$, how can we verify the condition defining $\nu$?

In addition, in Ustunel's notes, he first presented his definition of Wiener measure then introduced stochastic integral. So I am wondering if there is a way to begin with the definition of $\mu$, then to show $\mu$ satisfies the condition defining $\nu$ without using stochastic integral.

Of course I will still appreciate it if you help me show $\mu \implies \nu$ using stochastic integral.

Thank you!

Answer 1

For fixed $x \in \mathbb{R}$, we choose $f(x) := e^{\imath \, x \xi}$. By assumption,

$$(t,\omega) \mapsto e^{\imath \, \xi W_t(\omega)} - \frac{\xi^2}{2} \int_0^t e^{\imath \, \xi W_r(\omega)} \, dr$$

is a martingale, i.e.

$$\mathbb{E}\left( e^{\imath \, \xi W_t} + \frac{\xi^2}{2} \int_0^t e^{\imath \, \xi W_r} \, dr \mid \mathcal{B}_s \right) = e^{\imath \, \xi W_s}+ \frac{\xi^2}{2} \int_0^s e^{\imath \, \xi W_r} \, dr.$$

for any $s \leq t$. Multiplying both sides with $e^{-\imath \, \xi W_s}$ yields

$$\mathbb{E}(e^{\imath \, \xi (W_t-W_s)} \mid \mathcal{B}_s) = 1 - \frac{\xi^2}{2} \mathbb{E} \left( \int_s^t e^{\imath \, \xi (W_r-W_s)} \, dr \mid \mathcal{B}_s \right).$$

By Fubini's theorem, we can interchange the conditional expectation and integration:

$$\mathbb{E}(e^{\imath \, \xi (W_t-W_s)} \mid \mathcal{B}_s) = 1 - \frac{\xi^2}{2}\int_s^t \mathbb{E}(e^{\imath \, \xi (W_r-W_s)} \mid \mathcal{B}_s) \, dr.$$

This shows that $\varphi(t) := \mathbb{E}(e^{\imath \, \xi (W_t-W_s)} \mid \mathcal{B}_s)$ is a solution to the ordinary differential equation (ODE) $$\varphi'(t) = - \frac{\xi^2}{2} \varphi(t) \qquad \varphi(s)=1.$$

Obviously, the (unique) solution to this ODE is $$\varphi(t) = \exp \left( - (t-s) \frac{\xi^2}{2} \right).$$ In particular, we find that

$$\mathbb{E}(e^{\imath \, \xi (W_t-W_s)} \mid \mathcal{B}_s) = \mathbb{E}e^{\imath \, \xi (W_t-W_s)} = \exp \left(- (t-s) \frac{\xi^2}{2} \right).$$

We conclude that:

1. $W_t-W_s$ is Gaussian with mean $0$ and variance $t-s$.
2. $\mathcal{B}_s$ is independent from $W_t-W_s$. This implies that $(W_t)_{t \geq 0}$ has independent increments. In particular, $(W_t)_{t \geq 0}$ is a Markov process. Using the Markov property, one can easily obtain the finite-dimensional distributions from the distributions of $W_t$ for each $t \geq 0$.

Remarks:

• The proof shows that it suffices to have the martingale property for functions $f$ of the form $f(x) = e^{\imath \, x \xi}$, $\xi \in \mathbb{R}$.
• The definition of $\mu$ is chosen such that $(W_t)_{t \geq 0}$ is the unique solution to the martingale problem $$f(X_t)- \int_0^t Af(X_s) \, ds$$ for the (Laplace) Operator $Af := \frac{1}{2} f''$. Invoking certain theorems from this area, also yields the claim (but is overshoot in this particular case).
• As @LiuGang already mentioned in a comment, Lévy's characterization of Brownian motion can also be used to prove the claim. To this end, we have to overcome some technical issues since $f(x) := x$ and $f(x) := x^2$ are not bounded.