# How to show the following definition gives Wiener measure

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!

• Roughly speaking: Plug in $f(x) = e^{\imath \, x \xi}$ and then use that the martingale has constant expectation. This yields a differential equation for the characteristic function of $W_t$ which can be easily solved. Then, one has to conclude from the martingale property that $(W_t)_{t \geq 0}$ is a Markov process. This finally yields the correct result for the finite-dimensional distributions.
– saz
Commented Oct 16, 2014 at 17:41
• @saz Thank you! I don't see how to get $W_t$ is a Markov process. Could you explain a bit more? By the way, I just find maybe by some approximation arguments, we can say $W_t$ and $W_t^2 - t$ are martingales, then since $W_t$ is given continuous so we can apply Levy's theorem to say $W_t$ is Brownian motion. Commented Oct 16, 2014 at 18:29
• Yes, that's an alternative - but if you want to make this rigorous, it requires some effort, I would say. Concerning Markov-property: Well, it's not obvious. Basically the point is the following: The definition of $\mu$ shows 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 $Af := \frac{1}{2} f''$. Now one can show that whenever such a martingale problem has a unique solution $(X_t)_{t \geq 0}$, then it is a Markov process. If you are interested, I can (try to) add some more details as an answer... but not today.
– saz
Commented Oct 16, 2014 at 18:55
• @saz Yes I'm interested if that won't bother you too much. I find a proof of Levy's theorem in which one seems to prove the Markov property that we need here, maybe not rigorously since involving integral and differentiation of conditional expectations. Commented Oct 16, 2014 at 19:16
• Ah, right... thanks to your link I remembered some nicer way to prove the statement (which does not rely on martingale problems...).
– saz
Commented Oct 16, 2014 at 19:47

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.