# How can we prove that the generalized stochastic process induced by a real-valued Brownian motion is Gaussian?

Let $(B_t)_{t\ge 0}$ be a real-valued Brownian motion on a probability space $(\Omega,\mathcal A,\operatorname P)$, $\lambda$ be the Lebesgue measure on $[0,\infty)$ and $$\langle W,\phi\rangle:=\int\phi(t)B_t\;{\rm d}\lambda\;\;\;\text{for }\phi\in\mathcal D:=C_c^\infty([0,\infty))\;.$$

We can prove that the expectation $$\operatorname E[W](\phi):=\operatorname E\left[\langle W,\phi\rangle\right]\;\;\;\text{for }\phi\in\mathcal D$$ of $W$ is $0$ and the the covariance $$\rho[W](\phi,\psi):=\operatorname E\left[\langle W,\phi\rangle\langle W,\psi\rangle\right]\;\;\;\text{for all }\phi,\psi\in\mathcal D$$ of $W$ is $$\int\int\min(s,t)\phi(s)\psi(t)\;{\rm d}\lambda(s)\;{\rm d}\lambda(t)\;.$$

Now I want to prove, that $W$ is Gaussian, i.e. $$\alpha_1\langle W,\phi_1\rangle+\cdots+\alpha_n\langle W,\phi_n\rangle\text{ is normally distributed}$$ for all (linearly independent$^\ast$) $\phi_1,\ldots,\phi_n\in\mathcal D$ and $\alpha\in\mathbb R^n$. Unfortunately, I've no idea how we can do that.

[$^\ast$ I've found different notions of being Gaussian for a generalized stochastic process. Some of them state that $\phi_1,\ldots,\phi_n$ need to be linearly independent while the others omit this assumption.]

If we set

$$\phi := \sum_{j=1}^n \alpha_j \phi_j \in \mathcal{D},$$

then by the linearity of the integral

$$\alpha_1 \langle W,\phi_1 \rangle + \ldots + \alpha_n \langle W, \phi_n \rangle = \langle W,\phi \rangle.$$

Consequently, it suffices to show that $\langle W,\phi \rangle$ is Gaussian for all $\phi \in \mathcal{D}$.

Fix $\phi \in \mathcal{D}$. Since $\phi$ has compact support we can choose $R>0$ such that the support of $\phi$ is contained in $[0,R]$. Hence

$$\langle W,\phi \rangle = \int_{[0,R]} \phi(t) B_t \, d\lambda(t).$$

Since $t \mapsto \phi(t) B_t$ is continuous, hence Riemann-integrable, we may replace the Lebesgue integral by a Riemann integral:

$$\langle W,\phi \rangle = \int_0^R \phi(t) B_t \, dt.$$

Now define a sequence of partitions of the interval $[0,R]$ by $t_j^n := \frac{j}{n} R$ for $j=0,\ldots,n$ and $n \in \mathbb{N}$. Approximating the Riemann integral by Riemann sums, we get

$$\langle W,\phi \rangle = \lim_{n \to \infty} \sum_{j=0}^{n-1} B_{t_j^n} \phi(t_j^n) (t_{j+1}^n-t_j^n).$$

Note that

$$\sum_{j=0}^{n-1} B_{t_j^n} \phi(t_j^n) (t_{j+1}^n-t_j^n)$$

is Gaussian since $(B_t)_{t \geq 0}$ is a Gaussian process. Hence, $\langle W,\phi \rangle$ is Gaussian as a pointwise limit of Gaussian random variables (see e.g. this question and note that pointwise convergence implies convergence in distribution).

• I've got some problem to conclude that $\langle W,\phi\rangle$ is Gaussian and asked another question: math.stackexchange.com/questions/1634777/… – 0xbadf00d Jan 31 '16 at 17:31
• @0xbadf00d But I gave you the link to the other question, didn't I? So why ask another question? – saz Jan 31 '16 at 18:31
• Cause I don't understand the sketched proof and I hoped that we can make the conclusion without considering the specific means and variances. – 0xbadf00d Jan 31 '16 at 19:21

Here's another approach, based on an old result of M. Kac (If $X$ and $Y$ are independent random variables such that $X+Y$ is independent of $X-Y$, then $X$ and $Y$ are normally distributed with the same variance.). Let $B'$ be a second Brownian motion independent of $B$ (make a product space construction if necessary), and define $\langle W',\phi\rangle=\int\phi(t)B'_t\,dt$. As one easily checks, $(B+B')/\sqrt{2}$ and $(B-B')/\sqrt{2}$ are independent (Brownian motions). It follows that $\langle W,\phi\rangle+\langle W',\phi\rangle$ and $\langle W,\phi\rangle-\langle W',\phi\rangle$ are independent, hence Gaussian by Kac's theorem.

• I like your approach, but I don't know how I need to argue that such a second independent Brownian motion needs to exist. – 0xbadf00d Jan 31 '16 at 17:34
• Let $(\Omega'\mathcal A',\operatorname P')$ with Brownian motion $B'$ be a copy of your setup, and then form the product space $\bar\Omega:=\Omega\times\Omega'$ equipped with $\bar{\mathcal A}:=\mathcal A\otimes\mathcal A'$ and $\bar{\operatorname P}:=\operatorname P\otimes\operatorname P'$. Then $\bar B(\omega,\omega'):=B(\omega)$ and $\bar B'(\omega,\omega'):=B'(\omega')$ are independent Brownian motions defined on the product space. – John Dawkins Jan 31 '16 at 18:02
• Yes, but I would like that $W$ is Gaussian with respect to $\operatorname P$. Your approach would yield that $W$ is Gaussian with respect to $\operatorname P'$. – 0xbadf00d Jan 31 '16 at 19:18
• Because $\bar{\operatorname P}$ is product measure, the (marginal) distribution of $W$, under $\bar{\operatorname P}$, is $\operatorname P$. – John Dawkins Jan 31 '16 at 21:21
• Where can I find a proof the result you're using? – 0xbadf00d Feb 1 '16 at 10:19