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Let me remind first the construction of Brownian motion.

Fix a vector $x \in \mathbb{R}^n$ and define $p(t,x,y) := (2\pi t )^{-\frac{n}{2}} \cdot \exp{\left( - \frac{|x-y|^2}{2t} \right)},$ for $y \in \mathbb{R}^n$, $t >0$.

Then for $0 \leq t_1 \leq t_2 \leq \ldots \leq t_k$ we define a measure $\nu_{t_1, \ldots, t_k}$ on $\mathbb{R}^{nk}$ by $$\nu_{t_1, \ldots, t_k}(F_1 \times \ldots \times F_k)= \int \limits_{F_1 \times \ldots \times F_k}^{} p(t_1, x, x_1)\prod_{j=1}^{k-1}p(t_{j+1}-t_j,x_{j}, x_{j+1}) dx_1\ldots dx_k,$$ where the following conventions are used $dy=dy_1\ldots dy_k$ for Lebesgue measure and $p(0, x, y)dy=\delta_x(y)$ (the unit point mass at $x$).

By Kolmogorov's extension theorem applied to the probability measures $\nu_{t_1, \ldots, t_k}$ (which easily satisfies all the assumtions of that theorem) there exists a probability space $(\Omega, \mathcal{F}, P^x)$ and a stochastic process $\{B_t\}_{t \geq 0}$ on $\Omega$ such that the finite distributions of $B_t$ are given by $$ (\star) \ \ P^x(B_{t_1} \in F_1, \ldots, B_{t_k} \in F_k)= \int \limits_{F_1 \times \ldots \times F_k}^{} p(t_1, x, x_1)\prod_{j=1}^{k-1}p(t_{j+1}-t_j,x_{j}, x_{j+1}) dx_1\ldots dx_k.$$

Problem

I want to show that for the random variable $Z = (B_{t_1}, \ldots, B_{t_k}) \in \mathbb{R}^{nk}$ there exist a vector $M \in \mathbb{R}^{nk}$ and a non-negative definite matrix $C \in \text{M}_{nk \times nk}(\mathbb{R})$ such that $$E^x\left[\exp\left(i\left<u, Z \right> \right)\right]= \exp\left( -\frac{1}{2} \left<Cu, u\right> + i \left<u, M\right> \right),$$ for all $u=(u_1, \ldots, u_{nk}) \in \mathbb{R}^{nk}$ (left hand side stands for the characteristic function of the random variable $Z$). Moreover, $M$ is the mean value of $Z$ and $C$ is a covariance matrix of $Z$.

I was trying to calculate it explicitly by writing the left hand side by its definition and applying the ($\star$) formula, but the integral which I've got is not so nice and I cannot conclude what I want to.

This exercise is nothing else, but showing that $B_t$ is a Gaussian process (so it is not hard to guess what should be $M$ and $C$ in this case).

I hope you know some tricks how to compute such an integral in an easy way. Thanks in advance for any help.

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I think it suffices to show that the integral will indeed be of the form $\exp( -\frac{1}{2} \langle C u, u \rangle + i \langle u, M \rangle)$. It would then follow that $C$ and $M$ are covariance and mean. This is a trivial excercise in integration of a Gaussian. To carry it out, write $\langle u, Z \rangle = \sum_{i=1}^n u_i \sum_{k=1}^i X_k$ where $X_k$ are corresponding independent normal variates corresponding to $B_{t_i-t_{i-1}}$. It would then follow that c.f. is a product of exponentials of quadratics. –  Sasha Aug 25 '11 at 16:17
    
If I show the the characteristics function of $Z$ is of above form it implies that $B_t$ is a Gaussian process. By using the form of $M$ and $C$ we easily got that $E^x(B_t)=x$ and $E^x((B_t-x)(B_s-x))=n \mbox{min}\{s,t\}$ which gives us that $B_t$ has independent increments (equivalently, uncorrelated for normal distribution are independent). I think that your hint already involves that $B_t$ is normal distributed and it has independent increments, which is a further consequence of the fact I want to show. Am I right? –  Franz Aug 25 '11 at 17:27
    
I mean consider integration of $ \exp( i \sum_{k=1}^n u_k t_k )$. Because $t_i-t_{i-1}$ occur naturally in your measure, change variables, by letting $t_i = \sum_{m=1}^i s_m$. Your measure then becomes a product of independent Gaussians which you can carry out. Integration w.r.t. each $s_m$ will produce $e^{\text{quadratic in} u}$, hence their product will again be quadratic in $u$. –  Sasha Aug 25 '11 at 17:58
    
Ok, I understand. Probably, above you had some notation issue. We should denote $u_j := (u_1^j, \ldots, u_n^j) \in \mathbb{R}^n$ and now $$E(\exp(i \left< u, Z\right>) = \int_{\mathbb{R}^{nk}} \exp(i \sum_{j=1}^k u_jx_j) p(t_1, x, x_1) \prod_{j=1}^{k-1}p(t_{j+1}-t_j, x_j, x_{j+1})dx_1 \ldots dx_k.$$ Now, we should substitute $x_j = \sum_{m=1}^{j}\hat{x}_m$ and with this substitution we can easily separate the integrals. I think you can write this as the answer. –  Franz Aug 25 '11 at 20:46

1 Answer 1

up vote 3 down vote accepted

Denote $t_0 = 0$ and $x_0 = 0 $ for notational convenience. Then

$$ \mathbb{E}( \exp(i \langle u, Z \rangle ) = \int_{\mathbb{R}^{n k}} \exp( i \langle u, x \rangle ) \prod_{j=1}^k p(t_{j}-t_{j-1}, x_{j} - x_{j-1}) \mathrm{d}x_1 \cdots \mathrm{d} x_k $$ Now change variables $x_k = \sum_{i=1}^k s_k$. This has unit Jacobian, you get $$ \mathbb{E}( \exp(i \langle u, Z \rangle ) = \int_{\mathbb{R}^{n k}} \prod_{k=1}^k \exp( i \, s_j \sum_{m=j}^k u_q ) ) \prod_{j=1}^k p(t_{j}-t_{j-1}, s_j) \mathrm{d}s_1 \cdots \mathrm{d} s_k $$ Now integration with respect to each $s_j$ can be carried out independently and will produce $\exp( Q_j(u))$, where $Q_j$ is a quadratic multivariate polynomial in $u$. Hence the characteristic function of k-dimensional time-slice distribution of this Brownian motion process is going to be $\exp(Q(u))$.

Now $Q(u) = \langle C u, u \rangle + i \langle u, M \rangle $, where $C$ is covariance matrix and $M$ is mean vector. The free term will be missing, because c.f. is one for zero vector $u$.

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