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There is a purely probability theoretical argument in the proof of Lévy's characterization of Brownian motion, which I do not completely understand. I think it is rather easy. Suppose we know $$E[e^{iu^{tr}(X_t-X_s)}|\mathcal{F}_s]=e^{-\frac{1}{2}|u|^2(t-s)}$$

for all $u\in\mathbb{R}^d$. From this it should follow, that $X_t-X_s$ is independent of $\mathcal{F}_s$ and normally distributed with mean $0$ and covariance matrix $(t-s)Id_{d\times d}$, hence the $X^k$ should be independent Brownian Motion ($^k$ denotes the k-th coordinate). My suggestion is to take expectation:

$$E[e^{iu^{tr}(X_t-X_s)}]=e^{-\frac{1}{2}|u|^2(t-s)}$$

Hence I know that $(X_t-X_s)$ has the right distribution. Furthermore, by the structure of the convariance matrix, I know $(X^i_t-X_s^i)(X^k_t-X^k_t)$ are uncorrelated for $k\not=i$. Why should the $X^k$ be independent. I have independence of the product of increments, how do I get independence for $X^k$? I guess, this uses, that every coordinate is a normal distributed r.v. and for normal distributed r.v. "uncorrelated implies independent". Even more, I do not see how independence of $\mathcal{F}_s$ should follow. So any help would be appreciated. Thanks in advance!

math

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for normal distributed r.v. "uncorrelated implies independent" Hmmm... That depends very much on what you mean exactly by that. Note that there exists uncorrelated normal random variables that are not independent. –  Did Jul 17 '12 at 13:20
    
@did: You're right, I should write jointly normal. –  math Jul 17 '12 at 13:36

2 Answers 2

A very useful fact here is the following:

Lemma. Let $X,Y$ be random vectors, and consider the joint characteristic function $\phi(u,v) = E[e^{i u \cdot X + i v \cdot Y}]$. If we can factor $\phi$ as $\phi(u,v) = f(u) g(v)$, then $X,Y$ are independent.

Proof. Without loss of generality, assume $f(0) = g(0) = 1$. Then we see that $f(u) = \phi(u,0)$ is the chf of $X$, and likewise $g$ is the chf of $Y$. Thus $\phi$ is the joint chf of $(\tilde{X}, \tilde{Y})$ where $\tilde{X} \overset{d}{=} X$, $\tilde{Y} \overset{d}{=} Y$, and $\tilde{X}, \tilde{Y}$ are independent. Since the joint chf uniquely determines the joint distribution we have $(X,Y) \overset{d}{=} (\tilde{X}, \tilde{Y})$, which means $X,Y$ are independent.

Corollary. Let $X$ be a random vector, $\mathcal{G}$ a $\sigma$-field, and suppose the conditional chf $f(u) = E[e^{i u \cdot X} \mid \mathcal{G}]$ is deterministic. Then $X$ is independent of $\mathcal{G}$.

Proof. Let $Y$ be any $\mathcal{G}$-measurable random vector, with $g$ its chf. We compute the joint chf of $X,Y$: $$E[e^{i u \cdot X + iv \cdot Y}] = E[e^{i v \cdot Y} E[e^{i u \cdot X} \mid \mathcal{G}]] = E[e^{i v \cdot Y} f(u)] = g(v) f(u).$$ By the previous lemma, $X,Y$ are independent. $Y \in \mathcal{G}$ was arbitrary, so $X$ must be independent of $\mathcal{G}$.

Given what you know, this corollary tells us that $X_t - X_s$ is independent of $\mathcal{F}_s$.

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Eldrege : Thanks for your answer! This is exactly what I was looking for. If you would add an argument why $X_t^k$ and $X_t^l$ are independent, I will accept your answer. However, I'm very thankful for your help! –  math Jul 21 '12 at 8:40

The vector of increments are uncorrelated and the increments are multivariate normal and the covariance matrix for the increments is (t-s)I where I is a dxd identity matrix. So these coordinates are uncorrelated and normally distributed hence independent. Now if the increments from one coordinate are independent from the increments of the other by summing you will get that the coordinate sums are independent normal. summing the increments gives Xt$^k$ and Xt$^l$ for k not equal to l.

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Thanks for your answer. Actually I do not see how this answer my question. As I wrote, I know that $(X^k_t-X^k_s)$ and $(X_t^l-X_s^l)$ are independent. Why is $X_t^k$ and $X_t^l$ independent? How do I get the independence of $\mathcal{F}_s$? –  math Jul 17 '12 at 12:36
    
Sorry I misread. I think the answer is simple, just sum independent increments for one coordinate that are independent of the increments in the other coordinate and you get the answer. I will add it to my answer. –  Michael Chernick Jul 17 '12 at 12:42
    
What do you mean by summing increments. Since $(X_t^k-X_s^t) +(X_t^l-X_s^l)\not= X_t^k$ or $X_t^l$. And again, how do I get independence of $\mathcal{F}_s$? Sorry, if I'm completely wrong. –  math Jul 17 '12 at 12:50
    
Not obvious! Xt$^k$= (Xt$^k$-Xt-1$^k$)+(Xt-1$^k$ -Xt-2$^k$)+ ... and the same telescoping sum works for Xt$^l$. –  Michael Chernick Jul 17 '12 at 13:13

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