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A Brownian motion $\{W_t\}_{t\geq 0}$ is a stochastic process which satisfies four axioms, cf. The definition on Wikipedia.

I was reading a notes, and the lecturer wrote that, Because $W$ has independent Gaussian increments, the finite-dimensional distributions of $W$, i.e., the law of $(W_{t_1},··· ,W_{t_n})\,$ for any $0 \leq t_1 < ··· < t_n < \infty$, are multivariate normal distributions.

I have been learning analysis and know little about probability. Can anyone give me a reference/proof of his claim? I have no idea on how to go from the distributions of several independent random variables to the joint distribution of them. Many thanks for help!


If possible, I have one more question to ask. Is the reverse of his words true? i.e. If we have a stochastic process $\{B_t\}_{t\geq 0}$ such that $B_0=0$, $B$ is almost surely continuous, $B$ has independent increments, and the finite-dimensional distributions of $B$ are multivariate normal distributions, then is $B$ a Brownian motion? Thanks!

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  • $\begingroup$ Counter-example for the last question: $(2B_t)$. $\endgroup$ Commented Apr 17, 2023 at 7:20
  • $\begingroup$ One issue is whether your view of a Brownian motion is restricted to a standard Wiener process i.e. with $W_t-W_s\sim \mathcal{N}(0,t-s)$ for $0 \leq s \le t$ or whether some scaling of this or time-inhomogeneity may happen $\endgroup$
    – Henry
    Commented Apr 17, 2023 at 8:11

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The proof of the first claim follows easily from the following characterization of a multivariate normal distribution : a random vector is multivariate normal if and only if every linear combination of its components is normally distributed.

Now take $n$ real numbers $\alpha_1,\ldots,\alpha_n$. By solving the corresponding system of linear equations, you can find $\beta_1,\ldots,\beta_{n-1}$ such that $$\sum_{i=1}^n \alpha_i W_{t_i} = \sum_{k=1}^{n-1} \beta_k (W_{t_{k+1}}- W_{t_{k}}) $$ Hence we see that $\sum_{i=1}^n \alpha_i W_{t_i} $ is a sum of independent normally distributed random variables and thus must be normally distributed itself.

Regarding your second point : as geetha290krm's counterexample shows, the answer is no. What you have described is a Gaussian process with independent increments. However, if you require the increments to be stationary, that is $B_t - B_s \sim B_u - B_v $ whenever $t-s = u-v$, then you can prove that there exists $\mu\in\mathbb R $ and $\sigma^2 \ge 0$ such that $B_t \sim \mathcal N(\mu t,\sigma^2t) $. So in some sense, $B$ which satisfies these properties would be a "scaled" standard Brownian motion.

If you don't require stationarity though $B$ could get much wilder. See this blog post by George Lowther for a proof of this fact and an extended discussion : Continuous Processes with Independent Increments

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    $\begingroup$ Thanks Stratos! Your answer and all those useful external links help me a lot. Just a kind reminder for a typo: The summation for the $\beta_k$ term shall go from $k=2$ to $k=\color{red}{n}$. Thanks again. $\endgroup$
    – Sam Wong
    Commented Apr 18, 2023 at 6:34
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    $\begingroup$ Thank you for pointing out the typo ! It's fixed now ;) $\endgroup$ Commented Apr 18, 2023 at 18:28

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