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I just begins my self-study on Brownian motion. I got stuck on the part about random-vector and characteristic function. Here are my questions:

  1. I'm not quite get about how characteristic function of random vector gonna be like. I guess it would look like $\phi_{g}(t) = \mathbb{E}[e^{it\cdot g}]$. Is it?

  2. Then, given that $g$ is Gaussian $N(\mu,C)$, how can I derive that its characteristic function is $\exp(i t\cdot\mu - \frac{1}{2}t\cdot Ct)$? The formula looks similar to the single random variable case. This is my first dealing with random-vector, so if anyone can give intuition or some examples, I would really appreciate.

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What's your definition of a "Gaussian" random variable? The characterization using the density? –  saz Jun 21 '13 at 16:02

1 Answer 1

  1. If $\langle u,v\rangle$ denotes the inner product of two vectors of $\Bbb R^d$, then the characteristic function of the random vector $X=(X_1,\dots,X_d)$ is defined on $\Bbb R^d$ by $$\varphi_X(t):=E\left[e^{i\langle t,X\rangle}\right].$$

  2. You can first be reduced to the case $\mu=0$, considering a translation of $g$. Since $C$ is non-negative (and symmetric), we can write it as $B^tB$. As we know the formula of the density, we can derive the wanted formula when $B$ is invertible by a substitution. When it is not, approximate $B$ by $B+n^{-1}I$.

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