I have a simple question on Brownian motion.
On one hand in one of the definition one of the condition for a stochastic process $W_t(\omega)$ to be a Brownian motion is to verify:
1/ $W_0(\omega)=0$
2/ $W_t(\omega)$ to be a Gaussian process.
On the other hand according to the definition of a Gaussian process $\forall (t_k)_{1 \leq k \leq n} \Rightarrow (W_{t_1}(\omega);W_{t_2}(\omega);...;W_{t_n}(\omega))$ must be a Gaussian random vector.
If it is true $\forall t$ and finite $n$ it has to be true for the sequence of $t$ composed only of one $t_1=0$. But in such a case we have
$P(W_0(\omega) \leq \alpha)=1$ iff $\alpha \geq 0$
$P(W_0(\omega) \leq \alpha)=0$ iff $\alpha < 0$
And, so, obviously $W_0(\omega)$ is not a gaussian r. v.
Thank you.
In settings like this we usually broaden our definition of "Gaussian random vector / random variable" to include constants, viewed as a Gaussian distribution with variance zero. So $W_0 = 0$ is indeed a Gaussian random variable, having the distribution $N(\mu, \sigma^2)$ where $\mu = 0$ and $\sigma^2 = 0$.
Likewise, Gaussian random vectors are allowed to be degenerate, having covariance matrices that are singular, and constant vectors correspond to a zero covariance matrix.
It's true that the familiar statement "Gaussian random variables / vectors have a continuous distribution" now has exceptions, and needs to be limited to non-degenerate Gaussian random variables / vectors, i.e. those with nonzero variance / nonsingular covariance matrix.