# Stochastic process, Gaussian, with zero mean is a Wiener process

Let $(\Omega, \mathcal F , \mathbb P)$ be a probability space and let $\mathcal F = \{\mathcal F_t\}_{t\ge}$ a ﬁltration. Let $W=\{W_t;t ≥ 0\}$ be a stochastic process adapted to $\mathcal F$. Suppose that W is a zero-mean Gaussian process with covariance function $$E(W_s W_t) = \min(s, t), \quad ∀s,t ∈ [0, ∞).$$ Show that W is a standard Wiener process.

In other words, I have to prove that the following conditions are satisfied:

1. $W_0=0$
2. $W_t-W_s$ is Gaussian with zero mean and $Var=|t-s|$
3. increments of W corresponding to non-overlapping time intervals are independent
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What have you tried? –  saz May 23 '13 at 16:30
I got the 3. relationship between covariance and indipendence here –  user79144 May 23 '13 at 16:55
this may be picky, but the filtration should be that of the process, not just one to which W is adapted. For example, choosing $\mathcal F = \lbrace \mathcal F_{\infty} \rbrace$ does not affect covariances but does affect whether you want to call it a wiener process. –  mike May 23 '13 at 17:12
Which point(s) is (are) eluding you, in the end? –  Did May 23 '13 at 20:52

1. Choose $t=s=0$ in $\mathbb{E}(W_s \cdot W_t) = \min\{s,t\}$. (Note that $N(0,0)$ equals -by definition- $\delta_0$.)
2. Let $s<t$. By assumption, the random vector $X:=(W_s,W_t)$ is Gaussian with mean $0$ and covariance matrix $$C:=\begin{pmatrix} s & s \\ s & t \end{pmatrix}$$ Since $X$ is Gaussian, we know that $\ell^T \cdot X = \ell_1 \cdot W_s+\ell_2 \cdot W_t$ is Gaussian for $\ell \in \mathbb{R}^2$. There are known formulas how to calculate the mean and variance of $\ell^T \cdot X$. Find a suitable $\ell$.
3. Let $0:=t_0 < t_1 < \ldots < t_n$. Note that $$\Delta := \begin{pmatrix} W_{t_1}-W_{t_0} \\ \vdots \\ W_{t_n}-W_{t_{n-1}} \end{pmatrix} = \underbrace{\begin{pmatrix} -1 & 1 & 0 & 0 &\ldots & 0 \\ 0 &-1 & 1 & 0 & \ldots & 0 \\ \vdots & & \ddots & \ddots & & \\ 0 & 0 & 0 & \ldots & -1 & 1 \end{pmatrix}}_{=:M} \cdot \begin{pmatrix} W_{t_0} \\ \vdots \\ W_{t_n} \end{pmatrix}$$ Since $(W_{t_0},\ldots,W_{t_n})$ is Gaussian, by assumption, we conclude that $\Delta$ is Gaussian. Therefore, it suffices to show that the covariance matrix of $\Delta$ is a diagonal matrix. Again, there are known formulas how to calculate the covariance matrix of a linear transformation of a Gaussian random vector.
Remark In general, one requires that the paths $t \mapsto W(t,w)$ are continuous for almost all $w$. The theorem of Kolmogorov-Chentsov shows that, under the given assumptions, there exist always a modification $(\tilde{W}_t)_t$ of the process $(W_t)_t$ such that the sample paths $t \mapsto \tilde{W}(t,w)$ are continuous almost surely.