# Proving a Gaussian process w/ specified mean and covariance is Brownian motion

I'm struggling with a silly thing when trying to prove that a Gaussian process $$(X(t))_{t\geq 0}$$ with $$\mathbb{E}[X(t)] = 0$$ for all $$t$$ and $$\mathbb{Cov}[X(s),X(t)] = \min\{s,t\}$$ for all $$s,t$$ is Brownian motion. Specifically, I struggle to show that under these hypotheses, for each $$0=t_0, the increments $$(X(t_{k+1})-X(t_k)), k = 0, 1, \ldots, n-1$$ are independent random variables.

• You know that $X(t_0),\ldots,X(t_n)$ are independent? That is not even true. Commented May 22 at 9:23
• @geetha290krm Yeah, okay, I'm not sure where I got that from. Edited it out Commented May 22 at 9:29
• Compute $Cov (X(t_{k+1})-X(t_k),X(t_{j+1})-X(t_j))$. Commented May 22 at 9:34
• Oh, okay. I did not know that fact. Thank you @SassatelliGiulio. Do anyone of you wish to post this as an answer so I can close my question? Commented May 22 at 9:39

Not my solution, credit due to the people who commented on the question.

For multivariate normally distributed random vectors, uncorrelatedness implies independence (see for instance here).

Since the process $$X$$ is a Gaussian process, it follows for any two non-overlapping increments, $$X_{t_{2}} -X_{t_{1}}$$ and $$X_{t_{4}} - X_{t_{3}}$$, the increments are multivariate normal. Then by bilinearity of the covariance and the hypothesis we can write \begin{align*}cov(X_{t_{2}} -X_{t_{1}}, X_{t_{4}} - X_{t_{3}}) &= cov(X_{t_{2}}, X_{t_{4}}) - cov(X_{t_{2}}, X_{t_{3}}) + cov(X_{t_{1}}, X_{t_{4}}) - cov(X_{t_{1}}, X_{t_{3}})\\\\ &= t_{2}- t_{2} + t_{1} - t_{1} = 0. \end{align*} This extends to more than two increments, with analogous computations for each pair of increments.