Showing that $\text{Var}(X_t) = t$ if $(X_t)_{t > 0}$ is a gaussian process

Question

Let $(X_t)_{t > 0}$ be a stochastic process such that

1. $(X_t)_{t > 0}$ is a gaussian process,
2. it has continuous trajectories,
3. $\mathbb{E}(X_t) = 0$,
4. $\text{Cov}(X_t, X_s) = \min \{t, s \}$.

I would like to show that $(X_t)_{t > 0}$ is a Brownian motion. Continuity is obvious, $X_0 = 0$ also. I'm stuck with showing that $X_t - X_s \sim N(0, t-s), t > s$. Of course it's easy to show that $\mathbb{E}(X_t - X_s) = 0$ but I have some problems in computing the variance.

$$\text{Var}(X_t - X_s) = \text{Var}(X_t) + \text{Var}(X_s) - 2\text{Cor}(X_t, X_s) = \text{Var}(X_t) + \text{Var}(X_s) - 2s.$$

How can we show that $\text{Var}(X_t) = t$?

Answer 1

$var(X_t)$ is same as $cov(X_t,X_t)$ so it is $\min \{t,t\}=t$.