What is $Var(X_t-X_s)$ if $X_t = \sqrt{t} Z$ where $Z \sim N(0,1)$
The answer is given by $(\sqrt{t}-\sqrt{s})^2$. How do they get this?
My thoughts:
$X_t\sim N(0,t)$ and $X_s\sim N(0,s)$
I usually can simply say that $Var(X_t-X_s)=Var(X_t)-Var(X_s)=t-s$. But this works for independent variable only. So why are these not independent?
On the other hand, Brownian Motion $W_t \sim N(0,t)$ has the property that $W_t-W_s \sim N(0,t-s)$
The answer is given by A.S, i.e. $Var[X_t-X_s]=Var[X_t]+Var[X_s]-2Cov[X_t,X_s]=t+s-2\sqrt{ts}$.
To your question why $Var[X_t-X_s]\neq Var[W_t-W_s]$:
Note that indeed both $X_t$ and $W_t$ follow Gaussian process, $X_t\sim W_t\sim \mathcal{N}(0,t)$. However, $X_t-X_s=(\sqrt{t}-\sqrt{s})Z\neq (\sqrt{t-s})Z=X_{t-s}$, so it doesn't have stationary increments in contrast to Brownian motion.
Moreover, $X_t-X_s=(\sqrt{t}-\sqrt{s})Z$ is not independent from $X_s$, so it doesn't have independent increments in contrast to Brownian motion.
Brownian motion is a L\'evy process + a Gaussian process. While $X$ is only a Gaussian process.
Also note that $X$ is not an adapted process like Brownian motion.