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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 $W_t-W_s \sim N(0,t-s)$

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    $\begingroup$ $Var(X_t-X_s)=var(X_t)+var(X_s)-2cov(X_t,X_s)=t+s-2\sqrt{ts}$. $\endgroup$ – A.S. Dec 3 '15 at 11:43
  • $\begingroup$ @A.S what hypothesis am I missing? Why should the covariance be the geometric mean of the variances? $\endgroup$ – Justpassingby Dec 3 '15 at 12:01
  • $\begingroup$ @Jus I am interpreting $X_t\sim \sqrt tZ$ as a.s. equality - not just distributional equality - since otherwise we have no handle on covariance. $\endgroup$ – A.S. Dec 3 '15 at 12:05
  • $\begingroup$ @A.S. Thank you very much for your reply, I see where it comes from now. However I'm still not clear why $W_t \sim N(0,t)$ doesn't have the same property. In other words why is $Var(W_t-W_s)=t-s$ and not what we showed using co-variance? I am imagining that it's something to do with independence, but if both $X=\sqrt{t} Z \Rightarrow X \sim N(0,t)$ and $W_t \sim N(0,t)$ why don't they have the same Variance? $\endgroup$ – GRS Dec 3 '15 at 13:06
  • $\begingroup$ Because $X_t=\sqrt t Z$ is different from $W_t\sim \sqrt t Z$ (and $W_t-W_s$ independent of $W_s$ for $t>s$ or, equivalently, $cov(W_t,W_s)=|t-s|$). $\endgroup$ – A.S. Dec 3 '15 at 23:50
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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.

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