# Brownian motion

Suppose I have the process $X = X(t)$ for $t \ge 0$ given by $X(t) = \sqrt{t}*Z \,\forall t \ge 0$ where $Z$ is normally distributed with $N(0,1)$.

Is this a Brownian motion?

Solution yields:

$$X(t)-X(s)=Z\sqrt{t} - Z\sqrt{s} \sim N\left( 0,(\sqrt{t}-\sqrt{s})^2)\right) = N\left(0,t-2*\sqrt{s\,t}+s\right)$$

and now we must compare with $X(t-s)$, etc.

However this is not my question, my question is how does $(Z\sqrt{t}-Z\sqrt{s})$ become $N(0,(\sqrt{t}-\sqrt{s})^2)$?

Why is it not $N(0,\sqrt{t}-\sqrt{s})$ only, without squaring?

So basically what does $Z(\sqrt{t}-\sqrt{s})$ mean intuitively and mathematically?

Would be grateful for any answer.

• Also quite fundamental: does it even have independent increments? – Phil-ZXX Feb 16 '18 at 11:58
• Your notations need to be clarified. Is $Z$ a single Gaussian random variable, in which case $X(t)=\sqrt{t/s} X(s)$ ? Or are the $Z$ independent draws of a Gaussian distribution (a new one for each $t$) ? In either case the $X$ is not a martingale and is not a Brownian motion. – Antoine Conze Feb 16 '18 at 15:41

If $Z \sim N\left( 0,1\right)$ then

\begin{align*} E \left(Z\sqrt{t} - Z\sqrt{s} \right) = 0\\ \end{align*}

and

\begin{align*} Var\left(Z\sqrt{t} - Z\sqrt{s} \right) = Var\left(Z(\sqrt{t} - \sqrt{s} )\right)=(\sqrt{t} - \sqrt{s})^2Var(Z)=(\sqrt{t} - \sqrt{s})^2 \end{align*}

Hence, since then sum of normals is normal

\begin{align*} Z\sqrt{t} - Z\sqrt{s} \sim N\left( 0,(\sqrt{t}-\sqrt{s})^2\right) \end{align*}

What this means is that you are defining a new process $Z$ which has a volatility $(\sqrt{t}-\sqrt{s})^2$

Hope this helps