Scaling a Standard Normal and Testing Brownian Motion

Summary

The document explains why multiplying a standard normal variable by a constant scales its variance by the square of that constant. For the process defined as a single standard normal variable times the square root of time, the increment from s to t is that same random variable multiplied by the difference of the square roots. Its mean is zero and its variance is the squared difference, so the increment has a normal distribution with that variance.

This calculation alone does not establish that the process is Brownian motion. Its increments share the same random variable and therefore are not independent across disjoint time intervals; their variance also does not generally match the elapsed time. The answer's description of the process as having a volatility given by the squared difference is imprecise: that quantity is the increment variance, not a constant volatility. The exchange offers a useful variance-scaling explanation but leaves the full Brownian-motion check implicit.

Key ideas

• Multiplying a standard normal variable by a constant multiplies its variance by the square of that constant.
• The process increment is a single normal variable scaled by the difference of two square roots.
• The increment has zero mean and variance equal to the squared scaling factor.
• Matching an increment distribution alone does not establish that a process is Brownian motion.

Tags

# Brownian motion


# 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.

## Answer by Aldo Shumway (score 0, accepted)

https://quant.stackexchange.com/a/38271

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

Shown in full with attribution under the source's licence. Licence: CC BY-SA 4.0 (Stack Exchange)

This summary was written by Stratmill's research agent from the original; it is not a copy of the source.