# Math Help - Questions on Development of Brownian motion

1. ## Questions on Development of Brownian motion

I have been away from MHF for quite some time due to how busy my life has been due to work, personal life changes (house, job) and of course, my actuarial exams (in November).

I will not introduce Brownian motion here because I want someone familiar with it to answer a simple question or two for me...

The limit of the sum of infinitely small random movements over a period of time is the accumulation of continuous Random Walks. In other words, that is the bare bones version of Brownian motion. With the rules of Brownian motion set, it can officially be created.

Let there be n number of periods of total length T. The length of a period is then $h=\frac{T}{n}$. Stock price change, the difference of one stock price from a stock price the period before can be defined as: $S_{t+h}-S_t=Y_{t+h}\sqrt{h}$

The Y variable is the Random Walk effect. WHY THE SQUARE ROOT OF THE TIME PERIOD?

Let time start at zero, and instead of analyzing the individual periods (that are infinitely small), analyze the entire time period as a whole. Let’s be practical here; we’re interested in stock price movements over a course of real time, such as a day, week, month, quarter or year, not at a blink of an eye! The sum of these infinitely small periods is simply T, so a series can be developed.

$S_T-S_0=\sum_{k=1}^{n}Y_{hk}\sqrt{h}$

$=\sqrt{h}\sum_{k=1}^{n}Y_{hk}$

$=\sqrt{\frac{T}{n}}\sum_{k=1}^{n}Y_{hk}$

In Brownian motion, we want to find the expected stock price change, so some basic statistics have to be applied.

$E[S_T-S_0]=\sqrt{T}*E[\frac{\sum_{k=1}^{n}Y_{hk}}{\sqrt{n}}]$

The expected value of the Y variable in the series is zero as it is equally likely to go up as it is to go down. Therefore, the expected stock price change is zero!

This doesn’t mean that the model predicts no change in the stock. The dynamics of variance must be considered. The variance of the Y variable is simply one unit as the random walk will either go up 1 or go down 1.

$Var[S_T-S_0]=Var[\frac{\sum_{k=1}^{n}Y_{hk}}{\sqrt{n}}]$

$=Var[\frac{\sum_{k=1}^{n}1}{\sqrt{n}}]$

$=Var[\frac{1}{\sqrt{n}}]*n$

$=\frac{n}{n}=1$

This is why the square root of the time period is taken. It lets the sum of independent random variables approach a standard normal distribution! This is the Central Limit Theorem at work. This is the structure of Brownian motion.

These are some of the notes that I have been writing. As you can see, I have on simple question regarding the parameter of time.

2. Haven't you answered your question yourself (through the reference to the normalization in the central limit theorem)? By the way, you forgot the $T$ in the computation of the variance: one has $Var(S_T-S_0)=T$.

3. Originally Posted by Laurent
Haven't you answered your question yourself (through the reference to the normalization in the central limit theorem)? By the way, you forgot the $T$ in the computation of the variance: one has $Var(S_T-S_0)=T$.
Where do you see $Var(S_T-S_0)=T$?

Wait, there should be a T (square root of T squared). Yep, missed that. I'm still not sure why the square root of time is involved here.

4. ## Multiplication Rules

I also have a question regarding the multiplication rules used to apply Ito's Lemma.

$dt * dZ = 0$
$(dt)^2 = 0$
$(dZ)^2 = dt$
$dZ * dZ' = \rho dt$

My reasoning as I have nothing but these rules just given to me from my text.

$(dt)^2 = 0$
The infinitesimal nature of dt makes it very small, so when raised to a power, it effectively becomes zero (at least very close to).

$(dZ)^2 = dt$
$dZ * dZ' = \rho dt$
The multiplication of two standard Brownian motion processes together (dZ) will result in a change in time with a correlation factor ( ). However, why does $(dZ)^2 = dt$ ?

$dt * dZ = 0$

5. There are plenty of theoretical difficulties appearing when one wants to define stochastic integration. The "rules" in your textbook are an attempt to obviate those but, while convenient, these rules provide little understanding as you noticed.
In fact, what you write $dZ\ast dZ'$ or $(dt)^2$ are Stieltjes measures relative to quadratic covariations of the semi-martingales Z, Z', t. That's amusing you ask about $(dZ)^2=dt$ (which I would have written $\langle Z, Z\rangle_t = t$), because this is related to ${\rm Var} Z_t = t$, your first question.
$t$ is a bounded variation function, so that $\langle t, Z\rangle_t = 0$ and $\langle t, t\rangle_t =0$. Well, it probably doesn't help you much, but I don't have an intuitive answer to give. At least you know what to look for (in the Wikipedia or textbooks) if you want to learn more...