• Feb 26th 2007, 02:43 AM
chogo
hi all, Ive searched online for ages but even though i can find formulas and definitions i dont quite understand what quadratic variation is.

The reason i want to know this is, i'm looking to understand the importance of the fact that the quadratic variation of a Brownian motion process is t

B[0,t] = t

does anyone know how to explain this?
• Feb 26th 2007, 11:26 AM
ThePerfectHacker
Quote:

Originally Posted by chogo
hi all, Ive searched online for ages but even though i can find formulas and definitions i dont quite understand what quadratic variation is.

I know absolutely nothing about Brownian motion (:( I wish I did) but I looked at Wikipedia, it made some sense how they explain it.

Heir.

It looks like a variation of the Riemann integral to me.

Let me give an example.
Say we want to find the quadradic variation of f(t)=t on the interval [0,1]

So we partition the region (like in the Riemann integral), I am going to be using left-endpoints because that is what they (Wikipedia) use.

Thus, divide the region into "n" parts. Each part has 1/n width.

The initial value is
x_0=0
x_1=0+1/n
x_2=0+1/n+1/n
x_3=0+1/n+1/n+1/n
...
You get the idea.
Thus,
x_k = k/n
Where 0<=k<=n-1
(Because of left-endpoints).

Now,
[f(x_1)-f(x_0)]^2=(1/n-0/n)^2=1/n^2
[f(x_2)-f(x_1)]^2=(2/n-1/n)^2=1/n^2
[f(x_3)-f(x_2)]^2=(3/n-2/n)^2=1/n^2
...
[f(x_{n-1}-f_{n-2}]^2=[(n-1)/n-(n-2)/n]^2=1/n^2

Now, we sum them up,
<f>_1 = SUM (of that whole thing) = 1/n^2+1/n^2+...+1/n^2=(n-1)/n^2

Now, we take the limit as n --> infinity.
This clearly goes to zero.
Thus, the quadradic variation of f(t)=t of 1 (meaning on interval [0,1]) is zero.

Now, they mention an interesting theorem.