ito's lemma help

[baozi]

Hello everyone:
I'm not sure if it's the right place to ask.I'm reading some quant math books and get confused with Ito's Lemma.
dF =dF/dX*dX +1/2*(d^2 F)/dX^2*dX^2;(dX ~N(0,dt^2)),,, this part is straightforward.
The book told me dX^2 is defined by dX^2= dt as dt goes to 0?
How to prove this property?

 

[Polter]

Some books believe that using handwaving arguments talking about "dX^2= dt" instead of simply introducing quadratic variation is somehow helpful. I believe that these books are misguided.

What you're trying to prove is that [W](t) = t. Forget about raising differentials to a power, that's simply an informal notation to express the same thing (in other words, it won't guide you to the answer).

See:

• http://planetmath.org/quadraticvariation
• http://planetmath.org/quadraticvariationofbrownianmotion
• http://almostsure.wordpress.com/201...ations-and-integration-by-parts/#scn_ibp_sec2
• http://math.stackexchange.com/questions/92938/quadratic-variation-of-brownian-motion

Other relevant links (passing-by-reference to avoid repetition):
http://www.wilmott.com/messageview.cfm?catid=8&threadid=94565#672165

 

[curltron]

Hi baozi,

It should be that dX ~ N(0,dt) if dX^2 = dt.

 

[quotes]

Just suppose you have a time grid:
t1-t0, t2-t1, t3-t2, ..., tn-tn-1 (tn=T)
you want to generate a random motion W on it:
W1-0, W2-W1, W3-W2, ..., Wn-Wn-1 (Wn=W)
and you want the incremental W's to have the same independent distribution with expectation 0 and variance s*s
So you get the total variance of them, or the variance of W, as n*s*s, fix it as s(T)*s(T)=T

Now you can let n gets larger, the tim grid finer, but the variance of W the same.
Under this assumption,
If you do the sum of quadratic variation of incremental W's
you get
(W1-0)^2 + (W2-W1)^2 + (W3-W2)^2 +...+ (Wn-Wn-1)^2 ~ s(T)*s(T)
and the sum gets more stable as n gets larger
meaning dW*dW ~ dt=T/n