Ito's Lemma

Hello. Let me give some background.

The claim in Ito's Lemma is that dW^2 goes to dt as dt goes to zero and a partial justification is lim Q_n (n to infinity) = (b-a).

But Q_n is the same as integral (from a to b) dW^2. Letting a=0 and b=t, this becomes integral (0 to t) dW^2 = t and differentiation gives dW^2 = dt.

This can be written as :

E{(W(t + delta t)-W(t))^2} = E{((W(t+delta t)-W(t))/sqrt(delta t))^2 * delta t}= delta t.

I need to show that (W(t+delta t)-W(t))/(sqrt (delta t) equals N(0,1) or the standard Gaussian.

I can use the fact X is a random variable, E(X)=0 and finite variance, then I need to show V(X/c) = V(X)/c^2.

I'm having trouble showing that V(X/c) = V(X)/c^2.

Any help would be greatly appreciated.

Thanks.

Quote:

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The claim in Ito's Lemma is that dW^2 goes to dt as dt goes to zero and a partial justification is lim Q_n (n to infinity) = (b-a).

But Q_n is the same as integral (from a to b) dW^2. Letting a=0 and b=t, this becomes integral (0 to t) dW^2 = t and differentiation gives dW^2 = dt.

This can be written as :

E{(W(t + delta t)-W(t))^2} = E{((W(t+delta t)-W(t))/sqrt(delta t))^2 * delta t}= delta t.

I need to show that (W(t+delta t)-W(t))/(sqrt (delta t) equals N(0,1) or the standard Gaussian.

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No clue what's going on there, this stuff is out of my league at the moment but

Quote:

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I'm having trouble showing that V(X/c) = V(X)/c^2.

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I can do that
Variance is defined as
so
since expectation is linear



...wait v does stand for variance right?

Hello,

Quote:

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Originally Posted by badgerigar

No clue what's going on there, this stuff is out of my league at the moment but



I can do that
Variance is defined as
so
since expectation is linear



...wait v does stand for variance right?

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A definition that helps more (Hi) :







& the rest follows :)