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Math Help - Ito's Lemma

  1. #1
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    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.
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  2. #2
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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.
    No clue what's going on there, this stuff is out of my league at the moment but

    I'm having trouble showing that V(X/c) = V(X)/c^2.
    I can do that
    Variance is defined as var(X) = E((X-E(X))^2)
    so var(X/c) = E((X/c-E(X/c))^2)
    = E((X/c-E(X)/c)^2) since expectation is linear
    =E((1/c)^2(X-E(X))^2)
    =E((X-E(X))^2)/c^2
    =var(X)/c^2
    ...wait v does stand for variance right?
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  3. #3
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    Hello,

    Quote Originally Posted by badgerigar View Post
    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 var(X) = E((X-E(X))^2)
    so var(X/c) = E((X/c-E(X/c))^2)
    = E((X/c-E(X)/c)^2) since expectation is linear
    =E((1/c)^2(X-E(X))^2)
    =E((X-E(X))^2)/c^2
    =var(X)/c^2
    ...wait v does stand for variance right?
    A definition that helps more :

    var(X)=\mathbb{E}(X^2)-(\mathbb{E}(X))^2

    var\left(\frac Xc\right)=\mathbb{E}\left(\frac{X^2}{c^2}\right)-\left(\mathbb{E}\left(\frac Xc\right)\right)^2

    \forall \text{constant } a ~,~ \mathbb{E}(aX)=a\mathbb{E}(X)

    & the rest follows
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