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Here I have this question

(i) state Ito's formula

(ii) hence or otherwise show that

$\int^t_0B_s dB_s = \dfrac{1}{2}B^2_t -\dfrac{1}{2} t$

(iii) define the quadratic variation $Q(t)$ of Brownian motion over [0,t], given that $Q(t) = t$, use this result to prove (ii)

I can do everything up to the last bit of (iii), how can quadratic variation tell you this relationship?

disclaimer: this is not homework. I am trying to help a friend preparing for an exam and this was a past paper question.

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Please change the title of this "question" (currently being "Exercise on stochastic calculus" to meaningful (e.g. use words like "Ito formula" and "Quatratic Variation"). – Christian Fries May 11 '13 at 18:37

3 Answers 3

up vote 2 down vote accepted

"Like" Ito: $$d (B^2) = B dB + B dB + dB dB$$

That is $$B dB = \frac{1}{2} d (B^2) - \frac{1}{2} dB dB$$

Integrate. Last term is 1/2 the quadratic variation.

I understand the questions as follows: In iii) one has to define what $dB dB$ stands for and one has to "proof" the first line in my answer. In ii) one may use Ito to "know" that $dB dB = dt$.

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+1 sweet use of Ito product rule. – Veeken May 11 '13 at 19:33
how is this any different to the answer for part (ii) where you are asked to show that was the case using ito's formula? – Lost1 May 11 '13 at 23:45
Ito actually tells you that $d(B^2) = BdB + BdB + dt$. So ii) might mean: Proof it using Ito, while iii) means proof it in an elemantary by repeating the proof of Ito for the special case of $B^2$. – Christian Fries May 13 '13 at 13:53
You can either give a definition a quadratic variation or prove Ito's lemma for the Brownian motion from scratch and use it to define its quadratic variation. But you can't define the quadratic variation as the thing that makes Ito's lemma work. The right way to do it imho is to define the quadratic variation as the limit of a finite sum of squared increments. Otherwise you just forget about what this is really about and that's a recipe for disaster. For example, you would have no idea what the quadratic variation of a Poisson process is. – AFK Jun 6 '14 at 0:04

i picked this off from Shreve.

Start with the definition of sampled quadratic variation: (1) $\frac{1}{2}Q_\pi = \frac{1}{2}\sum\nolimits_{j=0}^{n-1} (W_{j+1}) - W_j)) ^2$ where $\pi$ = {0,1,2...,n} is a partition of $[0,T]$ (Note we took $\frac{1}{2}$ of both sides for reasons that will be clear in the next line.) Now we know (1) is equal to $\frac{T}{2}$, but we also know by simple algebra that
(1) =$\frac{1}{2}W_n^2 + \sum\nolimits_{j=0}^{n-1} W_j(W_j - W_{j+1})$.

All that remains to show the result is to make it rigorous in the sense that we're approximating a Brownian motion with a discretized version that converges in the limit as $n \to \infty$; we're also approximating the ito integral with sums, which also converge in the limit. Will leave this to you to iron out a bit further. Again, reference Shreve's notes if you don't have his excellent texts (google search :steve shreve notes)

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What i am asking is, how do you know the expression for part (ii) using knowledge of quadratic variation? – Lost1 May 11 '13 at 23:46

I think the question is asking to prove (ii) using (iii) alone, which precludes using Ito's lemma outright.

Indeed, for $T>0$ and $\Pi=\{t_{0}=0,t_{1},\ldots,t_{n}=T\}$, we get from Taylor's theorem applied to $f(w)=\frac{1}{2}w^{2}$, $$\frac{1}{2}(W(T))^{2}=\sum_{i=0}^{n-1}W(t_{i}))(W(t_{i+1})-W(t_{i}))+\frac{1}{2}\sum_{i=0}^{n-1}(W(t_{i+1})-W(t_{i}))^{2}.$$ Higher order terms vanish because $f^{(k)}\equiv0$ for $k\geq3$.

According to (iii), the second term is $\frac{1}{2}T$ and the first term is just an Ito sum which converges to the Ito integral $\int_{0}^{T}W(s)\;dW(s)$ as $||\Pi||=\max_{i}|t_{i+1}-t_{i}|\to0$.

Thus (as the LHS is unaffected by taking limits),


Rearranging terms we get $$\int_{0}^{T}W(s)\;dW(s)=\frac{1}{2}(W(T))^{2}-\frac{1}{2}T$$ which is what (i) asserts, but obtained from (more or less) first principles (i.e. (iii)).

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Please give me the argument that $f^k$ vanish for $k\geq 3$. – Zbigniew Dec 31 '14 at 2:36
Because $f=\frac{1}{2}w^{2}$, $f^{(1)}=w$, $f^{(2)}=1$, and therefore $f^{(k)}\equiv0$ for $k=3,4,\ldots$ – Taylor Martin Dec 31 '14 at 13:36

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