Using Brownian Quadratic Variation to Derive the Itô Integral Identity

# Quadratic variation question


# Quadratic variation question












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.

## Answer by Christian Fries (score 2, accepted)

https://quant.stackexchange.com/a/7961

"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 $dB dB = dt$.

## Answer by Vince (score 2)

https://quant.stackexchange.com/a/7962

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)

## Answer by Sargera (score 0)

https://quant.stackexchange.com/a/11573

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),

$\frac{1}{2}(W(T))^{2}=\int_{0}^{T}W(s)\;dW(s)+\frac{1}{2}T.$

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)).