Itô’s Lemma and the Brownian Integral of Brownian Motion

# Intergral of Brownian motion w.r.t. Brownian motion


# Intergral of Brownian motion w.r.t. Brownian motion












I don't understand why $S$ (highlight on picture), I learned $$\int_0^t W(s) dW(s) = \left. \frac{1}{2} (W^2(s)-s) \right \vert_0^t $$

everyone please explain for me. Thank you

## Answer by Richi Wa (score 5, accepted)

https://quant.stackexchange.com/a/25021

Apply Ito's lemma to $f(W_t) = W_t^2$ then $$ f(W_T) = f(W_0) + \int_0^T f'(W_t) dW_t + \frac{1}{2} \int_0^T f''(W_t) dt. $$ Thus $$ W_T^2 = 2 \int_0^T W_tdW_t + \frac12 2 T = 2 \int_0^T W_tdW_t + T. $$ If we rearrange terms then we get $$ \int_0^T W_tdW_t = (W_T^2-T)/2. $$

## Answer by Borun Chowdhury (score 6)

https://quant.stackexchange.com/a/25051

While Richard's answer is technically correct, just saying the result can be obtained using Ito's formula doesn't make the issue much clearer. So let me go into the microscopics of the issue.

The Ito integral is defined in the following way. Suppose we divide the time interval $[0,t]$ into $n$ pieces with $t_i = i~dt$ where $dt=\frac{t}{n}$ then we define the Ito integral as $$ \int_0^t G(t') dW(t') = {\text{ms-lim}}_{n\to \infty} \sum_{i=1}^n G(t_{i-1}) ( W_{t_i} - W_{t_{i-1}})~, $$ where the mean-square limit is defined in the following way: $$ {\text{ms-lim}}_{n\to \infty} X_n = X \Leftrightarrow \lim_{n \to \infty} \langle (X_n - X)^2 \rangle =0~. $$

Now let us evaluate the expression we would get on the RHS when evaluating $\int_0^t W(s) dW(s)$ from the above definition: $$ \begin{eqnarray} Y^{(1)}_n &=& \sum_{i=1}^n W_{t_{i-1}} (W_{t_i} - W_{t_{i-1}})~. \end{eqnarray} $$ Now noting each increment $(W_{t_i} - W_{t_{i-1}}) = \sqrt{dt} \mathcal N_{i-1}(0,1)$ is an independent normal distribution we get $$ \begin{eqnarray} Y^{(1)}_n &=& dt \sum_{i > j} \mathcal N_i(0,1) \mathcal N_j(0,1)~. \end{eqnarray} $$ The other expression we get from the discretized version of the RHS of the standard answer $\int_0^t W(s) dW(s)=\frac{1}{2} (W(t)^2 - t)$ is $$ \begin{eqnarray} Y^{(2)}_n &=& \frac{1}{2} \left(\sqrt{dt} \sum_{i =1}^n \mathcal N_i(0,1)\right)^2 - \frac{1}{2} \sum_{i=1}^n dt~. \end{eqnarray} $$

Now if we can just show that the two expressions are same in the mean-square limit we are done. We see that $$\begin{eqnarray} && \lim_{n \to \infty} \langle (Y^{(2)}_n - Y^{(1)}_n)^2 \rangle \\ &=& \lim_{n \to \infty} \langle \left( \frac{dt}{2} \left[ \left( \sum_{i,j} - \sum_{i>j} - \sum_{j>i} \right) \biggl( \mathcal N_i(0,1) \mathcal N_j(0,1) \biggr) - \sum_{i} 1 \right] \right)^2 \rangle \\ &=& \lim_{n \to \infty} \frac{dt^2}{4} \langle \left(\sum_{i=1}^n \mathcal N_i(0,1)^2 -n\right)^2 \rangle \\ &=& \lim_{n \to \infty} \frac{dt^2}{4} \langle (\chi_n^2 -n)^2 \rangle \\ &=& \lim_{n \to \infty} \frac{dt^2}{4} \operatorname{Var}(\chi_n^2)\\ &=& \lim_{n \to \infty} \frac{dt^2}{2} n \\ &=& \lim_{n \to \infty} \frac{t^2}{2n} \\ &=& 0 \end{eqnarray} $$ Here we used the results that the squares of $n$ standard normals is a chi-square with degree $n$ and that the same has mean $n$ and variance $2n$.

Hence we are able to prove in detail why $\int_0^t W(s) dW(s)=\frac{1}{2} (W(t)^2 - t)$.

## Answer by Arshdeep (score 1)

https://quant.stackexchange.com/a/78817

The surprise is not whether there's an extra "s" term, it is how you were able to approximate a very complex sum in the first place (dW is not small enough for normal approximations to work - riemann integrals rely on slicing the "dW" term in as small pieces as one needs). So Ito discovered you could still approximate (simplify) the integrals because you can make dW^2 as small as possible. That is Ito's lemma.

It is a neat coincidence that you can look at it as a quadratic adjustment to the normal case.