How to calculate $\int_{z=0}^T \int_{s=0}^z dW(s)dW(z)$

Question

We are given the following integral $$ \int_{z=0}^T \int_{s=0}^z dW(s)dW(z). $$

My approach was to discretize the integral, obtaining $$ \lim_{n \rightarrow \infty} \sum_{i = 1}^{n-1}\left[\lim_{m \rightarrow \infty} \sum_{j = 1}^{m - 1} W(t_j) - W(t_{j-1}) \right](W(t_i) - W(t_{i-1})) $$ where $t_i = i\frac{T}{n}$ and $t_j = i\frac{z}{m}$.

Then, by independence of Brownian Motion increments we can argue that the previous equals (not sure about this part) $$ \lim_{n \rightarrow \infty} \sum_{i = 1}^{n-1}(W(t_i) - W(t_{i-1}))^2. $$ for which I can show that it converges to $T$.

Am I solving this integral in the wrong way?

Answer 1

For example using the limit formula you presented you can see that

$$\int_{s=0}^zdW_s=W_z-W_0=W_z$$

where I normalized $W_0=0$. We then need to evaluate

$$\int_{z=0}^TW_zdW_z$$

You can again evaluate this using the limit formula. Alternatively apply Ito's lemma to $W_t^2$. You get

$$d(W_t^2)=2W_tdW_t+dt$$

i.e.

$$W_t^2=\int_{z=0}^T2W_tdW_t+T$$

and

$$\int_{z=0}^TW_tdW_t=0.5(W_T^2-T),$$

which is the final result.