The stochastic integral $\int W_t dW_t$

Question

I'm reading an introduction to Stochastic Calculus. I'm at the point where Ito integrals are developed and constrasted with the Stratonovich integral.

Below is a calculation of $\int_0^T W_t d W_t$.

How do I get from the first equation to the second? I presume ms-lim means $L^2$-limit, but even so, I can't seem to make sense of what the author's doing. I have other references for this integral, but I would like to understand this particular proof.

$W_t$ is a Wiener process and equation (4.29) just says:

$$ \mathbb{E}[(W_s-W_t)^2] = |s - t| $$

Answer 1

To use intuitive terms: basically, equation 4.29 is saying that, using a much simpler notation, the variance of the process, since the mean is 0, becomes $ \mathbb{E}[(W_{t+\Delta t}-W_{t})^2] =\Delta t$, and since there is independence, $ \mathbb{E}[(W_{t+\Delta t}-W_{t}) (W_{t+2 \Delta t}-W_{t+\Delta t})] =0$. He is taking $K$ to $\infty$ means making the increments $\Delta t$ smaller and smaller since the overall interval $[0,T]$ does not change.

Now the difference between Ito and Stratonovich lies in (once again, maximally simplifying notation): When you do a sum of a function of $W$, in one case you have "nonanticipation": $\sum_i^{\frac{n}{\Delta t}} f(W_{t+i \Delta t}) ( W_{t+(i+1) \Delta t}-W_{t+i \Delta t}),$ in the other $\sum_i^{\frac{n}{\Delta t}} f(W_{t+(i+1) \Delta t})( W_{t+(i+1) \Delta t}-W_{t+i \Delta t})$. You make as with the Riemann integral $\Delta t$ smaller and smaller and increase the number of steps.

Answer 2

The meaning is that if you take $\tau_k$ equal to $t_{k-1} $ you get the usual Ito integral, whereas if you take it in between $t_{k-1}$ and $t_k$ you don't. In particular if you always take exactly the midpoint then you get an additional term T/2 (check yourself). This might come as a surprise since for example for Riemann integrals (of continuous functions ) it doesn't matter which point you choose on each interval