Infinitesimal Increment of the Wiener Process (squared) - to then prove Ito integral

Question

I am a bit stuck trying to compute the infinitesimal increment for $X=W_{t}^{2}$.

I usually would do something like the following but a bit lost..

$x_{t} = W_{t}*W_{t}$

$x_{t+h} = (W_{t+h}-W_{t}) (W_{t+h}-W_{t})$

$x_{t+h} = (W_{t+h}^2)-2*(W_{t+h}W_{t})-(W_{t}^2)$

But I am sure this is not correct somewhere as I can't find any simplification to use in the second part....

I need to use the solution here to proof the Ito Integral for

$ \int_{0}^{u} W_{t}dW{t} = \frac{1}{2}(W_{u}^{2}-U)$

Any pointers/help would be appreciated.

Answer 1

The increment of $x_t=W_t^2$ is not your expression for $x_{t+h}\,$.

It is \begin{align} x_{t+h}-x_t&=W_{t+h}^2-W_t^2=W_{t+h}^2-2W_{t+h}W_t+W_t^2-2W_t^2+2W_{t+h}W_t\\ &=(W_{t+h}-W_t)^2+2W_t(W_{t+h}-W_t). \end{align} By the definition of the Ito integral and the quadratic variation this implies $$ x_t=W_t^2=\langle W\rangle_t+2\int_0^tW_s\,dW_s=t+2\int_0^tW_s\,dW_s $$ which is equivalent to the equation you wanted to prove.