Calculating a quadratic variation of a process

Question

I am self-studying stochastic calculus with several university lecture notes and encountered this example given as an exercise. If $Y_t = t(\int_0^tW_sdW_s)^2$ with $W$ being a Brownian motion, then what is its quadratic variation? It seems to me that some heuristic (using the definition of quadratic variation) or Ito's formula do not work for this case. I just guess that Ito's isometry would apply or a novel application of Ito's formula would work. Also, it would be enormously appreciated if you could recommend me some books or lecture notes containing several examples with which I may be accustomed to applications of several techniques of stochastic calculus like this case.

Answer 1

Using Ito's formula for a semi-martingale $(t,\int_{0}^{t}W_{s}{\rm d}W_{s})_{t \geq 0}$ and the function $f(x,y)=xy^{2}$, we obtain \begin{align*} t\left(\int_{0}^{t}W_{s}{\rm d}W_{s}\right)^{2} &=\int_{0}^{t}s2\left(\int_{0}^{s}W_{u}{\rm d}W_{u}\right){\rm d}(\int_{0}^{\bullet}W_{u}{\rm d}W_{u})_{s}+\int_{0}^{t}\left(\int_{0}^{s}W_{u}{\rm d}W_{u}\right)^{2}{\rm d}s \\ &\hspace{0.5cm}+\frac{1}{2}\int_{0}^{t}s2{\rm d}\langle \int_{0}^{\bullet}W_{u}{\rm d}W_{u}\rangle_{s} \\ &=2\int_{0}^{t}s\left(\int_{0}^{s}W_{u}{\rm d}W_{u}\right)W_{s}{\rm d}W_{s}+\int_{0}^{t}\left(\int_{0}^{s}W_{u}{\rm d}W_{u}\right)^{2}{\rm d}s \\ &\hspace{0.5cm}+\int_{0}^{t}sW_{s}^{2}{\rm d}s. \end{align*} Thus we obtain \begin{align*} \langle Y \rangle_{t} &=2\langle \int_{0}^{\bullet}s\left(\int_{0}^{s}W_{u}{\rm d}W_{u}\right)W_{s}{\rm d}W_{s} \rangle_{t} \\ &=2\int_{0}^{t}s^{2}\left(\int_{0}^{s}W_{u}{\rm d}W_{u}\right)^{2}W_{s}^{2}{\rm d}s =2\int_{0}^{t}sY_{s}W_{s}^{2}{\rm d}s. \end{align*}