Let the stochastic process given by $K=M^2-4\int_{0}^{t}W^2_s ds $ , where M is a stochastic process $(M_t)_{ t \geq 0}$ given by : $Μ_t=W^2_t-t$ and W is a standard Brownian Motion.
I need to calculate the predictable quadratic variation of K, < K >
My progress so far is proving that $4\int_{0}^{t}W^2_s ds=<X>$ where $X=W^2_t$ but i dont know how to move on to calculate < K >
The integral with respect to $ds$ is a finite variation process by definition, hence has zero quadratic variation. So the only contribution will be from the $M_t$ term.
Note that by definition, for a continuous local martingale $X$, the quadratic variation $[X]_t$ is the unique finite variation process such that $X^{2}_{t}-[X]_t$ is itself a continuous local martingale.
But you can check that $M_t = W_{t}^{2}-t$ is actually a true martingale, and so is $M_{t}^{2}-\mathbb{E}[M_{t}^{2}]$. You can verify this directly from the definitions (it holds for adapted processes with independent increments more general than Brownian motion, subject to integrability). So by uniqueness, $[M]_t = \mathbb{E}[M_{t}^{2}]$.
We can calculate : \begin{equation} \mathbb{E}[M_{t}^{2}] = \mathbb{E}[W_{t}^{4}]-2t\mathbb{E}[W_{t}^{2}]+t^2. \end{equation} By second and fourth moments of a normal distribution, $\mathbb{E}[W_{t}^{2}] =t$, $\mathbb{E}[W_{t}^{4}]=3t^2$. Hence, $[M]_t = 2t^2 = [K]_t.$