I need to campute the quadratic variation for a compound poisson process defined as $X_{t}=\sum_{i=1}^{N_{t}}{Y_{i}}$. I also know that the compensated poisson process can be expressed as $X_{t}-\lambda t E[Y]$ and that the quadratic variation is: $[X,X]_{t}=|x_{t}|^{2}-2\int_{0}^{T}{X_{u^{-}}dX_{u}}$.
My idea to show this, was to substitute inside the quadratic variation and I find:
$$ \begin{align} [X,X]_{t}&=|\sum_{i=1}^{N_{t}}{Y_{i}}|^{2}-2\int_{0}^{T}{X_{u^{-}}d(X_{u}-\lambda E[Y]u)}\\ &= \sum_{i=1}^{N_{t}}|{Y_{i}}|^{2}-2\left[\int_{0}^{T}{X_{u^{-}}d(X_{u})}-\lambda E[Y]\int_{0}^{t}{X_{u^{-}}du}\right] \end{align}$$ Now, can I say that $\int_{0}^{T}{X_{u^{-}}d(X_{u})}=\lambda E[Y]$? and that $\int_{0}^{t}{X_{u^{-}}du}=1$?
To compute the Quadratic covariation should I use the polarization identity?
The "square bracket" of $X$ is just $$ [X,X]_t=\sum_{s\le t}(\Delta X_s)^2=\sum_{i=1}^{N_t}Y_i^2, $$ and the associated "angle bracket" (predictable quadratic variation) is the compensator of $[X,X]$, namely $$ \langle X,X \rangle_t = \lambda t E[Y^2_1]. $$