Cfrgtkky

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: $[X,X]_{t}=|x_{t}|^{2}-2int_{0}^{T}{X_{u^{-}}dX_{u}}$.

My idea to show this, was to substitute inside the quadratic variation and I find:

$$ begin{align} [X,X]_{t}&=|sum_{i=1}^{N_{t}}{Y_{i}}|^{2}-2int_{0}^{T}{X_{u^{-}}d(X_{u}-lambda E[Y]u)}\
&= sum_{i=1}^{N_{t}}|{Y_{i}}|^{2}-2left[int_{0}^{T}{X_{u^{-}}d(X_{u})}-lambda E[Y]int_{0}^{t}{X_{u^{-}}du}right]
end{align}$$
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
$$
[X,X]_t=sum_{sle t}(Delta X_s)^2=sum_{i=1}^{N_t}Y_i^2,
$$
and the associated "angle bracket" (predictable quadratic variation) is the compensator of $[X,X]$, namely
$$
langle X,X rangle_t = lambda t E[Y^2_1].
$$