I am new about stochastic calculus but I would like to know if the following procedure for computing $E\left(L^{2}_{t}\right)$ and $E\left(L^{3}_{t}\right)$ if $L_{t}$ is a Lévy pure jump process is correct.
The assumptions considered are $\int^{t}_{0}E\left(\left|L^3_{u}\right|^2\right)du<+\infty$ and $E\left[L_1\right]=0$. Using integration by parts I get:
$$ L_{t}^{2}=2\int_{0}^{t}L_{u-}dL_{u}+\left[L,L\right]_{0}^{t}. $$
Recalling the Teugels martingales $\bar{L}_{t}^{\left(i\right)}=\sum_{0<h\leq t}\left(\Delta L_{h}\right)^{i}-m_{i}t,\ \ \ i\geq1$ with $m_1=0$ and $i\geq2 , m_{i}=\int x^{i}d\nu_{L}\left(x\right) $ ($\nu_{L}$ Lévy measure), the quadratic variation can be written as:
$$ \left[L,L\right]_{0}^{t}=\bar{L}_{t}^{\left(2\right)}+m_2 t, $$
then $$ L_{t}^{2}=2\int_{0}^{t}L_{u-}dL_{u}+\bar{L}_{t}^{\left(2\right)}+m_2 t. $$ Take the expectation of both sides of the equation and get: $$ E\left[L_{t}^{2}\right] = m_2 t. $$
For computing the quantity $E\left(L^{3}_{t}\right)$, I apply the integration by parts to the process $L^3_t=L_t*L_t^2$: $$ L_{t}^{3}=\int_{0}^{t}L_{u-}^{2}dL_{u}+\int_{0}^{t}L_{u-}dL_{u}^{2}+\left[2\int_{0}^{t}L_{u-}dL_{u}+\left[L,L\right]_{0}^{t},L\right]_{0}^{t} $$ Using the definition of quadratic covariation I get $$ \left[\left[L,L\right],L\right]_{0}^{t}=\sum_{0<h\leq t}\Delta\left[L,L\right]_{h}\Delta L_{h} $$ now $\Delta\left[L,L\right]_{h}=\sum_{0<s\leq h}\left(\Delta L_{s}\right)^{2}-\sum_{0<s<h}\left(\Delta L_{s}\right)^{2}=\left(\Delta L_{h}\right)^{2}$ then $$ \left[\left[L,L\right],L\right]_{0}^{t}=\sum_{0<h\leq t}\left(\Delta L_{h}\right)^{3}=\bar{L}_{t}^{\left(3\right)}+m_{3}t $$ where $\bar{L}_{t}^{\left(3\right)}$ is Teugels Martingales associated to the process $\sum_{0<h\leq t}\left(\Delta L_{h}\right)^{3}$. Rewrite $L^3_t$ as: $$ L_{1}^{3}=\int_{0}^{t}L_{u-}^{2}dL_{u}+2\int_{0}^{t}L_{u-}^{2}dL_{u}+\int_{0}^{t}L_{u-}d\bar{L}_{u}^{\left(2\right)}+m_{2}*\int_{0}^{t}L_{u-}du+2\int_{0}^{t}L_{u-}d\left[L,L\right]_{u}+\left[\left[L,L\right],L\right]_{0}^{t}= $$ $$ =3\int_{0}^{t}L_{u-}^{2}dL_{u}+3\int_{0}^{t}L_{u-}d\bar{L}_{u}^{\left(2\right)}+3m_{2}\int_{0}^{t}L_{u-}du+\bar{L}_{t}^{\left(3\right)}+m_{3}t $$ Computing the expectation of both sides I have $E\left(L_{1}^{3}\right)=m_{3}$.