Let $N = (N\_t)\_{t\geq 0}$ be a Poisson process and consider random variables $Z\_n$, $n\in N$. Compute the quadratic variations $[X]\_t$ where $X\_t = \sum\_{n=1}^{N\_t}Z\_n$.


What I did was plugging $X\_t$ into formula $V^2\_t(X:τ\_n=∑(X(t\_i)−X(t\_{i−1}))$, but I dont think it is completely right. Could you please help me with the solution?



Quadratic variation of sum of random variables
You can just use the definition of quadratic variation (see <https://en.wikipedia.org/wiki/Quadratic_variation>) to infer that 
 $[X]\_t = \sum\_{n=1}^{N\_t} Z\_n^2$. 


To see this, it may help to think of the points of the Poisson process $0<S\_1<S\_2<\ldots$ rather than the counting function $(N\_t)$. Recall that $S\_k$ is just the $k$'th discontinuity of the increasing step function $t \mapsto N\_t$. The process $(X\_t)$ is constant on the half-open intervals of the form $[S\_{k-1},S\_k)$ and on $[0,S\_1)$, and has a jump of magnitude $Z\_k$ at time $S\_k$, that is $X\_t=\sum\_{k \ge 1} Z\_k 1\_{S\_k \le t}$. 


Given a partition $\{t\_j\}\_{j=0}^L$ of $[0,t]$ (where $t\_0=0$ and $t\_L=t$) whose mesh is smaller than $\min\{S\_k-S\_{k-1} \, : k \, \le N\_t\}$ you will find that $\sum\_{j=1}^L (X\_{t\_j}-X\_{t\_{j-1}})^2= \sum\_{n=1}^{N\_t} Z\_n^2$.


