A question about the compound Poisson distribution In the context of compound Poisson distribution we have:

*

*$X \sim F$ a random variable;

*$\{X_n\}_{n\in \mathbb{N}}$ be a i.i.d. sequence of copies of $X$ - i.e. $X_j \sim F\,\, \forall \, j \in \mathbb{N}$ and $X_i, X_j$ independent for all $i,j \in \mathbb{N}$.

*$N\sim Po(\lambda)$ a Poisson random variable such that $N$ and $X_j$ are independent for all $j$.

Define the Poisson Sum as:
$$Y = \sum_{j=1}^N X_{j}$$
Accordingly to this (Properties section), when we calculate $E(Y)$ or $V(Y)$, I would like to know at what point it is used that $X_j$ and $N$ are independent?
I do not know the importance and use of the independence!
Help!
 A: I'm assuming

*

*$\{X_j\}_{j\in \mathbb{N}}$ is an iid sequence, i.e.  they are identically distributed and mutually--not pairwise as you have it-- independent.

*$N$ is independent of $\{X_j\}_{j\in \mathbb{N}}$.

By Adam's Law, the mean of $Y$ is
$$\small \begin{aligned} E\left[\sum_{j=1}^N X_j \right] &=E\left[E\left[\sum_{j=1}^N X_j\Bigg|N \right]\right]\\
&=E\left[\sum_{j=1}^NE\left[ X_j\big|N \right]\right]&&\qquad (\text{linearity of }E[\cdot ])\\
&=E\left[\sum_{j=1}^NE\left[ X_j \right]\right]&&\qquad (N,X_j\text{ indep.})\\
&=E\left[NE\left[ X_1 \right]\right]&&\qquad (\{X_j\}_{j\in \mathbb{N}}\text{ identically distrib.})\\
&=E\left[N\right]E\left[ X_1 \right],&&\qquad (\text{linearity of }E[\cdot ])\end{aligned}$$
which is a special case of Wald's equation.
By Eve's law, the variance of $Y$ is
$$ \small\begin{aligned}\text{Var}\left[\sum_{j=1}^N X_j \right]&=E\left[\text{Var}\left[\sum_{j=1}^N X_j\Bigg|N \right]\right]+\text{Var}\left[E\left[\sum_{j=1}^N X_j\Bigg|N \right]\right]\\
&=E\left[\text{Var}\left[\sum_{j=1}^N X_j\Bigg|N  \right]\right]+\text{Var}\left[NE[X_1]\right]&&(\text{see above})\\
&=E\left[\text{Var}\left[\sum_{j=1}^N X_j \right]\right]+\text{Var}\left[NE[X_1]\right]&&(N,\{X_j\}_{j\in \mathbb{N}} \text{ indep.})\\
&=E\left[\sum_{j=1}^N\text{Var}\left[ X_j \right]\right]+\text{Var}\left[NE[X_1]\right]&& (\{X_j\}_{j\in \mathbb{N}} \text{ indep.})\\
&=E\left[N\text{Var}\left[ X_1 \right]\right]+\text{Var}\left[NE[X_1]\right]&& (\{X_j\}_{j\in \mathbb{N}} \text{ identically distrib.})\\
&=E\left[N\right]\text{Var}\left[ X_1 \right]+\text{Var}\left[N\right](E[X_1])^2&& (\text{linearity of } E[\cdot ])\\
&= E\left[N\right]E\left[ X_1^2 \right]+\left(\text{Var}\left[N\right]-E\left[N\right]\right)(E[X_1])^2.
\end{aligned}$$
For $N$ Poisson with rate parameter $\lambda$, we have $E[N]=\text{Var}[N]=\lambda$, so the above simplifies to
$$E[Y]=\lambda E[X_1]\\
\text{Var}[Y]=\lambda E\left[ X_1^2 \right],$$
as obtained in the wiki link.
Note the variance computation required stronger assumptions than the mean computation, which is typically the case to get a nice variance expression.
