I know that the sum of two independent normal random variables is normal. Particulary, when one is copy of other, i.e., if $X_1, X_2 \sim \mathcal{N}(0,\sigma^2)$, independent, we have:
$$X_1 + X_2 \sim \mathcal{N}(0, 2\,\sigma^2) $$
Beside this, suppse that I want to sum a random number of copies. For example, let
- $N\sim Po(\lambda)$ be a Poisson random variable;
- $X\sim \mathcal{N}(0,\sigma^2)$ be a normal r.v. and;
- $\{X_n\}_{n\in \mathbb{N}}$ be a sequence of copies of $X$ - i.e. $X_j \sim \mathcal{N}(0,\sigma^2)\,\, \forall \, j \in \mathbb{N}$ and $X_i, X_j$ independent for all $i,j \in \mathbb{N}$.
and define the Poisson Sum as: $$Y = \sum_{j=1}^N X_{j}$$
If $N$ and $\{X_j\}_{j\in \mathbb{N}}$ are independent, we have that: \begin{aligned} \phi_Y(t) &= E\left[e^{i t Y }\right]&\\ =& \sum_{n} \,E\left[e^{i t \sum_{j=1}^N X_{j} }| N = n\right]\, P(N = n)&\\ =& \sum_{n} \,E\left[e^{i t \sum_{j=1}^n X_{j} }| N = n\right]\, P(N = n)&\\ =& \sum_{n} \,E\left[\Pi_{i=1}^n e^{i t X_{j} }| N = n\right]\, P(N = n)&\\ =&\sum_{n} \,E\left[\Pi_{i=1}^n e^{i t X_{j} }\right]\, P(N = n)&\hbox{($N$ and $\{X_j\}_{j\in \mathbb{N}}$ are independent)}\\ =&\sum_{n} \,\Pi_{i=1}^nE\left[ e^{i t X_{j} }\right]\, P(N = n)&\hbox{($X_j's$ mutually independent)}\\ =&\sum_{n} \,E\left[ e^{i t X }\right]^n\, P(N = n)&\hbox{($X_j's \sim X$ )}\\ =&\sum_{n} \,(\phi_X(t))^n\, P(N = n)&\\ =&E\big( \phi_X(t)^N \big) \end{aligned}
But now I want to characterize the case where $N$ and $\{X_j\}_{j\in \mathbb{N}}$ are NOT independent. My intuiton says that the distribution is a Gaussian Mixture. Is my intuition correct?
If Yes, I don't know how to properly formalize this.
How do I prove that something is a Gaussian mixture model using characteristic functions?
Some help?