# Why do we assume Gaussian margins in Gaussian mixture models [duplicate]

A Gaussian mixture model is a weighted sum of Gaussian densities, i.e.,

$L(\theta) = \sum_{i=1}^{m} \pi_{i} f(x_i)$

where $m$ is the number of the mixture component.

Hence, Gaussian mixture models is a sum of a finite mixture Gaussian distribution with unknown parameters. I read that, in Gaussian mixture models, each univariate margins are assumed to be normal. Is that correct and why?

We do not (need to) assume anything on the margins. Generally, in a mixture model setting we start with a single density $f(\cdot;\theta)$ and then we assume that we have random variables $(X_1, ..., X_N, Z_1, ..., Z_N)$ such that $(X_i, Z_i)$ are independent and for a parameter $\Theta = (\tau_1, ..., \tau_K, \theta_1, ..., \theta_k)$ we have that $X = (X_1, ..., X_N)$ and $Z = (Z_1, ..., Z_N)$ have a common density $p(x,z)$ and that $$p(x|z) = \prod_{i=1}^N p(x_i|z_i) = \prod_{i=1}^N f(x_i;\theta_{z_i})$$ and $$p(z) = \prod_{i=1}^N p(z_i) = \prod_{i=1}^N \tau_{z_i}$$ Let us assume that $N=1$ then we simply compute $$p(x) = \int_{\mathcal{Z}} p(x,z) dz = \int_{\mathcal{Z}} p(x|z)p(z) dz = \sum_{k=1}^K f(x;\theta_{k}) \tau_k$$
i.e. the marginal of each and every data variable $X_i$ is the same and it is this mixture expression that you have written above and not a single Gaussian!