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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?

Any help, please?

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  • $\begingroup$ The false premise of this question was comprehensively pointed out the last several times you posted it. Please do not post it again. $\endgroup$ – Glen_b -Reinstate Monica Aug 28 '18 at 12:00
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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!

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  • $\begingroup$ thank you so much for your answer. Hence, the margin of each variable is a mixture of Gaussian in the case of the Gaussian mixture model. Is that what you meant by "e. the marginal of each and every data variable Xi X is the same and it is this mixture expression that you have written above and not a single Gaussian!"? $\endgroup$ – Maryam Aug 28 '18 at 10:00
  • $\begingroup$ In addition, the margin of each component will be a Gaussian distribution. $\endgroup$ – Maryam Aug 28 '18 at 10:01

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