Suppose one has a multivariate Gaussian Mixture Model:

$$ \text{pdf}(\vec{x}) = \sum_{i=1}^N w_i \mathcal{N}(\mu^{(i)}, \Sigma^{(i)}) $$

Suppose $\vec{x} = \{\vec{a},\vec{b}\}$ and we marginalize out $\vec{b}$. For a single Gaussian component we would get:

$$ \mathcal{N}\left(\mu_a \mid \Sigma_a \right) $$

where $\mu_a$ is the corresponding part of $\mu = [\mu_a, \mu_b]'$ and $\Sigma = \begin{bmatrix}\Sigma_{a} & \Sigma_{ab} \\ \Sigma_{ba} & \Sigma_{b} \end{bmatrix}$.

For the marginalized mixture, does one need to update the weights?:

$$ \text{pdf}(\vec{a}) = \sum_{i=1}^N w'_i \mathcal{N}(\mu_a^{(i)}, \Sigma^{(i)}_a) $$

In other words, is $w_i' = w_i$ for all of the components?


Okay, I figured it out.

Marginal probability:

\begin{align*} pdf(a) & = \int_b pdf(a,b) \> db \\ & = \int_b \sum_i w_i \mathcal{N}(\mu^{(i)},\Sigma^{(i)}) \> db \\ & = \sum_i w_i \int_b \mathcal{N}(\mu^{(i)},\Sigma^{(i)}) \> db \\ & = \sum_i w_i \mathcal{N}(\mu_a^{(i)}, \Sigma_a^{(i)}) \end{align*}

The weights are unchanged.


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