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I have 2 independent random variables $X$ and $Y$ with gaussian mixture distribution like:

$$f(x) = \sum_{i=1}^{m} \phi_{X,i} \mathcal{N}(\mu_{X,i} , \sigma_{X,i}^{2})$$ $$f(y) = \sum_{i=1}^{m} \phi_{Y,i} \mathcal{N}(\mu_{Y,i} , \sigma_{Y,i}^{2})$$

With $\sum_{i=1}^{m}\phi_{X,i} =\sum_{i=1}^{m}\phi_{Y,i} =1$.

If I want to ponderate these 2 random variables like: $Z=\alpha\cdot X + (1-\alpha)Y$ with $\alpha\in (0,1)$, I could say that $f(z)=\sum_{i=1}^{m} \phi_{Z,i} \mathcal{N}(\mu_{Z,i} , \sigma_{Z,i}^{2})$?? If I could, so how I can find a relation between the parameters like:

$$\phi_{Z,i} = \phi_{Z,i}(\alpha,\phi_{X,i},\phi_{Y,i}) $$ $$\mu_{Z,i} = \mu_{Z,i}(\alpha,\mu_{X,i},\mu_{Y,i}) $$ $$\sigma_{Z,i}^{2} = \sigma_{Z,i}^{2}(\alpha,\sigma_{X,i}^{2},\sigma_{Y,i}^{2}) $$

for $i=1,...,m$. If I have another gaussian mixture $W$, I could extend this with $Z=\alpha_1 X + \alpha_2 Y + \alpha_3 W$ and $\sum_{j} \alpha_j =1$, $\alpha_j \in (0,1)$??

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  • $\begingroup$ It looks like there is a typo in the first equation: $\sigma_{Y,i} \rightarrow \sigma_{X,i}$. $\endgroup$ Dec 18 '19 at 15:08
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    $\begingroup$ Thanks, I already corrected it $\endgroup$
    – gabriel11
    Dec 18 '19 at 15:11
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Let $X_i \sim \mathcal{N}(\mu_{X,i}, \sigma_{X,i})$ and $Y_j \sim \mathcal{N}(\mu_{Y,j}, \sigma_{Y,j})$, where $i=1...m$ and $j=1...n$. Then variables $X$ and $Y$ can be expressed as \begin{equation} X = \sum_{i=1}^m \phi_{X,i} X_i, Y = \sum_{j=1}^n \phi_{Y,j} Y_j, \end{equation} whereas their sum is \begin{equation} Z = \alpha X + (1-\alpha)Y = \sum_{i=1}^m \alpha\phi_{X,i} X_i + \sum_{j=1}^n (1-\alpha)\phi_{Y,j} Y_j, \end{equation} and is distributed as a mixture of $m+n$ normal distributions: \begin{equation} f(z) = \sum_{i=1}^m \alpha\phi_{X,i} \mathcal{N}(\mu_{X,i}, \sigma_{X,i}) + \sum_{j=1}^n (1-\alpha)\phi_{Y,j} \mathcal{N}(\mu_{Y,j}, \sigma_{Y,j}) = \sum_{k=1}^{m+n} \phi_{Z,k} \mathcal{N}(\mu_{Z,k}, \sigma_{Z,k}), \end{equation} with the obvious re-definitions of the coefficients and the parameters.

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