I asked a similar question with two-component mixture variates, and I was wondering how it extends to a four-component mixture variate. In other words, I have a list of random variables, $X_1$, $X_2$, ..., $X_N$, associated with quaternary random variables $A_i$ such that $P(A_i=k) = p_k$ is known for all $k\in\{0,1,2,3\}$ and $\sum_{k=0}^3p_k=1$. I also know that, for all $i$ $$(X_i|A_i=0)\sim f(x)\\ (X_i|A_i=1)\sim g(x)\\ (X_i|A_i=2)\sim h(x)\\ (X_i|A_i=3)\sim j(x)\\$$ where $f$, $g$, $h$, and $j$ are known, and thus the distribution of the $X_i$ is a mixture given by $$p(x) = p_0\cdot f(x) + p_1\cdot g(x) + p_2\cdot h(x) + p_3\cdot j(x)$$ What's the general expression for the distribution of $Y=\sum_{i=1}^N X_i$? I expect a similar combinatorial argument can be applied to yield the right result, but I don't quite see it.
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Based on the form of the two-component case, I'd guess the final density would be
$$\sum_{c_0+c_1+c_2+c_3=N} {N\choose c_0,c_1,c_2,c_3} \cdot \left(\prod_{i=0}^3 p_i^{c_i}\right) \cdot \left(f^{c_0}*g^{c_1}*h^{c_2}*j^{c_3}\right)(x)$$
as per the multionimal theorem.
And for the general M-component mixture variates with $k\in\{0,...,M-1\}$ and $$\begin{aligned} P(A_i=k)&=p_k \\ \sum_{k=0}^{M-1}p_k&=1 \\ \left(X_i|A_i=k\right)&\sim f_k(x) \end{aligned} $$
the final distribution would be
$$\sum_{c_0+...+c_{M-1}=N} {N\choose c_0,...,c_{M-1}} \cdot \left(\prod_{i=0}^{M-1} p_i^{c_i}\right) \cdot \left(\prod_{i=0}^{M-1} f_i^{c_i}\right)(x)$$
where $$\left(\prod_{i=0}^{M-1} f_i^{c_i}\right)(x)$$ is the convolution of the $f_i(x)$ as previously defined.