# Parameterizing finite mixture distribution

Let's consider a finite mixture:

$$f(x) = \sum_{i=1}^{N}w_{i}p_{i}\left(x\right)$$

where:

• $$N$$ is the number of mixed distributions
• $$\left\{p_{1},\dots, p_{N}\right\}$$ is a finite set of one-dimensional probability density functions in $$\mathbb{R}^{+}$$
• $$\left\{w_{1},\dots, w_{N}\right\}$$ is a finite set of associated weights

Now, let's say that $$f$$ is a mixture over a parametric family, where distributions are parameterized by a set of $$\left\{\alpha^{1},\dots, \alpha^{n}\right\}$$ parameters (where the upper index is just an index and not an exponentiation):

$$f(x) = \sum_{i=1}^{N}w_{i}p\left(x;\alpha^{1}_{i},\dots, \alpha^{n}_{i}\right)$$

And let's say that these parameters can be themselves parameterized by a single parameter:

$$f(x, \lambda) = \sum_{i=1}^{N}w_{i}p_{i}\left(x;\alpha^{1}_{i}\left(\lambda\right),\dots, \alpha^{n}_{i}\left(\lambda\right)\right)$$

The question is: is there a distribution $$p$$ such that $$f$$ itself would be a member of the parametric family:

$$p\left(x;\alpha^{1}_{0},\dots, \alpha^{n}_{0}\right) = \sum_{i=1}^{N}w_{i}p\left(x;\alpha^{1}_{i}\left(\lambda\right),\dots, \alpha^{n}_{i}\left(\lambda\right)\right)$$

In other words, given:

• a mixture size $$N$$
• a parameter $$\lambda$$
• a set of parameters $$\left\{\alpha^{1}_{0},\dots, \alpha^{n}_{0}\right\}$$

would there be a distribution $$p$$ such that it would be possible to find a set of weights $$w_{i} \neq 0$$ and a set of set of parameters $$\left\{\left\{\alpha^{1}_{1}\left(\lambda\right),\dots, \alpha^{n}_{1}\left(\lambda\right)\right\}, \dots, \left\{\alpha^{1}_{N}\left(\lambda\right),\dots, \alpha^{n}_{N}\left(\lambda\right)\right\}\right\}$$ so that the equality above holds.

For example if $$p$$ is a normal distribution, can one find a set of $$N$$ normal distributions and their weights parameterized by $$\lambda$$ so that the mixture is itself a normal distribution?

• I would think that the circumstances would be bizarre. For example, if one adds sub-populations of normal distributions with the same means and variances the result is normal. However, there is no way to fit a mixture model to a normal distribution to recreate the sub-populations from which it arose. – Carl Oct 19 '18 at 0:11

If we ignore some trivial solutions such that $$w_1=1$$ and $$w_i=0$$ for the rest or $$a_i=a_j$$ for all $$i,j$$, then the only solution seems to be histograms with the same binning. With weights, you simply sum up the weights in particular bins. They guarentee the linearity.
• Added $w_{i} \neq 0$ – Vincent Oct 18 '18 at 20:36