# How many parameters are present in a (general) discrete mixture of five normal distributions?

What is the minimal amount of parameters that can be retained in a particular context?

• What exactly do you mean by "retained" in this case? – gung - Reinstate Monica Oct 5 '19 at 11:52
• For five independent normal distributions, you would need to have the population means and standard deviations of the five distributions, and the percentage of the mixture for each normal distribution. I guess that makes 15, and they're all essential. – BruceET Oct 5 '19 at 19:12
• @Bruce The five proportions are related by the sum-to-unity equation, leaving only 14 parameters. – whuber Oct 5 '19 at 20:08
• @whuber: Right about the 14 proportions, thanks. – BruceET Oct 5 '19 at 20:26

If the dimension is $$n$$, you need $$n$$ parameters for each mean vector and $$(n^2+n)/2$$ parameters for each covariance matrix in general. Additionally, if the number of mixture components is $$m$$, $$m-1$$ parameters are needed for mixture composition weights since they need to sum up to $$1$$. Special constraints over covariance matrices reduce this number of parameters needed.
• are the $m-1$ parameters you mention the mixture weights themselves, or are they other supporting parameters? wouldn't there have to be $m$ mixture weights if there are $m$ mixture components because each weight is assigned to a different component? – develarist Jul 13 at 13:45
• @develarist yes, they're mixture weights. $m-1$ is enough because they'll add up to $1$, and the last one can be found from others. – gunes Jul 13 at 13:47
• but the expectation-maximization algorithm will in fact actively solve for all $m$ mixture weights, right – develarist Jul 13 at 13:49