Suppose $F$ and $G$ are two cumulative distribution functions on $\mathbb{R}$. Let $\alpha F + (1 - \alpha) G$ be their convex combination for some $\alpha \in [0, 1]$. We know it is also a cdf.

Is there any well known parametric family of distributions that is "closed" under mixture? That is, if both $F$ and $G$ belong to this family, then so is any $\alpha F + (1 - \alpha) G$ for $\alpha \in (0, 1)$?

Thank you!


1 Answer 1


I don't have a complete answer to your question, but clearly, one of the consequences of such closure would be that any member of this parametric family would need to be expressible as $$F = \sum_{i=1}^\infty \alpha_i F_i$$ subject to the constraint $$\forall i, \alpha_i \ge 0, \quad \sum_{i=1}^\infty \alpha_i = 1.$$

In fact, we can use this to construct a family from a single parametric distribution; e.g., use the standard normal and construct the family of distributions that are mixtures of the standard normal, parametrized by the mixture weights themselves. Then this family is closed under additional mixing.

For further thought. Suppose $X_i \sim \operatorname{Bernoulli}(p_i)$ with CDF $F_i(x)$. Then what is the mixture of $F_1$ with $F_2$? What is the $n$-fold mixture of $F_1, F_2, \ldots, F_n$?

  • $\begingroup$ Thank you very much for your suggestions! So Bernoulli family is closed under mixture? $\endgroup$
    – user295959
    Jan 3, 2021 at 6:11
  • $\begingroup$ @user295959 Well, what did your calculations suggest? It is worth noting that the Bernoulli distribution has many special properties that arise out of the fact that it only has support on $\{0, 1\}$, and it is completely characterized by the parameter $p$. So when we take a mixture of Bernoulli variables, the support does not change. What about a categorical distribution? Would this family also be closed under mixture? Why or why not? $\endgroup$
    – heropup
    Jan 3, 2021 at 7:15

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