Weibull Mixture question

Question

Is it possible that a mixture of Weibull RVs is also Weibull distributed, and if yes, what are the necessary conditions?

Answer 1

The Weibull survival function with shape parameter $k$ and scale parameter $\lambda$ (both positive) has the form

$$S(x; \lambda, k) = \exp\left(-(x/\lambda)^k\right)$$

for $x \gt 0.$ A finite mixture of $n$ such distributions is determined by positive mixture weights $p_i$ (necessarily summing to unity) and corresponding parameters and has survival function

$$S = \sum_{i=1}^n p_i \exp\left(-(x/\lambda_i)^{k_i}\right).$$

Equating these two expressions and some straightforward analysis show the following:

1. By studying the asymptotic behavior of $\log S$ for large $x,$ conclude that

   $k = k_1 = k_2 = \ldots = k_n.$

2. Again by studying this asymptotic behavior assuming all the $k_i$ are equal to $k,$ conclude that

   $\lambda = \lambda_1 = \ldots = \lambda_n.$

These are necessary and sufficient conditions.

Consequently

$$\eqalign{ S &= \sum_{i=1}^n p_i \exp\left(-(x/\lambda_i)^{k_i}\right) \\ &= \sum_{i=1}^n p_i \exp\left(-(x/\lambda)^k\right) \\ &= \left(\sum_{i=1}^n p_i\right) \exp\left(-(x/\lambda)^k\right) \\ &= \exp\left(-(x/\lambda)^k\right) }$$

isn't really a mixture at all.

For an example of how such asymptotic investigations may be carried out rigorously, see this answer to the same question about Normal distributions.