Coefficients of Gaussian mixture

Question

This is in context of Gaussian mixtures $$p(\boldsymbol{x}) = \sum_{k=1}^K \pi_k\cal{N}(\boldsymbol{x}|\boldsymbol{\mu_k},\boldsymbol{\Sigma_k})$$

Bishop mentions on Page-111

Also, the requirement that $p(\boldsymbol{x})\ge 0$, together with $\cal{N}(\boldsymbol{x}|\boldsymbol{\mu_k},\boldsymbol{\Sigma_k})\ge$ $0$, implies $\pi_k\ge0$ for all $k$

How do we prove this?

A trivial counter-example that comes is ($k=2$), with a single Gaussian taken twice (i.e., $\boldsymbol{\mu_1}=\boldsymbol{\mu_2}$, $\boldsymbol{\Sigma_1}=\boldsymbol{\Sigma_2}$) and $\pi_1=1.2$, $\pi_2=-0.2$.

Answer 1

You are misinterpreting the quote. The requirement follows from the fact that we need $p(\boldsymbol x)$ to be a proper probability distribution (non-negative, that integrates to 1), so we use a convex combination of the mixture components. Moreover, $\pi_k$ weights are interpreted as probabilities, which makes the model have an intuitive interpretation.

See also Intuition for why sum of gaussian RVs is different from gaussian mixture