# Coefficients of Gaussian mixture

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 $$\pi_1=1.2$$, $$\pi_2=-0.2$$.

• What do you want to prove? That if one of the elements in the sum is negative, the total can be negative?
– Tim
Commented Oct 23, 2022 at 7:13
• As mentioned, I want to prove the cited statement . And I don't know if the claim itself is correct -- the "counter-example" is for the same. Commented Oct 23, 2022 at 7:15
• It's a requirement, what about it you want to prove?
– Tim
Commented Oct 23, 2022 at 7:16
• I want to prove the "implies $\pi_k \ge 0$" part. Commented Oct 23, 2022 at 7:17

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.

• I had earlier edited my question to correct the counter-example. Commented Oct 23, 2022 at 8:59
• I agree to the probabilistic interpretation of $\pi_k$. As a requirement, it is fine (but the author didn't put it that way). Commented Oct 23, 2022 at 9:53
• @abs regardless of the interpretation, the first part of the answer applies, the result needs to be proper distribution. Maybe the wording was unfortunate, but the author just lists a set of conditions that make it true.
– Tim
Commented Oct 23, 2022 at 10:15
• But is it true that a concave combination necessarily does not give a proper distribution? Commented Oct 23, 2022 at 10:18
• As an aside, the author imposes $0\le\pi_k\le1$ on page-431 (9.8). This, with the cited statement in the question, seems confusing. Commented Oct 23, 2022 at 10:21