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

  • $\begingroup$ What do you want to prove? That if one of the elements in the sum is negative, the total can be negative? $\endgroup$
    – Tim
    Commented Oct 23, 2022 at 7:13
  • $\begingroup$ 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. $\endgroup$
    – muser
    Commented Oct 23, 2022 at 7:15
  • $\begingroup$ It's a requirement, what about it you want to prove? $\endgroup$
    – Tim
    Commented Oct 23, 2022 at 7:16
  • $\begingroup$ I want to prove the "implies $\pi_k \ge 0$" part. $\endgroup$
    – muser
    Commented Oct 23, 2022 at 7:17

1 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

  • $\begingroup$ I had earlier edited my question to correct the counter-example. $\endgroup$
    – muser
    Commented Oct 23, 2022 at 8:59
  • $\begingroup$ I agree to the probabilistic interpretation of $\pi_k$. As a requirement, it is fine (but the author didn't put it that way). $\endgroup$
    – muser
    Commented Oct 23, 2022 at 9:53
  • $\begingroup$ @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. $\endgroup$
    – Tim
    Commented Oct 23, 2022 at 10:15
  • $\begingroup$ But is it true that a concave combination necessarily does not give a proper distribution? $\endgroup$
    – muser
    Commented Oct 23, 2022 at 10:18
  • $\begingroup$ 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. $\endgroup$
    – muser
    Commented Oct 23, 2022 at 10:21

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