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How statement and proof provenance work

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Literature-sourced: the exact statement appears in a cited source; only wording and notation differ.
AI-adapted: a semantically identical restatement of literature-sourced material, modulo indexing, notation, and boundary cases adopted by the library.
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A two-point sample has unbounded Gaussian-mixture likelihood

Example

Assume AC. For the observations $x_{1} = 0, x_{2} = 1 \in R$ , use two Gaussian components of weights $1 / 2, 1 / 2$ , both with mean $0$ , with variances $ε^{2}$ and $1$ for $ε > 0$ . Write $ϕ_{1} (x; 0, s^{2}) = (2 π s^{2})^{- 1 / 2} \exp (- x^{2} / (2 s^{2}))$ , $f_{ε} (x) = \frac{1}{2} ϕ_{1} (x; 0, ε^{2}) + \frac{1}{2} ϕ_{1} (x; 0, 1)$ , and $L_{ε} = f_{ε} (0) f_{ε} (1)$ . Every parameter in this family has positive weights and positive-definite one-dimensional covariances. The likelihood $L_{ε}$ is unbounded above as $ε ↓ 0$ ; hence its log likelihood is unbounded above as well.

Example references

Finite Gaussian mixture with positive-definite component covariances — the finite-mixture likelihood and Gaussian density formula.

Positive-definite multivariate Gaussian density is normalized — the component functions are normalized probability densities for every $ε > 0$ .

Facts & Assumptions

Given: Assume AC; take observations $0, 1$ , two components with weights $1 / 2, 1 / 2$ , common mean $0$ , and variances $ε^{2}$ and $1$ with $ε > 0$ .

[A1]

The Axiom of Choice: AC is the stated assumption of the finite-Gaussian definition and the local component-density normalization theorem.

[F1]

Finite Gaussian mixture with positive-definite component covariances: the one-dimensional component density is $ϕ_{1} (x; μ, s^{2}) = (2 π s^{2})^{- 1 / 2} e^{- (x - μ)^{2} / (2 s^{2})}$ and the sample likelihood is the product of the mixture densities at the observations.

[F2]

Positive-definite multivariate Gaussian density is normalized: for each $s^{2} > 0$ , the displayed component function is a normalized Gaussian density and is finite and positive at every real observation.

[F3]

The natural logarithm as the inverse of the exponential function and Order, continuity, range, and the product, quotient, and reciprocal laws for the natural logarithm: $\log$ is the increasing inverse of $\exp$ , so $\log (\exp R) = R$ for every real $R$ .

Proof

1.1A1F1F2algebra

Let $f_{ε}$ be the mixture density. From [F1], $f_{ε} (0) = \frac{1}{2} ϕ_{1} (0; 0, ε^{2}) + \frac{1}{2} ϕ_{1} (0; 0, 1) \geq \frac{1}{2 \sqrt{2 π} ε} .$ Also $f_{ε} (1) = \frac{1}{2} ϕ_{1} (1; 0, ε^{2}) + \frac{1}{2} ϕ_{1} (1; 0, 1) \geq \frac{e^{- 1 / 2}}{2 \sqrt{2 π}} > 0.$ These bounds hold for every $ε > 0$ , for which both variances are positive; [F2] supplies the stated density interpretation under [A1].

2.1F1F3step 1.1algebra∎

The sample likelihood $L_{ε} = f_{ε} (0) f_{ε} (1)$ therefore satisfies $L_{ε} \geq \frac{e^{- 1 / 2}}{8 π ε} .$ Given any real $R$ , choose $0 < ε < e^{- 1 / 2} / (8 π e^{R})$ ; then step 1.1 gives $L_{ε} > e^{R}$ . By [F3], $\log L_{ε} > \log (e^{R}) = R$ . Thus both the likelihood and its log are unbounded above.

Depends on

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Sources

Christopher M. Bishop, Pattern Recognition and Machine Learning, §9.2.2 (standard reference, not scraped)
Hastie, Tibshirani, and Friedman, The Elements of Statistical Learning, 2nd ed., corrected 12th printing, §14.3.7 (standard reference, not scraped)