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Synthetic likelihood and indirect inference — Examples

1 · Prerequisites

2 · Summary

Examples

A Bernoulli summary has a non-Gaussian two-point law even though its mean and variance define a positive-variance Gaussian synthetic likelihood. A finite model shows how a non-injective summary binding map can lose parameter identification and how the declared tie rule still selects a target. A two-dimensional Rademacher summary has a positive-definite population covariance, while its two-simulation covariance is random and always singular.

3 · Logical flowchart

4 · Definitions, theorems and proofs

None yet.

5 · Examples, counterexamples and false statements

ExampleConstruction: AI-generatedVerification: AI-generatedOpen item page →

A Gaussian synthetic likelihood for a Bernoulli summary is not exact

Example

Let $Θ = (0, 1)$ , let $Y ∣ p$ be Bernoulli with $P_{p} (Y = 1) = p$ and $P_{p} (Y = 0) = 1 - p$ , and use the identity summary $S (Y) = Y$ .

Facts & Assumptions

Given: The Bernoulli model and statistic above, with $p \in (0, 1)$ .

[A1]

AC is assumed only to use the Gaussian synthetic-likelihood density interface. The Axiom of Choice

[F1]

The synthetic likelihood uses the Gaussian density with the exact summary mean and covariance when that covariance is positive definite. Ideal and finite-simulation synthetic likelihood

[F2]

A positive-definite multivariate-normal law has the stated Lebesgue density. A positive-definite multivariate normal law has its Gaussian density

[F3]

A statistic is a measurable map, and its law is the pushforward law. Statistics and estimators Law or distribution of a random element

[F4]

Means and variances are expectations and centered second moments. Expectation of a nonnegative or integrable random variable Moments, variance, and covariance on a probability space

Verification

1.1F1F2F3F4givenalgebra

The identity summary has mean $p$ and variance $p (1 - p) > 0$ , so its ideal Gaussian synthetic likelihood is the density $g_{p} (x) = (2 π p (1 - p))^{- 1 / 2} \exp ⁣ (- \frac{(x - p)^{2}}{2 p (1 - p)})$ .

2.1A1F1F2step 1.1algebra

For $x \in (1 / 4, 3 / 4)$ and $p \in (0, 1)$ , $∣ x - p ∣ < 1$ . Hence $g_{p} (x) \geq c_{p} : = (2 π p (1 - p))^{- 1 / 2} \exp ⁣ (- \frac{1}{2 p (1 - p)}) > 0$ . The Gaussian law therefore assigns $(1 / 4, 3 / 4)$ probability at least $c_{p} / 2 > 0$ , since this interval has Lebesgue length $1 / 2$ .

3.1F3step 2.1algebra∎

The true summary law is supported on ${0, 1}$ , so it assigns $(1 / 4, 3 / 4)$ probability zero. It differs from the Gaussian law in step 2.1. Thus the Gaussian synthetic likelihood is not the exact likelihood of this non-Gaussian summary model.

ExampleConstruction: AI-generatedVerification: AI-generatedOpen item page →

A non-injective binding map cannot recover a sign

Example

Let $Θ = {- 1, 1}$ , $X = {- 1, 1}$ with its full sigma-algebra, and let $P_{θ}$ be the point mass at $Y = θ$ . Take the statistic $S (y) = y^{2}$ .

Facts & Assumptions

Given: The finite experiment, statistic and moment-binding target above.

[F1]

The binding function is $b (θ) = E_{θ} [S]$ , and the target minimizes $∥ b (θ) - s ∥_{2}^{2}$ with the declared lexicographic tie-break. Binding function and indirect-inference target

[F2]

The compact-domain minimum and its lexicographic selector exist under the definition’s stated continuity hypotheses. The compact binding target exists and has a unique lexicographic tie-break

[F3]

The parameter identity is identifiable exactly when distinct parameter labels give distinct sampling laws. Identifiability of an estimand

[F4]

In a parametric model, identifiability is equivalent to constancy on observational-equivalence classes; for the identity target this is injectivity of the model map. Identifiability is constancy on observational classes

[F5]

The summary law assigns to each Borel set $B$ the probability $P_{θ} (S^{- 1} (B))$ . Law or distribution of a random element

Verification

1.1F1givenalgebra

Under either point-mass law, $S (Y) = Y^{2} = 1$ surely. Thus $b (- 1) = b (1) = 1$ .

1.2F3F4givenalgebra

The original laws differ: $P_{- 1} ({- 1}) = 1$ but $P_{1} ({- 1}) = 0$ . Therefore the parameter identity is identifiable in the original experiment.

2.1F1F2step 1.1algebra

For the observed summary $s = 1$ , $q_{1}$ is the constant continuous function zero on the finite compact set $Θ$ , so $M_{1} = {- 1, 1}$ . The target’s lexicographic rule selects $- 1$ .

3.1F4F5step 1.1algebra∎

By the pushforward-law definition [F5], both summary laws assign probability one to ${1}$ , so the summary-law map is not injective and the parameter identity is not identifiable from this summary. The moment target therefore does not recover the sign, even though the raw experiment distinguishes it.

ExampleConstruction: AI-generatedVerification: AI-generatedOpen item page →

Two simulated square summaries give a random singular covariance

Example

Let the summary vector $Z = (Z^{(1)}, Z^{(2)})$ be uniform on the four points ${(- 1, - 1), (- 1, 1), (1, - 1), (1, 1)} \subset R^{2}$ . For one parameter value, simulate $Z_{1}, Z_{2}$ independently from this law and use the two-simulation covariance formula of Finite-simulation plug-in synthetic-likelihood evaluation.

Facts & Assumptions

Given: Two independent draws from the stated four-point summary law.

[A1]

AC is inherited through the procedure’s Gaussian-density branch; it is not used in the finite probability calculations. The Axiom of Choice

[F1]

IID means a common law and mutual independence. Identical distribution and IID families

[F2]

The covariance matrix uses centered second moments; the finite plug-in covariance is the Bessel-corrected matrix in the algorithm. Moments, variance, and covariance on a probability space Finite-simulation plug-in synthetic-likelihood evaluation

[F3]

For any two two-dimensional vectors, that sample covariance is not positive definite. A two-simulation covariance in two dimensions is singular

[F4]

The algorithm returns unavailable when its plug-in covariance is not positive definite. Finite-simulation plug-in synthetic-likelihood evaluation

[F5]

The ideal synthetic likelihood uses the population summary mean and covariance, when the latter is positive definite. Ideal and finite-simulation synthetic likelihood

Verification

1.1F2F5givenalgebra

Each coordinate has mean zero and variance one, while the cross moment is zero. Hence the population summary covariance is $I_{2}$ , positive definite.

1.2F1F2givenalgebra

By independence and the four equally likely support points, $P (Z_{1} = Z_{2}) = 4 (1 / 4)^{2} = 1 / 4$ . On this event the sample covariance is the zero matrix.

2.1F1F2step 1.2algebra

The event $Z_{1} = (1, 1)$ and $Z_{2} = (- 1, - 1)$ has probability $1 / 16$ . On it, ${\overset{ˉ}{Z}}_{2} = (0, 0)$ and ${\hat{Σ}}_{2} = (\begin{array}{cc} 2 & 2 \\ 2 & 2 \end{array})$ , which differs from the zero matrix in step 1.2. Thus the sample covariance is a nonconstant random matrix.

3.1A1F3F4step 1.1algebra∎

By [F3], every two-simulation covariance here is not positive definite, so [F4] makes the finite plug-in synthetic likelihood unavailable for every outcome despite the positive-definite population covariance in step 1.1.

Synthetic likelihood and indirect inference — Examples

1 · Prerequisites

2 · Summary

Examples

3 · Logical flowchart

4 · Definitions, theorems and proofs

5 · Examples, counterexamples and false statements

A Gaussian synthetic likelihood for a Bernoulli summary is not exact

Example

Facts & Assumptions

Verification

A non-injective binding map cannot recover a sign

Example

Facts & Assumptions

Verification

Two simulated square summaries give a random singular covariance

Example

Facts & Assumptions

Verification

Sources