ExampleConstruction: AI-generatedVerification: AI-adaptedPipeline-generatedprecheck passjudge pass (gpt-6-luna)

How statement and proof provenance work

The first chip identifies the source of the statement or construction; the second identifies the source of its local proof or verification.

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.
AI-generated: a genuinely novel statement formulated by AI, with no source for the claim itself.

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Exact two-time scalar Rauch–Tung–Striebel smoothing calculation

Example

Consider the scalar random walk $X_{0} \sim N (0, 1)$ , $X_{j} = X_{j - 1} + q_{j}$ , $Y_{j} = X_{j} + r_{j}$ for $j = 1, 2$ , with mutually independent $q_{j}, r_{j} \sim N (0, 1)$ . Fix the data $y_{1} = 1, y_{2} = 2$ . Every innovation and one-step forecast covariance is positive, so both forward filtering and the RTS backward recursion are defined.

Facts & Assumptions

Given: The scalar model and observations in the Example.

[A1]

AC is used by the multivariate-normal model and conditional-law suppliers. The Axiom of Choice.

[F1]

The model has mutually independent Gaussian initial, process, and measurement variables with the displayed state and observation equations. Linear-Gaussian state-space models with independent noises.

[F2]

The Kalman filter computes the prediction, innovation, gain, and update by the stated recursions. Exact linear-Gaussian Kalman filter.

[F3]

The RTS procedure initializes at the final filter and applies the backward gain, mean, and covariance updates. Rauch–Tung–Striebel fixed-interval Gaussian smoother.

[F4]

Under positive-definite innovation and forecast covariances, the RTS outputs are fixed-interval conditional-law versions. The Rauch–Tung–Striebel recursion gives fixed-interval Gaussian smoothing laws.

Verification

Given: $X_{0} \sim N (0, 1)$ , unit process and measurement variances, unit transition and observation coefficients, and $(y_{1}, y_{2}) = (1, 2)$ .

1.1A1F1F2givenalgebra

At time $1$ , the filter predicts $m_{1 ∣ 0} = 0$ and $P_{1 ∣ 0} = 2$ . Hence $S_{1} = 3$ , $v_{1} = 1$ , and $K_{1} = 2 / 3$ , so $m_{1 ∣ 1} = 2 / 3$ and $P_{1 ∣ 1} = 2 - (2 / 3) 3 (2 / 3) = 2 / 3$ . In particular $S_{1} > 0$ .

2.1F2step 1.1algebra

At time $2$ , the prediction is $m_{2 ∣ 1} = 2 / 3$ and $P_{2 ∣ 1} = 5 / 3$ . Thus $S_{2} = 8 / 3$ , $v_{2} = 4 / 3$ , and $K_{2} = 5 / 8$ , giving $m_{2 ∣ 2} = 2 / 3 + (5 / 8) (4 / 3) = 3 / 2$ and $P_{2 ∣ 2} = 5 / 3 - (5 / 8) (8 / 3) (5 / 8) = 5 / 8$ . Both $S_{2}$ and $P_{2 ∣ 1}$ are positive.

3.1F3F4step 1.1step 2.1algebra∎

The first backward gain is $C_{1} = (2 / 3) / (5 / 3) = 2 / 5$ . Therefore $m_{1 ∣ 2} = 2 / 3 + (2 / 5) (3 / 2 - 2 / 3) = 1$ and $P_{1 ∣ 2} = 2 / 3 + (2 / 5) (5 / 8 - 5 / 3) (2 / 5) = 1 / 2$ . The next gain is $C_{0} = 1 / 2$ , so $m_{0 ∣ 2} = 0 + (1 / 2) (1 - 0) = 1 / 2$ and $P_{0 ∣ 2} = 1 + (1 / 2) (1 / 2 - 2) (1 / 2) = 5 / 8$ . The theorem [F4] identifies the resulting smoothing kernels as $N (1 / 2, 5 / 8)$ for $X_{0}$ , $N (1, 1 / 2)$ for $X_{1}$ , and $N (3 / 2, 5 / 8)$ for $X_{2}$ at the selected data history.

Depends on

Used by

Nothing in the library uses this result yet.

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Sources

Simo Särkkä, Bayesian Filtering and Smoothing, Chapter 8, §8.2 (standard reference, not scraped)
Kody Law, Andrew M. Stuart, and Konstantinos Zygalakis, Data Assimilation: A Mathematical Introduction, Chapter 3, §3.1 (standard reference, not scraped)