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Metric approximation error decomposes along a finite reference chain

Statement

Let $(M, d)$ be a metric space, let $m \geq 0$ be an integer, and take $T_{0}, T_{1}, \dots, T_{m}, \hat{T} \in M$ . Then $d (T_{0}, \hat{T}) \leq \sum_{j = 1}^{m} d (T_{j - 1}, T_{j}) + d (T_{m}, \hat{T}),$ where the sum is empty when $m = 0$ . If $m \geq 1$ and $T_{1}$ is the exact target from Exact inferential targets and approximation chains, the first term is the declared reference-to-target discrepancy and subsequent terms quantify adjacent approximation changes in the same metric. The scalar identity and absolute bound of Pathwise error decomposition along a reference chain are exact telescoping statements. No independence or default variance-addition rule follows; the MSE identity in MSE of a sum of error components includes covariance terms retains total bias and covariance terms.

Facts & Assumptions

Given: A metric space $(M, d)$ , an integer $m \geq 0$ , and points $T_{0}, \dots, T_{m}, \hat{T} \in M$ .

[F1]

A metric is symmetric and satisfies $d (x, z) \leq d (x, y) + d (y, z)$ . Metric space: $d (x, y) = 0$ iff $x = y$ , symmetry, and the triangle inequality; pseudometric and ultrametric

[F2]

Scalar ledger stages are adjacent signed differences on one stated scale; the definition alone asserts no independence, centering, or unbiasedness. Successive statistical and computational error targets

[F3]

A finite scalar ledger chain satisfies the signed telescoping identity and its absolute-value bound without an independence assumption. Pathwise error decomposition along a reference chain

[F4]

For square-integrable scalar error terms, the MSE of their sum includes squared total bias and covariance terms; summing component variances alone is generally not the identity. MSE of a sum of error components includes covariance terms

Proof

Given: The hypotheses in the Statement.

1.1givenF1inductionalgebra

We prove by induction on $m$ that $d (T_{0}, T_{m}) \leq \sum_{j = 1}^{m} d (T_{j - 1}, T_{j})$ . For $m = 0$ both sides are zero; for $m = 1$ this is equality. If it holds through $m$ , then the triangle inequality [F1] gives $d (T_{0}, T_{m + 1}) \leq d (T_{0}, T_{m}) + d (T_{m}, T_{m + 1}) \leq \sum_{j = 1}^{m + 1} d (T_{j - 1}, T_{j})$ .

2.1F2F3F4step 1.1algebra∎

Apply the triangle inequality once more to $T_{0}, T_{m}, \hat{T}$ and then step 1.1 to bound $d (T_{0}, \hat{T})$ by the displayed sum. For a scalar chain, [F2] specifies the same-scale adjacent errors and [F3] gives their exact signed telescope; when the terms are square-integrable, [F4] explains why squaring them retains bias and covariance terms rather than giving an automatic variance sum.

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Sources

Anthony Davison and David Hinkley, Bootstrap Methods and Their Application (1997), Chapter 2 §2.5 (standard reference, not scraped)
Alison L. Gibbs and Francis E. Su, On Choosing and Bounding Probability Metrics, International Statistical Review 70(3), 419–435 (2002) (standard reference, not scraped)