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IRLS equals the Fisher-scoring step when information is invertible

Statement

Under the admissibility hypotheses of One minimum-norm IRLS step, assume also that each map $η_{i} \mapsto θ_{i}$ is differentiable, each map $η_{i} \mapsto μ_{i}$ is differentiable with finite nonzero derivative ${\dot{μ}}_{i}$ , and $V_{i} : = b^{''} (θ_{i}) > 0$ . Each response has finite second moment, its density has the fixed common support and locally uniform integrable envelopes for its first and second $θ_{i}$ derivatives required by Explicit exponential-family response contracts, and each component score has finite second moment. If $I (β) = X^{T} W X$ is invertible, let $U (β)$ and $I (β)$ be the score and expected score outer-product information from General-link score and Fisher information. Then the weighted least-squares output satisfies

$β_{n e w} = β + I (β)^{- 1} U (β) .$

Facts & Assumptions

Given: The algorithm's finite admissible current $β$ , common scale, finite nonzero link derivative, and finite positive variance values; the differentiable natural-parameter map; finite component response second moments; fixed common support and locally integrable first- and second-derivative envelopes; finite component score second moments; and invertible $I (β)$ .

[F1]

General-link score and Fisher information: $U = X^{T} ({\dot{μ}}_{i} (y_{i} - μ_{i}) / (a V_{i}))_{i}$ and $I = X^{T} diag ({\dot{μ}}_{i}^{2} / (a V_{i})) X$ .

[F2]

One minimum-norm IRLS step: $z_{i} = η_{i} + (y_{i} - μ_{i}) / {\dot{μ}}_{i}$ , $w_{i} = {\dot{μ}}_{i}^{2} / (a V_{i})$ , and $β_{n e w}$ minimizes $\sum_{i} w_{i} (z_{i} - x_{i}^{T} γ)^{2}$ .

[F3]

For a linear map $T : V \to W$ between finite-dimensional inner-product spaces, $x$ minimises $∥ T x - b ∥$ if and only if $T^{*} (T x - b) = 0$ , equivalently $T^{*} T x = T^{*} b$ ; minimisers exist and any two differ by an element of $\ker T$ : a minimizer of $∥ T γ - b ∥$ satisfies $T^{*} (T γ - b) = 0$ ; minimizers exist and differ by $\ker T$ .

Proof

1.1F1F2algebra

For every $i$ , [F2] gives $w_{i} (z_{i} - η_{i}) = {{\dot{μ}}_{i}^{2} / (a V_{i})} {(y_{i} - μ_{i}) / {\dot{μ}}_{i}} = {\dot{μ}}_{i} (y_{i} - μ_{i}) / (a V_{i})$ . Since $η = X β$ , multiplying by $X^{T}$ and using [F1] yields $X^{T} W (z - X β) = U (β)$ .

2.1F1F2F3step 1.1givenalgebra∎

Apply [F3] to the weighted least-squares map $T = W^{1 / 2} X$ and target $W^{1 / 2} z$ . Its normal equation is $X^{T} W (X β_{n e w} - z) = 0$ , so $I (β) (β_{n e w} - β) = X^{T} W (z - X β) = U (β)$ by step 1.1. The assumed invertibility of $I (β)$ now gives $β_{n e w} = β + I (β)^{- 1} U (β)$ .

Depends on

Used by

Dependency tree · two levels

16 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources

Philippe Rigollet, MIT 18.650 Statistics for Applications, Lecture 10: Generalized Linear Models, PDF pp. 47–52 (standard reference, not scraped)
Ryan Tibshirani, Generalized Linear Models, CMU Advanced Methods for Data Analysis, PDF p. 5 (standard reference, not scraped)