advanced30 min studyLesson 737 in Mathematics

Doubly robust estimation and cross-fitting

Combine outcome regression and assignment weighting while stating what robustness and rate conditions actually guarantee.

On this page

What you’ll understand

  • Derive the augmented inverse-probability score and its two correctness branches.
  • Interpret the product-error remainder under overlap.
  • Explain cross-fitting without claiming that it repairs identification failures.

Outcome regression and inverse weighting estimate the same intervention contrast using different nuisance functions. Augmented inverse probability weighting combines them so that a particular population bias vanishes if either the assignment model or both outcome regressions are correct. This double robustness is a precise algebraic property, not a guarantee against every modeling mistake. Valid causal interpretation still rests on consistency, exchangeability, and positivity, and valid root-n inference requires stronger rate and moment conditions than consistency alone.

Augment predictions with weighted residuals

Let mₐ(L) be the true observed outcome mean in arm a and let e(L) be the true propensity. For candidate functions g₁,g₀ and q, start with the predicted mean contrast g₁ minus g₀. Add the treated residual divided by q and subtract the control residual divided by one minus q. Averaging this score gives the augmented estimator. The candidates must be measurable baseline functions, and q must stay strictly between zero and one where evaluated; otherwise the displayed expression is undefined.

If q equals e, conditional expectation cancels the assignment probabilities, so the residual terms correct both prediction errors exactly. If both g functions equal their true m functions, each arm’s residual has conditional mean zero, leaving the correct predicted contrast even when q is misspecified. These are statements for a fixed candidate score at the population level. Estimated functions introduce dependence on training data, so turning the identity into a sample consistency theorem requires convergence and an appropriate treatment of that dependence.

ϕ(O;g,q)=g1(L)−g0(L)+A{Y−g1(L)}q(L)−(1−A){Y−g0(L)}1−q(L).\phi(O;g,q)=g_1(L)-g_0(L)+\frac{A\{Y-g_1(L)\}}{q(L)}-\frac{(1-A)\{Y-g_0(L)\}}{1-q(L)}.
The augmented ATE estimate averages phi; the two outcome functions and propensity are nuisance quantities.↔ Scroll wide equations horizontally.

The remainder is a product, not a universal shield

Subtract the true conditional effect from the conditional expectation of the score. The remaining bias is the propensity error times a weighted combination of outcome-regression errors. If q is bounded away from zero and one, Cauchy–Schwarz bounds its population magnitude by a constant times the propensity L² error multiplied by the sum of the two outcome L² errors. This is the source of first-order insensitivity near the true nuisance functions and the reason good estimation in one component can compensate for slower estimation in the other.

For an ordinary asymptotically linear estimator, a useful sufficient requirement is that this product be smaller in probability than n to the negative one half. If both nuisance errors are smaller than n to the negative one quarter, that requirement follows, but that symmetric rate is not necessary. Mere double-robust consistency with one correct limit does not automatically yield a centered normal interval when the other nuisance remains badly misspecified. Distinguish the zero-bias population branches from the conditions needed for a particular inference theorem.

∣Eϕ−τ∣≤Cε∥q−e∥2(∥g1−m1∥2+∥g0−m0∥2),product=op(n−1/2).|\mathbb E\phi-\tau|\le C_\varepsilon\|q-e\|_2\bigl(\|g_1-m_1\|_2+\|g_0-m_0\|_2\bigr),\qquad \text{product}=o_p(n^{-1/2}).
The bound assumes q∈[epsilon,1−epsilon]; the product rate is one ingredient in a cross-fitted inference result.↔ Scroll wide equations horizontally.

Cross-fitting separates training from evaluation

Partition independent observations into a fixed number of folds. For each fold, fit all nuisance functions using observations outside that fold and evaluate the augmented score only on the held-out observations. Aggregate those scores across folds. A unit’s own outcome is therefore not used to train its score functions. This reduces empirical-process restrictions and overfitting interactions in the proof; it does not make folds’ fitted functions jointly independent or create unmeasured confounder information.

One convenient sufficient regime uses iid observations, bounded outcomes and bounded outcome predictions, true and estimated propensities uniformly bounded away from the endpoints, L²-consistent nuisances, the stated product rate, and a positive limiting influence-score variance. With a fixed fold count these conditions support asymptotic normal inference using the score variance. Clustered data, time series, weak overlap, or adaptive reuse of observations require additional analysis. The method improves estimation of an already identified effect; it cannot turn an unidentified observational contrast into a causal one.

Check both robustness branches in one stratum

In a single baseline stratum let true e=0.25, m₁=0.8 and m₀=0.2, so the effect is 0.6. Evaluate the population augmented contrast for candidate q=0.5,g₁=0.6,g₀=0.1, then correct either nuisance branch.

  1. With both branches wrong, the augmented treated mean is 0.6+(0.25/0.5)(0.8−0.6)=0.7. The augmented control mean is 0.1+(0.75/0.5)(0.2−0.1)=0.25, giving contrast 0.45.

  2. Replace q by the true 0.25 while keeping those predictions. The corrected treated mean becomes 0.6+(0.25/0.25)·0.2=0.8, and the control mean becomes 0.1+(0.75/0.75)·0.1=0.2.

  3. Alternatively keep q=0.5 but set g₁=0.8 and g₀=0.2. Both conditional residual means vanish, so their difference is again 0.6. Correcting just one outcome regression would not generally suffice for this branch.

The result

Either correct propensity or both correct outcome regressions removes population bias here; when neither branch is right the bias is visible.

Explore the model

Interactive lab

Interactive lab · network

Residual correction remains sensitive to weak overlap

Reset, set input −2 and weight 2. The fixed-bias sigmoid gives q=σ(−4.5), about 0.011; predict that a treated residual of 0.1 would contribute about 9.10 after division by q. Reset, set input 2 and weight 2; q≈0.971 gives correction about 0.103. Compute the residual division on paper. The instrument only supplies a toy sigmoid probability, not a cross-fitted causal estimate. Changes keep time; Reset restores both controls and zero time.

Residual correction remains sensitive to weak overlapOne inspectable sigmoid unit, σ(wx − 0.5). Playback traces the computation graph. There is no training, attention mechanism or hidden deep network. Current readings: Product · wx: 1.2; Logit · wx − 0.5: 0.7; Sigmoid output: 0.6682. All parameters are adjustable with the labeled controls below.One input. One weight. One sigmoid.x1.20× w1.20+ b0.70σ0.668output0.668w = 1.00b = −0.50σ(z) = 1 / (1 + exp(−z))
Computation graphSignal trace
Product · wx
1.2
Logit · wx − 0.5
0.7
Sigmoid output
0.6682
t = 0 s

One inspectable sigmoid unit, σ(wx − 0.5). Playback traces the computation graph. There is no training, attention mechanism or hidden deep network. Parameter changes pause playback and recompute the model at the current elapsed time. Reset restores both parameters and returns time to zero. Playback starts paused. Adjust either parameter, then choose Play or Step.

Look forOutcome augmentation can reduce error without removing the variance cost of dividing a nonzero residual by a tiny propensity.

Check your understanding

Which is a sufficient double-robust population correctness branch for this ATE score?

Try it yourself

For one evaluated observation, A=1,Y=1,q=0.25,g₁=0.6,g₀=0.2. Compute the augmented score phi. Must each score lie between negative one and one when Y is binary?

Take a moment to make a prediction or work it out on paper. Open a hint when you need one.

Give me a hint

The control residual term is zero; add the treated weighted residual to the predicted contrast.

Compare with the solution

The predicted contrast is 0.6−0.2=0.4. The treated correction is (1−0.6)/0.25=1.6, so phi=2.0. Individual augmented scores need not lie in the binary ATE range [−1,1]; they are weighted estimating quantities, not individual treatment effects. Under the identifying and nuisance conditions their population mean targets the ATE. A finite-sample average can also leave that range, so boundedness of the causal parameter is not automatic boundedness of this estimator.

Take it with you

  • Double robustness is an algebraic two-branch property of a specified augmented score.
  • Root-n inference requires overlap, moments, nuisance convergence, and a small product remainder.
  • Cross-fitting separates training and evaluation but does not repair hidden confounding or unsupported treatment alternatives.

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