Why Product Rates Matter
Longitudinal estimators combine several estimated treatment probabilities and outcome regressions. Even when each nuisance estimate is imperfect, their errors can have a smaller joint effect: a treatment-probability error is multiplied by an outcome-regression error. Product-rate conditions make this idea precise. They specify when the resulting second-order drift is negligible compared with the \(n^{-1/2}\) sampling fluctuation.
The indexing matters. At time \(t\), the relevant regression error is the error in the regression used at that time, \(\bar Q_t\), not the error in the next regression. In the one-time case, the remainder must involve both the treatment-probability error and the error in \(\bar Q_1\). The known terminal outcome \(Y\) does not make that product disappear.
Longitudinal Setup and Notation
Consider \(K\) treatment times. Write \(H_t\) for the history just before treatment \(A_t\), including baseline covariates and the preceding treatments and covariates. Fix a static regime \(\bar a=(a_1,\ldots,a_K)\). Let \(g_t(a_t\mid H_t)\) be the true conditional probability of receiving \(a_t\), and define the sequential outcome regressions by
Here and below, a regression evaluated at a regime history uses the treatment values prescribed by \(\bar a\). The identified regime mean is \(\psi=E[\bar Q_1(H_1)]\). Its causal interpretation requires the usual consistency, sequential exchangeability, and positivity assumptions; the rate argument below concerns estimation of this identified mean.
Let \(\hat g_t\) and \(\hat{\bar Q}_t\) denote nuisance fits, with \(\hat{\bar Q}_{K+1}=Y\) exactly. For a prefix of the regime, define
where \(I(\bar A_t=\bar a_t)\) indicates that all treatments through time \(t\) followed the regime. The efficient influence function for this fixed regime mean is
The corresponding estimated expression \(D_{\hat\eta}\) replaces the true regressions and treatment probabilities by their fitted versions and centers the first term at \(P\hat{\bar Q}_1\), where \(Pf=E[f(O)]\). Cross-fitting means that each observation is evaluated using nuisance fits trained without that observation, usually in a fixed number of folds.
Deriving the Correctly Indexed Drift
First consider the cross-fitted one-step estimator whose validation-fold contribution is
For fixed fitted nuisance functions, its population value is \(T=E[\hat{\bar Q}_1]+\sum_{t=1}^K E[W_t(\hat{\bar Q}_{t+1}-\hat{\bar Q}_t)]\). Put \(\Delta_t=\hat{\bar Q}_t-\bar Q_t\) for \(t\le K\), and \(\Delta_{K+1}=0\). The conditional-mean definition of \(\bar Q_t\) makes the weighted expectation of each true residual \(\bar Q_{t+1}-\bar Q_t\) equal to zero, even with the fitted weight: the extra factor \(\prod_{j\le t}g_j/\hat g_j\) is measurable in the history at time \(t\), where the residual has conditional mean zero. The remaining terms telescope to give the exact drift identity
In this display the indicator in \(W_{t-1}\) is redundant if \(W_{t-1}\) is defined as above; it emphasizes that the expectation is over histories that followed the regime prefix. At \(t=1\), the prefix weight is one. The factor \(g_t/\hat g_t\) arises because averaging the time-\(t\) treatment indicator conditional on \(H_t\) replaces it by its true probability \(g_t\).
Proof. Write \(W_{t-1}=W^{\,0}_{t-1}R_{t-1}\), where \(W^{\,0}_{t-1}=I(\bar A_{t-1}=\bar a_{t-1})/\prod_{j<t}g_j\) is the true prefix weight and \(R_{t-1}=\prod_{j<t}g_j/\hat g_j\) on that prefix. The ratio bound gives \(|R_{t-1}|\le C_0\), uniformly in \(n\), since \(K\) is fixed. Also \[ \left|1-\frac{g_t}{\hat g_t}\right| =\frac{|\hat g_t-g_t|}{\hat g_t} \le c^{-1}|\hat g_t-g_t|. \] The true prefix weight changes expectation over observed histories into expectation under \(P_t^a\). Applying these bounds to the drift identity, then applying Cauchy–Schwarz at each time, gives \[ |T-\psi| \le C_0c^{-1}\sum_{t=1}^K E_{P_t^a}[|\hat g_t-g_t|\,|\Delta_t|] \le C_0c^{-1}\sum_{t=1}^K\delta_{g,t}\delta_{Q,t}. \] This proves the bound. \(\square\)
From the Drift Bound to Asymptotic Linearity
For the cross-fitted one-step estimator, the decomposition is
with fold-specific nuisances understood in the empirical terms. Thus two distinct requirements are involved: the population drift must be \(o_p(n^{-1/2})\), and the centered empirical difference must be \(o_p(n^{-1/2})\).
Proof. The drift bound makes \(\sqrt n(T-\psi)=o_p(1)\). For a held-out fold of size \(m\), condition on its training data. The fitted influence-function difference is then fixed relative to that fold and has conditional mean \(P(D_{\hat\eta}-D)\). Consequently the centered fold average has conditional variance at most \[ m^{-1}P[(D_{\hat\eta}-D)^2]. \] Since \(m\) is proportional to \(n\), the assumed \(L_2(P)\) convergence and conditional Chebyshev inequality imply that the fold’s centered difference, multiplied by \(\sqrt n\), is \(o_p(1)\). Summing over a fixed number of folds preserves this conclusion. This controls \(\sqrt n(P_n-P)(D_{\hat\eta}-D)\), not the uncentered quantity \(\sqrt nP_n(D_{\hat\eta}-D)\). The latter includes a drift and need not be negligible from \(L_2\) convergence alone. Substitution into the decomposition leaves \(n^{-1/2}\sum_iD(O_i)+o_p(1)\). The stated normal limit follows from the independent, identically distributed central limit theorem applied to the mean-zero influence function \(D\). \(\square\)
For a longitudinal TMLE, a targeting step is intended to control its empirical influence-function equation, but a successful-looking update alone does not establish this theorem. One must also have an estimator expansion with a second-order remainder controlled by the relevant nuisance products (or another proved remainder bound), and a negligible targeting or estimating-equation error. The product-rate argument supplies the drift part; it does not replace those estimator-specific conditions.
Worked Example: Checking the One-Time Remainder
Let \(K=1\), \(A\sim\mathrm{Bernoulli}(0.5)\), the regime be \(A=1\), and \(Y=A\). Then \(g_1=0.5\), \(\bar Q_1=1\), \(\bar Q_2=Y\), and \(\psi=1\). Take constant fits \(\hat g_1=0.25\) and \(\hat{\bar Q}_1=0\). The one-step population value is \[ T=E[\hat{\bar Q}_1]+E\left[\frac{I(A=1)Y}{0.25}\right] =0+ \frac{0.5}{0.25}=2, \] so \(T-\psi=1\). The exact remainder identity gives \[ \left(1-\frac{0.5}{0.25}\right)(0-1)=(-1)(-1)=1. \] Thus the treatment-probability error pairs with the error in \(\bar Q_1\). Although \(\bar Q_2=Y\) is known exactly, its zero error does not remove this remainder.
Worked Example: Rates Across Three Treatment Times
Suppose \(K=3\), and for every \(t\), \(\delta_{g,t}=O_p(n^{-1/3})\) while \(\delta_{Q,t}=O_p(n^{-1/4})\). Each stage product is \(O_p(n^{-1/3-1/4})=O_p(n^{-7/12})\). Since \(7/12>1/2\), \(n^{-7/12}=o(n^{-1/2})\). With three fixed stages, the sum of the three products is still \(o_p(n^{-1/2})\). These rates therefore make the stated drift negligible, provided the influence-function convergence and estimator-expansion conditions also hold.
Worked Example: A Boundary Rate Is Not Enough
Suppose there are two stages and, at each stage, both nuisance errors are \(O_p(n^{-1/4})\). Their product is only \(O_p(n^{-1/2})\), not necessarily \(o_p(n^{-1/2})\). For example, if one error is exactly \(n^{-1/4}\) and the other is exactly \(n^{-1/4}\), their product is \(n^{-1/2}\), and multiplying the drift by \(\sqrt n\) leaves a constant-sized term. The theorem’s condition is therefore not verified by these rates alone. One could instead have \(o_p(n^{-1/4})\) for one factor and \(O_p(n^{-1/4})\) for the other, giving an \(o_p(n^{-1/2})\) product.
Diagnostics, Alternatives, and Limits
In practice, the rates are rarely known exactly. They are theoretical sufficient conditions, not quantities that can generally be certified by inspecting one fitted model. Useful diagnostics include checking that folds are formed as intended, that nuisance predictions are evaluated on held-out observations, and that estimated treatment probabilities and weights are not extreme. Poor overlap can make both the product bound and influence-function variance unstable; a small estimated product does not repair a failure of positivity.
The one-step estimator and TMLE are alternative ways to use the same influence-function structure. The one-step derivation above makes the population drift explicit. TMLE instead updates a plug-in estimate through a targeting step and can preserve the parameter’s natural range, but its asymptotic expansion and remainder still need justification. An untargeted sequential-regression plug-in does not automatically have the same influence-function expansion. These approaches can have different finite-sample behavior; the rate condition alone does not guarantee that one will be numerically stable.
The displayed product bound is sufficient, not necessary. It is conservative because it bounds each stage separately and uses Cauchy–Schwarz. Special cancellations can yield smaller drift, while weak positivity, growing follow-up length, or unbounded ratios can invalidate the fixed-constant bound. For a different target, stochastic intervention, or estimator construction, the remainder and its paired nuisance errors must be derived for that specific setting rather than copied by analogy.
Check Your Understanding
Answer each question, then compare with the solutions.
- For \(K=1\), which outcome-regression error is paired with the treatment-probability error in the drift bound?
- If at every fixed stage \(\delta_{g,t}=O_p(n^{-0.30})\) and \(\delta_{Q,t}=O_p(n^{-0.25})\), does the product condition hold?
- Why does cross-fitting with \(L_2\) convergence control the centered empirical difference but not necessarily its uncentered version at the \(\sqrt n\) scale?
- Does an \(O_p(n^{-1/4})\) rate for both factors automatically give a negligible drift at the \(\sqrt n\) scale?
Solutions. (1) It is \(\hat{\bar Q}_1-\bar Q_1\), because the time-\(t\) treatment error pairs with the regression error at that same stage. (2) Yes: the product is \(O_p(n^{-0.55})=o_p(n^{-1/2})\), and a fixed sum over stages remains of that order. (3) Conditional on training data, the held-out centered difference has mean zero and variance bounded by the squared \(L_2(P)\) error divided by fold size. The uncentered empirical difference also contains its population mean, which \(L_2\) convergence alone need not make \(o_p(n^{-1/2})\). (4) No. Their product is only \(O_p(n^{-1/2})\), which need not be little-o of \(n^{-1/2}\); a stronger rate or another argument is needed.