TMLE Variance Estimation and Asymptotic Inference

Why TMLE Needs an Influence-Function Standard Error

A targeted maximum likelihood estimator (TMLE) produces an estimate of a causal parameter, but an estimate alone does not describe its sampling uncertainty. For the average treatment effect (ATE), the efficient influence function provides a principled way to estimate that uncertainty: evaluate its empirical version for each observation, measure the variation across those values, and scale by the sample size. This tutorial develops that procedure and explains the large-sample conditions that make the resulting confidence intervals meaningful.

Let the observed data be independent and identically distributed observations $O_i=(W_i,A_i,Y_i)$, where $A_i$ is binary, $W_i$ is a vector of measured covariates, and $Y_i$ is the outcome. Write $Q(a,W)=E(Y\mid A=a,W)$ for the outcome regression and $g(W)=P(A=1\mid W)$ for the treatment mechanism. The causal interpretation of the ATE requires the usual identification conditions: consistency, conditional exchangeability given $W$, and positivity. In particular, treatment probabilities must be bounded away from zero and one over the covariate values relevant to the target population.

Definition: For the ATE $\psi=E\{Q(1,W)-Q(0,W)\}$, the efficient influence function at nuisance functions $Q$ and $g$ is $$ D(O;Q,g,\psi)=\frac{A}{g(W)}\{Y-Q(1,W)\}-\frac{1-A}{1-g(W)}\{Y-Q(0,W)\}+Q(1,W)-Q(0,W)-\psi. $$ An influence function has mean zero at the true data distribution. The empirical efficient influence function (EIF) for a fitted TMLE is obtained by substituting its targeted outcome regression, estimated treatment mechanism, and ATE estimate into this expression for each observation.

Denote the targeted outcome regression by $\hat Q^*$, the estimated treatment mechanism by $\hat g$, and the TMLE by $\hat\psi$. For observation $i$, the empirical EIF is

\hat D_i= \frac{A_i}{\hat g(W_i)}\{Y_i-\hat Q^*(1,W_i)\} -\frac{1-A_i}{1-\hat g(W_i)}\{Y_i-\hat Q^*(0,W_i)\} +\hat Q^*(1,W_i)-\hat Q^*(0,W_i)-\hat\psi.

For a standard ATE TMLE, $\hat\psi$ is the sample average of the targeted predicted contrasts. The targeting equation sets the empirical average of the weighted residual terms to zero (up to numerical tolerance). Consequently, the average of the empirical EIF values is zero, also up to numerical tolerance. This property is a useful computational check; it does not mean that the individual EIF values are zero.

From Empirical EIF Values to a Standard Error

The large-sample variance of an asymptotically linear estimator is determined by the variance of its influence function. If the estimator has the expansion

\hat\psi-\psi=\frac{1}{n}\sum_{i=1}^{n}D(O_i;Q_0,g_0,\psi)+o_p(n^{-1/2}),

then, under the regularity conditions for a central limit theorem, $\sqrt{n}(\hat\psi-\psi)$ is approximately normal with mean zero and variance $E\{D(O;Q_0,g_0,\psi)^2\}$. The plug-in estimate uses the empirical EIF values to estimate that variance.

Result: Let $\bar{\hat D}=n^{-1}\sum_{i=1}^n\hat D_i$, and let $$ \hat\sigma_D^2=\frac{1}{n-1}\sum_{i=1}^{n}(\hat D_i-\bar{\hat D})^2. $$ The influence-function standard error is $\widehat{SE}(\hat\psi)=\hat\sigma_D/\sqrt{n}$. Equivalently, one may use $n^{-1}\sum_i\hat D_i^2$ to estimate the EIF variance when the empirical EIF mean is zero; the two variance conventions are asymptotically equivalent.

The sample-variance formula is often convenient because it explicitly centers the values. When the targeting equation is solved exactly and $\bar{\hat D}=0$, its numerator is simply $\sum_i\hat D_i^2$. Small numerical differences in the targeting equation can make centering useful in practice. Each EIF value and the resulting standard error have the units of the outcome; the EIF variance has squared outcome units.

Worked Example: Computing the EIF Variance From Targeted Predictions

Suppose a hypothetical four-person data set has the fitted values and outcomes below. For illustration, the estimated treatment probability is $0.50$ for each person, and the targeted prediction contrasts average to $\hat\psi=2$ outcome units. The fitted values satisfy the empirical targeting equation.

Person	$A_i$	$Y_i$	$\hat Q^*(1,W_i)$	$\hat Q^*(0,W_i)$	Contrast
1	1	6.5	6	3	3
2	0	4.5	5	4	1
3	1	7	7	5	2
4	0	4	6	4	2

The plug-in contrast estimate is $(3+1+2+2)/4=2$. For person 1, the weighted residual is $1(6.5-6)/0.50=1$, so $\hat D_1=1+3-2=2$. For person 2, the weighted residual is $-1(4.5-4)/0.50=-1$, so $\hat D_2=-1+1-2=-2$. For person 3, it is $1(7-7)/0.50=0$, giving $\hat D_3=0+2-2=0$. For person 4, it is $-1(4-4)/0.50=0$, giving $\hat D_4=0+2-2=0$. Thus the empirical EIF values are $2,-2,0,0$, whose average is zero, as expected.

Their sample variance is $$ \hat\sigma_D^2=\frac{(2-0)^2+(-2-0)^2+(0-0)^2+(0-0)^2}{4-1} =\frac{8}{3}=2.667. $$ Therefore $\widehat{SE}(\hat\psi)=\sqrt{(8/3)/4}=\sqrt{2/3}=0.8165$ outcome units. The arithmetic illustrates the estimator; with only four observations, a normal-approximation interval should not be treated as reliably calibrated.

Wald Confidence Intervals and Tests

When the estimator is asymptotically normal and the EIF variance estimate is consistent, a two-sided approximate $100(1-\alpha)\%$ Wald confidence interval is

\hat\psi \ \pm\ z_{1-\alpha/2}\widehat{SE}(\hat\psi),

where $z_{1-\alpha/2}$ is the corresponding standard normal quantile. For a 95% interval, $z_{0.975}$ is approximately 1.96. The interval procedure has approximately 95% coverage over repeated samples under the stated conditions; it does not assign a 95% probability to the fixed parameter being inside a particular calculated interval.

Worked Example: A 95% Interval and a Test of the Null

In a separate hypothetical analysis, a TMLE estimates the ATE in systolic blood pressure to be $1.20$ mmHg from $n=400$ independent participants. The empirical EIF sample variance is $9.00$ $(\text{mmHg})^2$. Assume the identification and asymptotic conditions described below are reasonable for this illustration.

The standard error is $\sqrt{9.00/400}=0.150$ mmHg. The approximate 95% confidence interval is $$ 1.20\pm 1.96(0.150)=1.20\pm0.294=[0.906,\ 1.494]\ \text{mmHg}. $$

In repeated samples under the model and regularity conditions, intervals constructed this way would cover the population ATE about 95% of the time. Here, the interval is entirely above zero. For the two-sided test $H_0:\psi=0$ versus $H_a:\psi\ne0$, the Wald statistic is $Z=1.20/0.150=8.00$. Its standard-normal two-sided p-value is less than $0.0001$. This is strong evidence against a zero ATE under the assumptions; it is not the probability that the null hypothesis is true.

Conditions Supporting Asymptotic Inference

The formula for a standard error does not, by itself, guarantee that a confidence interval is valid. Its justification relies on an asymptotically linear expansion, a central limit theorem for the EIF, and consistent estimation of the EIF variance. The following conditions summarize the main requirements for the ATE TMLE setting.

Sampling and identification: observations are independent and identically distributed from the target population, and consistency, conditional exchangeability, and positivity support identification of the ATE.
Regularity and moments: the parameter is regular in the statistical model under consideration, the EIF has finite, nonzero variance, and the outcome and inverse treatment-probability weights have sufficient moment control for the relevant central limit and variance-estimation arguments. Bounded outcomes and treatment probabilities bounded away from zero and one are common sufficient safeguards.
Nuisance estimation: the fitted nuisance functions approach their true values in appropriate norms, and their estimation errors satisfy a product-rate condition that controls the second-order remainder.
Empirical-process control: if nuisance functions are estimated and evaluated on the same observations, their complexity must be controlled by suitable empirical-process conditions, such as appropriate Donsker conditions. Cross-fitting can relax this requirement by evaluating nuisance estimates on observations not used to fit them, while still requiring the rate and moment conditions.

For cross-fitted nuisance estimates, a common sufficient rate condition is

\|\hat Q-Q_0\|\,\|\hat g-g_0\|=o_p(n^{-1/2}),

where the norms are suitable population $L_2$ norms and $Q_0,g_0$ denote the true nuisance functions. Both nuisance estimates are ordinarily required to be consistent for the EIF to converge to its true counterpart and for its variance estimate to be consistent. If both errors are $o_p(n^{-1/4})$, their product is $o_p(n^{-1/2})$, which is a convenient sufficient condition. The product condition itself allows different rates: one nuisance can converge more slowly if the other converges sufficiently quickly.

Worked Example: Checking Nuisance-Rate Conditions

Suppose, hypothetically, the outcome-regression error in $L_2$ norm is $O_p(n^{-0.30})$, and the treatment-mechanism error is $O_p(n^{-0.25})$. Their product is $O_p(n^{-0.55})$. Since $n^{-0.55}/n^{-0.50}=n^{-0.05}$ tends to zero, the product is $o_p(n^{-1/2})$, so these rates meet the stated product-rate requirement.

By contrast, if both errors are only $O_p(n^{-0.20})$, their product is $O_p(n^{-0.40})$. The ratio to $n^{-1/2}$ is $n^{0.10}$, which grows rather than tending to zero. These rates do not establish the required product condition. This comparison is about the order of the nuisance errors, not about whether a particular sample estimate happens to look stable.

A Short Justification of the Standard Error Formula

The standard error scaling follows directly from the asymptotic linear representation. If $D_i=D(O_i;Q_0,g_0,\psi)$ are independent with mean zero and finite variance $\sigma_D^2$, then the central limit theorem gives

\frac{1}{\sqrt n}\sum_{i=1}^n D_i \ \overset{d}{\longrightarrow}\ N(0,\sigma_D^2).

Multiplying the estimator expansion by $\sqrt n$ shows that $\sqrt n(\hat\psi-\psi)$ has this same limiting distribution. Thus $\hat\psi$ is approximately normal with variance $\sigma_D^2/n$. If the empirical EIF variance $\hat\sigma_D^2$ converges to $\sigma_D^2$, replacing the unknown variance gives the estimated standard error $\sqrt{\hat\sigma_D^2/n}$. The plug-in step is justified by consistency of the EIF values and their variance, not merely by calculating a sample variance.

Practical Checks and a Common Pitfall

Before reporting inference, check that the EIF formula matches the estimator and treatment coding, that each fitted treatment probability is strictly between zero and one, and that the empirical EIF mean is close to zero when the targeting equation should enforce it. Inspect the magnitude of the EIF values as well: extreme inverse-probability weights can produce an unstable variance estimate. Such checks can reveal computational problems, but they cannot prove the identification assumptions or the nuisance-rate conditions.

Common pitfall: A small standard error is not evidence by itself that a TMLE confidence interval is valid. The incorrect reasoning is, “The empirical EIF variance is small, so the interval must be trustworthy.” The correction is that the standard error estimates sampling variability only when the estimator has the required asymptotic linearity and the empirical EIF variance consistently estimates the true EIF variance. Poor positivity, inadequate nuisance rates, or uncontrolled empirical-process behavior can invalidate that justification even when the arithmetic is correct.

The empirical EIF therefore plays two related roles: its spread estimates uncertainty, and its observed behavior offers a diagnostic check on the targeting computation. Reliable asymptotic inference requires both correct calculation and the regularity conditions that connect the fitted EIF values to the true influence function.

Check Your Understanding

Answer each question by referring to the definitions and conditions above.

Which three fitted quantities are substituted to compute each empirical EIF value for an ATE TMLE?
Why does the standard error divide the estimated EIF standard deviation by $\sqrt n$?
A hypothetical estimate is $0.8$ outcome units with a standard error of $0.2$. Write the 95% Wald interval formula and calculate its endpoints.
If two nuisance errors are each $O_p(n^{-0.30})$, what is the order of their product, and does it satisfy the stated product-rate condition?
Why does a small calculated standard error not, by itself, establish that an asymptotic confidence interval is valid?

Person	\(A_i\)	\(Y_i\)	\(\hat Q^*(1,W_i)\)	\(\hat Q^*(0,W_i)\)	Contrast
1	1	6.5	6	3	3
2	0	4.5	5	4	1
3	1	7	7	5	2
4	0	4	6	4	2