Influence Functions for Means and Variances

Why These Influence Functions Matter

The population mean and variance are familiar summaries, but their role in semiparametric theory is more specific: each describes how a target parameter changes under small changes to the data distribution. That local change is represented by an influence function. In a fully nonparametric model, the influence function also gives the canonical gradient, whose variance is the smallest asymptotic variance available to regular estimators on the usual square-root-of-sample-size scale.

The calculations use the score of a differentiable path through a distribution, as in the earlier material on score operators and semiparametric efficiency bounds. The key point here is to derive the gradients for two particular functionals and verify that they have mean zero and finite variance. These checks determine both whether the gradients are valid and what efficiency bounds they imply.

Setup and Definitions

Let \(Y\) have distribution \(P\), and write \(E_P\) for expectation under \(P\). The target parameters are the population mean \(\mu=E_P[Y]\) and population variance \(\sigma^2=E_P[(Y-\mu)^2]\). The sample consists of independent observations from \(P\). We consider a fully nonparametric model: locally, the distribution is allowed to vary without restrictions such as a specified parametric family.

A regular one-dimensional path \(P_t\), with \(P_0=P\), describes a small distributional perturbation indexed by \(t\). Its score \(S\) at \(t=0\) has \(E_P[S]=0\). The set of possible scores has closure \(L^2_0(P)\), the space of square-integrable, mean-zero functions under \(P\). This is the tangent space in the fully nonparametric model.

For the mean, finite second moment, \(E_P[Y^2]<\infty\), makes the gradient square-integrable. For the variance, we require a finite fourth moment, \(E_P[Y^4]<\infty\), so that the squared-deviation gradient is square-integrable. As usual, the pathwise derivatives below are understood for regular paths along which the relevant expectations are differentiable.

Canonical Gradient for the Mean

Along a path with score \(S\), the derivative of the mean is \(\left.\frac{d}{dt}E_{P_t}[Y]\right|_{t=0}=E_P[YS]\). Since every score has mean zero, \(E_P[YS]=E_P[(Y-\mu)S]\). Thus \(Y-\mu\) represents the derivative in every score direction. It is itself mean zero, and its second moment is finite when \(Y\) has finite variance.

The bound is expressed in squared units of \(Y\). It does not assert that every estimator has this variance: it is a lower bound for regular estimators under the model and moment conditions. For the sample mean, the usual variance calculation gives \(\operatorname{Var}(\bar{Y})=\sigma^2/n\), matching the bound when the variance exists.

Canonical Gradient for the Variance

The variance depends on the distribution both through its second moment and through its mean. Write \(\sigma^2=E_P[Y^2]-\mu^2\). Differentiating both terms along a path gives \(\left.\frac{d}{dt}\sigma_t^2\right|_{t=0}=E_P[Y^2S]-2\mu E_P[YS]\). Because \(E_P[S]=0\), the right-hand side can be written as \(E_P[(Y^2-2\mu Y)S]\). Subtracting any constant from the factor multiplying \(S\) leaves this expectation unchanged. In particular, subtracting its expectation produces a mean-zero representative: \[ Y^2-2\mu Y-E_P[Y^2-2\mu Y] =(Y-\mu)^2-\sigma^2. \] This also shows why the mean’s change must be accounted for: differentiating only the second moment would omit the term \(-2\mu E_P[YS]\).

The fourth-moment condition ensures the gradient has finite variance; the bound has units equal to the fourth power of the units of \(Y\), as appropriate for the variance of an estimator of a variance. A finite population variance alone does not guarantee a finite bound. If the fourth moment is infinite, the displayed gradient is not square-integrable, so it cannot serve as a canonical gradient in \(L^2(P)\).

Interpretation and a Common Pitfall

The mean gradient records how a single observation’s deviation from the population mean contributes to local changes in the mean. The variance gradient instead records whether its squared deviation is above or below the population variance. Observations far from the mean can therefore have a large effect on the variance gradient, which is why its efficiency bound depends on a fourth moment.

• Pitfall: treating the variance gradient as just \((Y-\mu)^2\). That function has expectation \(\sigma^2\), not zero, and is not in the mean-zero tangent space. The correct gradient is \((Y-\mu)^2-\sigma^2\).
• Pitfall: differentiating only \(E_P[Y^2]\). The variance is \(E_P[Y^2]-\mu^2\), so its derivative includes the change in \(\mu^2\). Omitting it generally gives the wrong gradient.
• Pitfall: confusing the bound with a guarantee about a particular estimator. The bound concerns regular estimators under the stated model and conditions. It is not a claim that every estimator attains it.
• Pitfall: assuming finite variance is enough for both gradients. It suffices for the mean gradient, but the variance gradient requires a finite fourth moment to be square-integrable.

In the fully nonparametric model, these calculations are especially direct because every square-integrable mean-zero score direction is available. The canonical gradients are therefore the centered representers of the two pathwise derivatives, and their variances quantify the local information limits for estimating the corresponding population summaries.