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Semiparametric Efficiency · Tutorial 798 of 952

The Quantile Influence Function

Derive how a population quantile responds to small distributional changes and see why a sparse density near the target quantile raises its efficiency bound.

Advanced 10 min read

What You'll Learn

  • Define a population quantile and state the regularity conditions needed for its influence function.
  • Derive the quantile influence function using contamination and differentiable submodels.
  • Interpret the sign and units of the influence function at observations above and below the quantile.
  • Calculate the semiparametric efficiency bound and connect it to the asymptotic variance of a regular estimator.
  • Explain how the density at the target quantile determines sensitivity and precision.

Why the Density at a Quantile Matters

A population quantile identifies a location in a distribution: the median, for example, divides the population probability into two halves. But a quantile can be stable or highly sensitive to small changes in the distribution. The key is how much probability is concentrated near that location. If the distribution has little probability density near the quantile, a small change in cumulative probability can require a comparatively large movement in the quantile itself.

The influence function measures the first-order change in a statistical functional under a small perturbation of the distribution. For a quantile, it reveals both the direction and the scale of the response to an observation. Its squared average also gives the semiparametric efficiency bound in the unrestricted model, connecting local sensitivity to the best possible asymptotic precision.

Quantiles and Regularity Conditions

Let \(Y\) have distribution \(P\) and cumulative distribution function \(F(y)=P(Y\leq y)\). For a fixed probability level \(\tau\), with \(0<\tau<1\), define the population \(\tau\)-quantile by

$$ q_\tau=\inf\{y:F(y)\geq \tau\}. $$

The analysis below assumes that \(F\) is continuous and strictly increasing in a neighborhood of \(q_\tau\), so \(F(q_\tau)=\tau\) and the quantile is locally unique. Assume also that \(F\) has a density \(f\) that is continuous near \(q_\tau\), with \(f(q_\tau)>0\). Positivity is essential: it lets us translate a small change in cumulative probability into a small change in the quantile by dividing by the local density.

Definition: The influence function of a functional \(T(P)\), evaluated at a value \(y\), is its first-order response to an infinitesimal point-mass contamination at \(y\). For the quantile, this response is considered at values \(y\ne q_\tau\); the value assigned at the single point \(q_\tau\) does not affect the influence function as an element of \(L^2(P)\), because \(P(Y=q_\tau)=0\) under the assumptions above.

For a small contamination fraction \(\varepsilon\), the perturbed distribution is \(P_\varepsilon=(1-\varepsilon)P+\varepsilon\Delta_y\), where \(\Delta_y\) places all probability at \(y\). Write \(F_\varepsilon\) and \(q_{\tau,\varepsilon}\) for its cumulative distribution function and quantile. The influence function is the derivative of \(q_{\tau,\varepsilon}\) at \(\varepsilon=0\), when that derivative exists.

Worked Example: Direction of a Median’s Response

Suppose a hypothetical outcome has population median \(q_{0.5}=70\) kilograms, and its density there is \(f(70)=0.04\) per kilogram. Calculate the influence function for contamination at 80 kilograms and at 60 kilograms.

The formula derived below is \(\operatorname{IF}(y;q_\tau,P)=[\tau-\mathbf{1}\{y\leq q_\tau\}]/f(q_\tau)\). At 80 kilograms, the observation is above the median, so the indicator is zero:

$$ \operatorname{IF}(80;q_{0.5},P) =\frac{0.5-0}{0.04\ {\rm kg}^{-1}} =12.5\ {\rm kg}. $$

The influence is positive: adding a small amount of probability above the median moves the median up to first order. At 60 kilograms, the indicator is one:

$$ \operatorname{IF}(60;q_{0.5},P) =\frac{0.5-1}{0.04\ {\rm kg}^{-1}} =-12.5\ {\rm kg}. $$

Contamination below the median moves it down to first order. For example, a contamination fraction of \(0.01\) gives first-order changes of approximately \(0.01(12.5)=0.125\) kilograms above the median and \(0.01(-12.5)=-0.125\) kilograms below it. These are local approximations, not exact changes for an arbitrary finite contamination fraction.

Deriving the Influence Function

For \(y\ne q_\tau\), the contamination point lies either below or above the quantile. In a sufficiently small neighborhood of the original quantile, the contamination indicator is therefore constant: it is one if \(y<q_\tau\) and zero if \(y>q_\tau\). The perturbed quantile satisfies \(F_\varepsilon(q_{\tau,\varepsilon})=\tau\) locally. Differentiating this equation at \(\varepsilon=0\) yields the influence function.

1
Write the perturbed CDF.
\(F_\varepsilon(x)=(1-\varepsilon)F(x)+\varepsilon\mathbf{1}\{y\leq x\}\). At the perturbed quantile, \(F_\varepsilon(q_{\tau,\varepsilon})=\tau\).
2
Differentiate at zero.
Let \(\dot q\) denote the derivative of \(q_{\tau,\varepsilon}\) at zero. The derivative of \(F(q_{\tau,\varepsilon})\) is \(f(q_\tau)\dot q\); the derivative from the mixture weight is \(-F(q_\tau)+\mathbf{1}\{y\leq q_\tau\}\).
3
Solve for the quantile’s response.
The total derivative must equal zero because the quantile equation remains equal to \(\tau\). Using \(F(q_\tau)=\tau\), solve for \(\dot q\).
$$ 0=f(q_\tau)\dot q-\tau+\mathbf{1}\{y\leq q_\tau\}, \qquad \dot q= \frac{\tau-\mathbf{1}\{y\leq q_\tau\}}{f(q_\tau)}. $$

Thus, under the stated conditions, the quantile influence function is

Result: For \(y\ne q_\tau\),
$$ \operatorname{IF}(y;q_\tau,P) = \frac{\tau-\mathbf{1}\{y\leq q_\tau\}}{f(q_\tau)}. $$
The numerator is dimensionless, while \(f(q_\tau)\) has units inverse to those of \(Y\). Consequently, the influence function has the same units as the outcome and the quantile.

The influence function takes two values, apart from its immaterial value at \(q_\tau\). Below the quantile it is \((\tau-1)/f(q_\tau)\), which is negative. Above the quantile it is \(\tau/f(q_\tau)\), which is positive. The farther the target quantile lies in a region with low density, the larger the absolute influence values.

Derivation Along Smooth Submodels

The same derivative can be expressed in the language of semiparametric efficiency. Consider a differentiable one-dimensional submodel \(P_t\) through \(P=P_0\), with score \(s(Y)\) at \(t=0\). A score has mean zero, \(E[s(Y)]=0\), and finite second moment. Suppose the submodel is regular enough that the CDF and quantile can be differentiated at zero, and write \(q_\tau(t)\) for its quantile.

The derivative of the CDF at a fixed point \(q_\tau\) is \(E[\mathbf{1}\{Y\leq q_\tau\}s(Y)]\). Differentiating the quantile equation \(F_t(q_\tau(t))=\tau\) therefore gives

$$ 0=E[\mathbf{1}\{Y\leq q_\tau\}s(Y)] +f(q_\tau)\left.\frac{d q_\tau(t)}{dt}\right|_{t=0}. $$

Since \(E[s(Y)]=0\), the derivative can be rewritten as an inner product with the influence function:

$$ \left.\frac{d q_\tau(t)}{dt}\right|_{t=0} =-\frac{E[\mathbf{1}\{Y\leq q_\tau\}s(Y)]}{f(q_\tau)} =E\left[ \frac{\tau-\mathbf{1}\{Y\leq q_\tau\}}{f(q_\tau)}s(Y) \right]. $$

This identity verifies that the influence function represents the derivative in every regular score direction. It is centered because \(P(Y\leq q_\tau)=\tau\), and it is square-integrable when \(f(q_\tau)>0\). In the unrestricted nonparametric model, the tangent space is the space of mean-zero square-integrable functions. The displayed influence function belongs to that space and represents the derivative, so it is the canonical gradient.

The Semiparametric Efficiency Bound

For a regular asymptotically linear estimator of \(q_\tau\), its influence function must represent the derivative in the score directions. In the unrestricted model, the canonical gradient has the smallest variance among such influence functions: any additional component orthogonal to the tangent space can only add variance. The variance of this gradient is therefore the semiparametric efficiency bound for one observation.

To calculate it, note that \(\mathbf{1}\{Y\leq q_\tau\}\) is a Bernoulli random variable with success probability \(\tau\). Its variance is \(\tau(1-\tau)\), and the influence function has mean zero. Hence

$$ \mathcal{I}^{-1}_{q_\tau} = E[\operatorname{IF}(Y;q_\tau,P)^2] = \frac{\tau(1-\tau)}{f(q_\tau)^2}. $$

Here \(\mathcal{I}^{-1}_{q_\tau}\) denotes the variance bound per observation, not a probability or an interval. For a sample of \(n\) independent, identically distributed observations, the corresponding lower bound on asymptotic variance is \(\tau(1-\tau)/[n f(q_\tau)^2]\). A regular estimator that attains the bound is called semiparametrically efficient.

Worked Example: Efficiency Bound for a Lower Quantile

Suppose a hypothetical duration outcome has a population 25th percentile \(q_{0.25}=4.0\) years and density \(f(q_{0.25})=0.20\) per year. Calculate the efficiency bound per observation and the resulting asymptotic standard-error bound for \(n=400\) independent observations.

Substitute \(\tau=0.25\) and \(f(q_\tau)=0.20\ {\rm year}^{-1}\) into the bound:

$$ \mathcal{I}^{-1}_{q_{0.25}} = \frac{0.25(1-0.25)}{(0.20\ {\rm year}^{-1})^2} = \frac{0.1875}{0.04\ {\rm year}^{-2}} = 4.6875\ {\rm year}^2. $$

For \(n=400\), the asymptotic variance bound is \(4.6875/400=0.01171875\ {\rm year}^2\), so the standard-error bound is

$$ \sqrt{\frac{0.25(0.75)}{400(0.20)^2}} = \sqrt{0.01171875}\ {\rm years} \approx 0.1083\ {\rm years}. $$

This is a lower bound on asymptotic standard error under the stated model and regularity conditions, not a guaranteed standard error for every estimator or finite sample. The density is measured per year, so the variance and standard error have units of years squared and years, respectively.

Using the Influence Function to Understand Precision

For an estimator with the quantile influence function, the asymptotic linear representation is

$$ \widehat q_\tau-q_\tau = \frac{1}{n}\sum_{i=1}^n \operatorname{IF}(Y_i;q_\tau,P) +o_P(n^{-1/2}). $$

The leading term is an average of mean-zero influence values. The central limit theorem then gives an asymptotic variance equal to the influence-function variance divided by \(n\), namely \(\tau(1-\tau)/[n f(q_\tau)^2]\). This explains why the bound is not merely an abstract property of a model: it describes the first-order sampling variability that a regular efficient estimator can attain.

Worked Example: Comparing Density and Standard Error

Consider a hypothetical sample of \(n=1{,}000\) independent cholesterol measurements, with target median estimated as \(180\) mg/dL. Suppose a suitable local density estimate at that value is \(0.025\) per mg/dL. Using the median efficiency formula, calculate the asymptotic standard-error benchmark.

For the median, \(\tau=0.5\), so \(\tau(1-\tau)=0.25\). Substitution gives

$$ \operatorname{SE}_{\rm bound} = \sqrt{\frac{0.5(0.5)} {1{,}000(0.025\ {\rm (mg/dL)}^{-1})^2}} = \sqrt{\frac{0.25}{0.625}} = \sqrt{0.4} \approx 0.6325\ {\rm mg/dL}. $$

The value \(180\) mg/dL identifies the location at which the density is evaluated; the density value, rather than the numerical size of the median itself, controls this first-order precision calculation. If the density at the median were half as large, the standard-error bound would be twice as large, all else equal, because the bound depends on the inverse square of the density and the standard error on its inverse.

Why Density Controls Sensitivity

The quantile equation is \(F(q_\tau)=\tau\). Near \(q_\tau\), a small movement \(\Delta q\) changes the CDF by approximately \(f(q_\tau)\Delta q\). If a perturbation changes cumulative probability by an amount \(\Delta F\), restoring the quantile equation requires approximately \(\Delta q=-\Delta F/f(q_\tau)\). A large density means probability accumulates quickly as the outcome changes, so only a small location movement is needed. A small positive density means the same probability change requires a larger movement.

The efficiency bound makes the same point quantitatively: \(f(q_\tau)\) appears squared in the denominator. Quantiles in sparse regions of a distribution are therefore less precisely estimable, even when the sample size and quantile level are unchanged. This relationship is especially important when comparing estimates across populations or quantile levels: their densities at the target quantiles may differ substantially.

Common Mistake

A frequent error is to reverse the sign of the influence function. One might reason that an observation above the median pulls the median down because it increases the upper tail, or that an observation below pulls it up. That is incorrect. The influence function is positive above the quantile, so adding a small amount of probability there moves the quantile up to first order; it is negative below the quantile, so contamination there moves it down. The sign follows directly from \(\tau-\mathbf{1}\{y\leq q_\tau\}\), not from a vague intuition about changing tail mass.

Another error is to interpret a large density as producing a large variance. The bound is proportional to \(1/f(q_\tau)^2\), so a larger density lowers the bound. Also, the efficiency bound is a lower bound on asymptotic variance for regular estimators in the specified model; it is not a claim that an estimator’s finite-sample variance must equal that value.

Key takeaway: Under a unique interior quantile and a continuous, positive density at that quantile, the influence function is \([\tau-\mathbf{1}\{Y\leq q_\tau\}]/f(q_\tau)\). Its variance, \(\tau(1-\tau)/f(q_\tau)^2\), is the semiparametric efficiency bound per observation. Lower density at the target quantile means greater sensitivity and a higher bound on sampling variance.

Check Your Understanding

Answer each question using the formulas and assumptions developed above.

  1. For a quantile level \(\tau=0.75\), what is the influence-function numerator for an observation below the quantile? What is its sign?
  2. Suppose two populations have the same target quantile level, but one has half the density at its target quantile. How do their efficiency bounds compare, all else equal?
  3. Why must the density at the target quantile be positive for the influence-function formula derived here?
  4. A hypothetical median is 50 units, with density \(0.10\) per unit there. Calculate the influence function for contamination at 55 units and interpret its sign and units.
  5. Explain why the efficiency bound is divided by \(n\) when describing the asymptotic variance for a sample of \(n\) independent observations.
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