How exactly does a "1-step" influence function-based estimator estimate a target functional (like average treatment effect) for an unknown distribution?

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As described in Aaron Fisher and Edward H. Kennedy's tutorial (2019), the aim is to find an unbiased estimate of a functional $T(P)$ and its variance for an unknown distribution $P$ by approximating with a smoothed parametric estimate (or possibly nonparametric estimate using machine learning models) $T(\tilde{P})$ from the observed data.

The plots above are from Aaron Fisher and Edward H. Kennedy's tutorial (2019) on influence functions. Panel B illustrates how a Taylor series approximation is used to estimate a hypothetical $T(P)$ by a linear extrapolation from the slope (calculated using an influence function) at $T(\tilde{P})$ (that is, the $y$-intercept), where the solid line shows the path of the target functional value $P_{\epsilon}$ as we vary the weight $\epsilon \in [0,1]$.

The Taylor series expansion is then:

$$T(P_0)=T(P_1)+\left(\frac{\partial}{\partial \epsilon}T(P_{\epsilon})\Big{|}_{\epsilon =1}\right)\times (0-1)-R_2$$

where $R_2$ is the second order term of the Taylor expansion.

The plots seem a little too convenient, where a step size of exactly $\Delta \epsilon=1$, from 1 to 0, magically gets you from $T(P_1)=T(\tilde{P})$ to $T(P_0)=T(P)$ even though you don't know in advance what the $T(P_{\epsilon})$ path looks like.


1 Answer 1


About the convenient plot and the Taylor approximation

The plots seem a little too convenient, where a step size of exactly $\Delta \epsilon=1$, from 1 to 0, magically gets you from $T(P_1)=T(\tilde{P})$ to $T(P_0)=T(P)$ even though you don't know in advance what the $T(P_{\epsilon})$ path looks like.

$$T(P_0)=T(P_1)+\left(\frac{\partial}{\partial \epsilon}T(P_{\epsilon})\Big{|}_{\epsilon =1}\right)\times (0-1)-R_2$$

This remainder term $R_2 = -\frac{1}{2} \frac{\partial^2}{\partial \epsilon^2}T(P_{\epsilon})\Big{|}_{\epsilon = \bar{\epsilon}}$ is a consequence from Taylor's theorem.

It is not the second order term of the Taylor expansion, instead it is the remainder term. Note that for this term $\epsilon \neq 1$, but $\epsilon = \bar{\epsilon}$.

The value $\bar{\epsilon}$ is undetermined but it must be some value between the boundaries (0 and 1).

You can see it alternatively as following: the remainder term $R_2$ of the taylor approximation is limited by the lower and upper limits of the second derivative $T^{\prime\prime}(P_\epsilon)$

$$ \frac{1}{2} \min_{0\leq\epsilon\leq1}T^{\prime\prime}(P_\epsilon) \leq R_2 \leq \frac{1}{2} \max_{0\leq\epsilon\leq1} T^{\prime\prime}(P_\epsilon) $$

and so there is some value $\bar\epsilon$ for the real value $T^{\prime\prime}(P_\bar{\epsilon})$ which is somewhere in between.

$$ \frac{1}{2} \min_{0\leq\epsilon\leq1}T^{\prime\prime}(P_\epsilon) \leq \frac{1}{2} T^{\prime\prime}(P_\hat\epsilon) \leq \frac{1}{2} \max_{0\leq\epsilon\leq1} T^{\prime\prime}(P_\epsilon) $$

The exact path, the thick line remains unknown. We do not magically get it.

But we can know that the remainder term $R_2$, the difference between our linear estimate (the broken thin line thin) and the exact path, is relatively small (an error term that doesn't grow faster than the second derivative, which is for most smooth functions not so large).

About an intuitive view of the robustness due to correction with influence curves

Hoping to see a clear, intuitive explanation of how efficient influence curves are applied to initial probability distribution estimates (perhaps using nonparametric machine learning models) to arrive at an unbiased estimate of a target function.

It is the first time that I read about 1-step estimators, to my intuition it seems like some form of scoring algorithm where the score and fisher information are based the influence functions and an empirical distribution (how a change in the observation changes the parameter estimate can be inverted to how a change in the parameter changes the probability of observations, and related the likelihood function).


The practical example and computation below may provide some intuition:

In this example the target is to estimate for the population distribution function $f(x)$ the integrated squared density:

$$T(f(x)) = E[f(x)] = \int_{-\infty}^{\infty} f(x)^2 dx$$

In the appendix C of the article from Fisher and Kennedy it is stated that in this case the influence function is

$$IF(x,f) = 2(f(x)-T(f))$$

In the code below we first estimate the distribution with a normal distribution in which case the initial estimate is $T(\tilde{f}(x)) = 1/\sqrt{4 \pi \hat{\sigma^2}}$. See in the image below that this estimate with a normal distribution is not a good one if the data is not normal distributed (in this case we generate the data according to a geometric distribution).

normal distribution is not a good approximation

So we use the influence functions to correct the biased normal distribution estimate and shift that distribution with a first order approximation to the empirical distribution (a sum of delta functions).

The effect is a reduction of the bias from using the normal distribution as an approximation for the distribution. We get an estimate that is more robust than our estimate with a (potentially biased) parameterized distribution.

In this case the plugin solution $\sum \hat{f}(x)^2$ is actually doing pretty well, and even better than the 1-step estimator. This is because the computation is done with a sample size of $n=100$ in which case the mass distribution $f(x)$ can be well estimated. But for a small sample there will be only a small amount of cases in each bin and we would have $\sum \hat{f}(x)^2 \approx \sum (1/n)^2 = 1/n$ and that's when the approximation with a parametric distribution (and the 1-step estimator for robustness) are useful.

the effect of the first order step adjustment


trueval <- sum(dgeom(0:300,0.1)^2)

onestep <- function(n = 20, plotting = FALSE) {
  x <- rgeom(n,0.1)
  # estimating distribution with normal distribution
  # using method of moments
  mu <- mean(x)
  var <- var(x)*n/(n-1)
  Test <- 1/sqrt(4*pi*var)
  # computing influence functions 
  # for T(f(x)) = integral of f(x)^2 dx
  IF <- 2*(dnorm(x,mu,var^0.5)-Test)
  # making corrections
  T1step <- Test + sum(IF)/n  

  if (plotting) {
    h<-hist(x, breaks = c(0:200)-0.5, xlim = c(-10,40), freq = FALSE,
         main = "geometric data and normal estimate")
    xs <- seq(-10,40,0.01)
    lines(xs, dnorm(xs,mu,var^0.5))
  } else {
    h<-hist(x, breaks = c(0:200)-0.5, xlim = c(-10,40), freq = FALSE,
            main = "geometric data and normal estimate", plot = FALSE)
  plugin <- sum(h$density^2)
  # return

onestep(n=1000, plotting=TRUE)


trials <- replicate(10^3,onestep(100))


h1 <- hist(trials[1,], breaks = seq(0,0.40,0.0025), plot = FALSE)
h2 <- hist(trials[2,], breaks = seq(0,0.40,0.0025), plot= FALSE)
h3 <- hist(trials[3,], breaks = seq(0,0.40,0.0025), plot = FALSE)

plot(h1$mids,h1$density, type = "l", col = "gray", xlim = c(0,0.1),ylim=c(0,100),
     xlab = "estimated T", ylab = "density")
lines(h2$mids, h2$density )
lines(h3$mids, h3$density, lty = 3)

lines(rep(trueval,2),c(0,100), lty = 2)
text(trueval,85,"true value", pos=4, srt=-90, cex = 0.7)

legend(0.065,100,c("estimate with normal dist", "1-step improvement", "plugin estimate"),
       cex = 0.7, col = c(8,1,1), lty = c(1,1,3))
title("comparing sample distribution of estimates")

In simple words

You could see the 1-step estimator as a sort of blend between two estimators of the population distribution: a parametric estimate $\tilde{f}(x)$, and an empirical estimate $\hat{f}(x)$ (the empirical estimate being a mass distribution with weight 1/n for each data point).

$$\epsilon \tilde{f}(x) + (1-\epsilon) \hat{f}(x)$$

When $\epsilon =1$ you have the parametric estimate which may be biased, but the empirical estimate (when $\epsilon=0$) may be too sparse to correctly describe the true distribution function. Then the 1st order approximation, by using a Taylor approximation, is being used to blend the two together.

The influence functions describe the derivative of the functional as function of $\epsilon$.

  • $\begingroup$ Thanks very much, this is helpful. I'll have to read through this response again tonight to understand more thoroughly (and review theory on Taylor expansion remainder terms). $\endgroup$
    – RobertF
    May 26, 2020 at 20:49
  • $\begingroup$ One initial question: the plot in Panel B of Fisher & Kennedy suggests that the unknown "true" distribution $T(P)$ is stationed at $\epsilon =0$ while $T(\tilde{P})$ might be a smoothed parametric estimation. However in your response you state that the $T(P)$ distribution at $\epsilon =0$ can be seen as an empirical estimate $\hat{f}(x)$ (for example a weighted average) & therefore is not unknown but very much known, and that rather than seeking $\epsilon =0$ as the final destination we're instead hunting for some optimal value of $\epsilon$ between $0$ and $1$, is that correct? $\endgroup$
    – RobertF
    May 26, 2020 at 20:54
  • 1
    $\begingroup$ @RobertF I would have to read that article once more again. But to me it seemed like $\tilde{P}$ is some initial (or even fixed) estimate. And the one-step is a bias correction based on the observations (or you could say based on the experimental distribution). It is not like you are gonna get exactly the value $T(P)$ (and in my example/demonstration the bias remains indefinitely even when you increase the sample size $n\to \infty$ but something like that is also explained in that article, the bias remains related to $P-\tilde{P}$ and you should better use a decent $\tilde{P}$). $\endgroup$ May 26, 2020 at 21:31
  • 1
    $\begingroup$ First question: you are correct about my interpretation, but maybe it is wrong. I believe that the article explains it nicely but what is missing (a lot) is a practical example that shows how it works. My answer is how I interpret their work. I indeed believe that they try to make a 1st order approach, not to the true $T(P)$ but to the empirical $T(\hat{P})$. And the reason to do this is because the exact $T(\hat{P})$ doesn't work for small sample sizes or for continuous distributions (when the distribution/density is badly estimated by a finite sample). $\endgroup$ May 26, 2020 at 21:40
  • 1
    $\begingroup$ Second question: I do not know about Gateaux derivatives, but the expression $\epsilon \tilde{f}(x) + (1-\epsilon) \hat{f}(x)$ seems intuitive to me. The integrated squared density/mass would be in simple words (maybe a bit hand waving): $$ \epsilon^2 \int \tilde{f}(x)^2 dx + (1-\epsilon)^2\sum \hat{f}(x)^2$$or$$ \epsilon^2 T(\tilde{f}) + (1-\epsilon)^2\sum \hat{f}(x)^2$$ ie. a sum of an integral over the probability density and a summation over the probability mass. $\endgroup$ May 26, 2020 at 21:51

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