Inverse probability weighting

${\displaystyle {\hat {\mu }}_{a,n}^{IPWE}={\frac {1}{n}}\sum _{i=1}^{n}Y_{i}{\frac {\mathbf {1} _{A_{i}=a}}{{\hat {p}}_{n}(A_{i}|X_{i})}}}$

1. ${\displaystyle \mu _{a}=\mathbb {E} \left[{\frac {\mathbf {1} _{A=a}Y}{p(A|X)}}\right]}$ where ${\displaystyle p(a|x)={\frac {P(A=a,X=x)}{P(X=x)}}}$
2. construct ${\displaystyle {\hat {p}}_{n}(a|x)}$ or ${\displaystyle p(a|x)}$ using any propensity model (often a logistic regression model)
3. ${\displaystyle {\hat {\mu }}_{a,n}^{IPWE}=\sum _{i=1}^{n}{\frac {Y_{i}1_{A_{i}=a}}{n{\hat {p}}_{n}(A_{i}|X_{i})}}}$

With the mean of each treatment group computed, a statistical t-test or ANOVA test can be used to judge difference between group means and determine statistical significance of treatment effect.

Recall the joint probability model ${\displaystyle (X,A,Y)\sim P}$ for the covariate ${\displaystyle X}$, action ${\displaystyle A}$, and response ${\displaystyle Y}$. If ${\displaystyle X}$ and ${\displaystyle A}$ are known as ${\displaystyle x}$ and ${\displaystyle a}$, respectively, then the response ${\displaystyle Y(X=x,A=a)=Y(x,a)}$ has the distribution

${\displaystyle {\begin{aligned}Y(x,a)\sim {\frac {P(x,a,\cdot )}{\int P(x,a,y)\,dy}}.\end{aligned}}}$

We make the following assumptions.

• (A1) Consistency: ${\displaystyle Y=Y^{*}(A)}$
• (A2) No unmeasured confounders: ${\displaystyle \{Y^{*}(0),Y^{*}(1)\}\perp A|X}$. More formally, for each bounded and measurable functions ${\displaystyle f}$ and ${\displaystyle g}$,
  $${\displaystyle {\begin{aligned}\qquad \mathbb {E} _{(A,Y)}\left[f(Y(X,a))\,g(A)\,|\,X\right]=\mathbb {E} _{Y}\left[f(Y(X,a))\,|\,X\right]\,\mathbb {E} _{A}\left[g(A)\,|\,X\right].\end{aligned}}}$$

  
  This means that treatment assignment is based solely on covariate data and independent of potential outcomes.
• (A3) Positivity: ${\displaystyle P(A=a|X=x)=\mathbb {E} _{A}[\mathbf {1} (A=a)\,|\,X=x]>0}$ for all ${\displaystyle a}$ and ${\displaystyle x}$.

Under the assumptions (A1)-(A3), we will derive the following identities

${\displaystyle {\begin{aligned}\mathbb {E} \left[Y^{*}(a)\right]=\mathbb {E} _{(X,Y)}\left[Y(X,a)\right]=\mathbb {E} _{(X,A,Y)}\left[{\frac {Y\mathbf {1} (A=a)}{P(A=a|X)}}\right].\qquad \cdots \cdots (*)\end{aligned}}}$

The first equality follows from the definition and (A1). For the second equality, first use the iterated expectation to write

${\displaystyle {\begin{aligned}\mathbb {E} _{(X,Y)}\left[Y(X,a)\right]=\mathbb {E} _{X}\left[\mathbb {E} _{Y}\left[Y(X,a)\,|\,X\right]\right].\end{aligned}}}$

By (A3), ${\displaystyle \mathbb {E} _{A}[\mathbf {1} (A=a)\,|\,X]>0}$ almost surely. Then using (A2), note that

${\displaystyle {\begin{aligned}\mathbb {E} _{Y}\left[Y(X,a)\,|\,X\right]&={\frac {\mathbb {E} _{Y}\left[Y(X,a)\,|\,X\right]\,\mathbb {E} _{A}[\mathbf {1} (A=a)\,|\,X]}{\mathbb {E} _{A}[\mathbf {1} (A=a)\,|\,X]}}\\&={\frac {\mathbb {E} _{(A,Y)}\left[Y(X,a)\mathbf {1} (A=a)\,|\,X\right]}{\mathbb {E} [\mathbf {1} (A=a)\,|\,X]}}\qquad \cdots \cdots (A2)\\&=\mathbb {E} _{(A,Y)}\left[{\frac {Y(X,a)\mathbf {1} (A=a)}{\mathbb {E} [\mathbf {1} (A=a)\,|\,X]}}\,{\bigg |}\,X\right].\qquad \cdots \cdots ({\text{denominator is a function of X}})\end{aligned}}}$

Hence integrating out the last expression with respect to ${\displaystyle X}$ and noting that ${\displaystyle Y(X,a)\mathbf {1} (A=a)=Y(X,A)\mathbf {1} (A=a)=Y\,\mathbf {1} (A=a)}$ almost surely, the second equality in ${\displaystyle (*)}$ follows.

The Inverse Probability Weighted Estimator (IPWE) is known to be unstable if some estimated propensities are too close to 0 or 1. In such instances, the IPWE is dominated by a small number of subjects with large weights. Recently developed smoothed IPW estimators by employing Rao-Blackwellization, however, reduce the variance of IPWE by up to 7-fold and can also protect the augmented inverse probability weighted estimator from model misspecification. [5]

${\displaystyle {\begin{aligned}{\hat {\mu }}_{a,n}^{AIPWE}&={\frac {1}{n}}\sum _{i=1}^{n}{\Biggl (}{\frac {Y_{i}1_{A_{i}=a}}{{\hat {p}}_{n}(A_{i}|X_{i})}}-{\frac {1_{A_{i}=a}-{\hat {p}}_{n}(A_{i}|X_{i})}{{\hat {p}}_{n}(A_{i}|X_{i})}}{\hat {Q}}_{n}(X_{i},a){\Biggr )}\\&={\frac {1}{n}}\sum _{i=1}^{n}{\Biggl (}{\frac {1_{A_{i}=a}}{{\hat {p}}_{n}(A_{i}|X_{i})}}Y_{i}+(1-{\frac {1_{A_{i}=a}}{{\hat {p}}_{n}(A_{i}|X_{i})}}){\hat {Q}}_{n}(X_{i},a){\Biggr )}\\&={\frac {1}{n}}\sum _{i=1}^{n}{\Biggl (}{\hat {Q}}_{n}(X_{i},a){\Biggr )}+{\frac {1}{n}}\sum _{i=1}^{n}{\frac {1_{A_{i}=a}}{{\hat {p}}_{n}(A_{i}|X_{i})}}{\Biggl (}Y_{i}-{\hat {Q}}_{n}(X_{i},a){\Biggr )}\end{aligned}}}$

With the following notations:

1. ${\displaystyle 1_{A_{i}=a}}$ is an indicator function if subject i is part of treatment group a (or not).
2. Construct regression estimator ${\displaystyle {\hat {Q}}_{n}(x,a)}$ to predict outcome ${\displaystyle Y}$ based on covariates ${\displaystyle X}$ and treatment ${\displaystyle A}$, for some subject i. For example, using ordinary least squares regression.
3. Construct propensity (probability) estimate ${\displaystyle {\hat {p}}_{n}(A_{i}|X_{i})}$. For example, using logistic regression.
4. Combine in AIPWE to obtain ${\displaystyle {\hat {\mu }}_{a,n}^{AIPWE}}$

The later rearrangement of the formula helps reveal the underlying idea: our estimator is based on the average predicted outcome using the model (i.e.: ${\displaystyle {\frac {1}{n}}\sum _{i=1}^{n}{\Biggl (}{\hat {Q}}_{n}(X_{i},a){\Biggr )}}$). However, if the model is biased, then the residuals of the model will not be (in the full treatment group a) around 0. We can correct this potential bias by adding the extra term of the average residuals of the model (Q) from the true value of the outcome (Y) (i.e.: ${\displaystyle {\frac {1}{n}}\sum _{i=1}^{n}{\frac {1_{A_{i}=a}}{{\hat {p}}_{n}(A_{i}|X_{i})}}{\Biggl (}Y_{i}-{\hat {Q}}_{n}(X_{i},a){\Biggr )}}$). Because we have missing values of Y, we give weights to inflate the relative importance of each residual (these weights are based on the inverse propensity, a.k.a. probability, of seeing each subject observations) (see page 10 in [7]).

The "doubly robust" benefit of such an estimator comes from the fact that it's sufficient for one of the two models to be correctly specified, for the estimator to be unbiased (either ${\displaystyle {\hat {Q}}_{n}(X_{i},a)}$ or ${\displaystyle {\hat {p}}_{n}(A_{i}|X_{i})}$, or both). This is because if the outcome model is well specified then its residuals will be around 0 (regardless of the weights each residual will get). While if the model is biased, but the weighting model is well specified, then the bias will be well estimated (And corrected for) by the weighted average residuals.[7][8][9]

The bias of the doubly robust estimators is called a second-order bias, and it depends on the product of the difference ${\displaystyle {\frac {1}{{\hat {p}}_{n}(A_{i}|X_{i})}}-{\frac {1}{{p}_{n}(A_{i}|X_{i})}}}$ and the difference ${\displaystyle {\hat {Q}}_{n}(X_{i},a)-Q_{n}(X_{i},a)}$. This property allows us, when having a "large enough" sample size, to lower the overall bias of doubly robust estimators by using machine learning estimators (instead of parametric models).[10]