Set-up

In the following discussion, we are interested about the causal effect of an action $A$ (or treatment, in biomedical literatures) on an outcome $Y$.

Define the action space to be $\mathcal{A}$ which can be discrete or continuous. Correspoondingly, define a measure $\mu$ on $\mathcal{A}$, e.g. the counting measure if $\mathcal{A}$ is finite or Lebesgue measure if $\mathcal{A}$ is continuous. Let $Y(a)$ denote the real-valued counterfactual outcome that would be observed if the action was set to $a\in\mathcal{A}.$ Then the observed outcome $Y$ is consistent with the actual action $A$, i.e. $Y=Y(A).$ Furthermore, let $X\in\mathcal{X}\subseteq\mathbb{R}^d$ be a collection of observed covariates.

Generalized Average Causal Effect

Definition 1 (Generalized average causal effect, GACE). Given a contrast function $\pi:\mathcal{A}\times\mathcal{X}\to\mathbb{R},$ the corresponding GACE is defined as

$$J := \mathbb{E}\left[\int_\mathcal{A}Y(a)\pi(a|X)\mathrm{d}\mu(a)\right]. \tag{1}$$

Some illustrative examples are given below.

  • (Counterfactual mean). Let $\mathcal{A}$ be discrete and finite with counting measure $\mu,$ and define $\pi(a|x)=\mathbb{1}{\lbrace a=a_0\rbrace}.$ Then the counterfactual mean can be derived as

    $$\mathbb{E}[Y(a_0)] = \sum_{a\in\mathcal{A}}\mathbb{E}\left[Y(a)\mathbb{1}{\lbrace a=a_0\rbrace}\right] =: J_{a_0} $$

  • (Average treatment effect, ATE). Let $\mathcal{A}=\lbrace 0,1\rbrace.$ If we are interested in the effect of treatment $A=1$ compared with control $A=0,$ we can define $\pi(a\vert x)=2a-1,$ then we can obtain the average treatment effect $J=\mathbb{E}[Y(1) - Y(0)].$
  • (Policy evaluation). If $\pi(a\vert x)$ is a density on $\mathcal{A}$ for each $x\in\mathcal{X}$ with respect to $\mu,$ then $J$ is the average outcome when we follow a policy that assigns an action drawn from $\pi(\cdot\vert X)$ to an individual with covariate $X.$

In real scenario, the observed covariates $X$ are not sufficient to account for confounding. Hence, assume there exist additional unmeasured confounders $U\in\mathcal{U}\subseteq\mathbb{R}^{p_u},$ so that

\[Y(a)\not\perp A\ \vert\ X,\ \text{but}\ Y(a)\perp A\ \vert\ U,X\ \ \forall a\in\mathcal{A}.\]

We could easily identify $J$ by adjusting for both $X$ and $U$ if they were observed.

Identification of GACE under NUCA

In ideal setting, no unobserved confounding assumption is admitted, i.e. both $X$ and $U$ were observed. We suggest three commonly-used estimators below.

  • (Inverse probability weighting, IPW). Define the generalized propensity score $f(a\vert u,x)$ as the conditional density of the distribution $A\ \vert\ U, X$ relative to the base measure $\mu.$

    $$\phi_\mathrm{IPW}(y,a,u,x;f) = \frac{\pi(a|x)}{f(a|u,x)}y. $$

  • (Regression adjustment). To account for confoundings, covariates $X$ and $U$ can be included in a regression model. We define the regression function as $k_0$ and assume that it is correct, i.e. $k_0(a,u,x)=\mathbb{E}[Y\vert A=a,U=u,X=x].$ The regression-based estimator is obtained by taking the expectation of

    $$\phi_\mathrm{REG}(y,a,u,x;k_0) = \int k_0(a',u,x)\pi(a'|x)\mathrm{d}\mu(a').$$

  • (Doubly robust estimation) Based on the generalized propensity score $f$ and regression function $k_0,$ the dobly robust estimator is obtained by taking the expectation of

    $$\phi_\mathrm{DR}(y,a,u,x;k_0,f) = \frac{\pi(a|x)}{f(a|u,x)}\left(y - k_0(a,u,x)\right) + \int k_0(a',u,x)\pi(a'|x)\mathrm{d}\mu(a'). $$

Lemma 1. If $Y(a)\perp A\ \vert\ U,X$ and $\vert\pi(A\vert X)/f(A\vert X,U)\vert < \infty,$ then

$$J = \mathbb{E}\phi_\mathrm{IPW}(Y,A,U,X;f) = \mathbb{E}\phi_\mathrm{REG}(Y,A,U,X;k_0) = \mathbb{E}\phi_\mathrm{DR}(Y,A,U,X;k_0,f).$$

However, since $U$ are unobserved, we may resort to some alternative methods with extra data or stronger assumptions.

Negative Control

Framework

The negative control framework is proposed to handle the problem of unmeasured confounding. Two additional types of observed variables are introduced in this framework: negative control actions $Z\in\mathcal{Z}\subseteq \mathbb{R}^{p_z}$ and negative control outcomes $W\in\mathcal{W}\subseteq\mathbb{R}^{p_u}.$

The idea of negative control is borrowed from biomedical experiments. The negative actions $Z$ do not directly affect the outcome $Y$, and neither the negative actions $Z$ nor the primary action $A$ can affect the negative outcomes $W.$ Meanwhile, both $Z$ and $W$ are relevant to the unmeasured confounders, hence they can be viewed as proxies for $U$ to some degree. A typical causal DAG for this setting is shown below.

Let $W(a,z)$ and $Y(a,z)$ denote the corresponding counterfactuals with respect to $(a,z)\in\mathcal{A}\times\mathcal{Z},$ then we can formalize the negative control assumptions as follows.

Assumption 1 (Negative controls).

  1. Consistency: $Y=Y(A,Z),\ W=W(A,Z)$;
  2. Negative control actions: $Y(a,z)=Y(a),\ \forall a\in\mathcal{A}$;
  3. Negative control outcomes: $W(a,z)=W,\ \forall a\in\mathcal{A},z\in\mathcal{Z}$;
  4. Latent unconfoundedness: $(A,Z)\perp(Y(a),W)\ \vert\ U,X,\ \forall a\in\mathcal{A}$;
  5. Overlap: $\left\vert\frac{\pi(a\vert x)}{f(a\vert x,u)}\right\vert < \infty, \forall a\in\mathcal{A},x\in\mathcal{X},u\in\mathcal{U}$.

For brevity, we use $O:=(Z,X,W,A,Y)$ to denote the observable variables, since $U$ is unobserved. In general, our observations contains $n$ independent and identically distributed realizations of $O.$

Bridge Functions

We inherit the notations from the NUCA setting, i.e. $f(a\vert u,x)$ is the generalized propensity score and $k_0(a,u,x)=\mathbb{E}[Y\vert A=a,U=u,X=x]$ the regression function. When $U$ is unobserved, neither $f$ nor $k_0$ are estimable. In the negative control framework, two types of bridge functions are proposed for compensation.

Definition 2 (Bridge functions). $h_0$ is called a outcome bridge function, if $h_0 \in L_2(W, A, X)$ and

$$\mathbb{E}[h_0(W,A,X)|A,U,X] = k_0(A,U,X)$$

holds almost surely.

$q_0$ is called an action bridge function, if $\pi q_0\in L_2(Z,A,X)$ and

$$\mathbb{E}[\pi(A|X)q_0(Z,A,X)|A,U,X] = \frac{\pi(A|X)}{f(A|U,X)}$$

holds almost surely.

These bridge functions are not necessarily unique, and we define the classes of bridge functions as follows:

$$\begin{align} \mathcal{H}_0 &= \lbrace h\in L_2(W,A,X):\ \mathbb{E}[Y - h(W,A,X)|A,U,X] = 0\ \ \mathrm{a.s.}\rbrace, \\ \mathcal{Q}_0 &= \left\lbrace q:\ \pi q\in L_2(Z,A,X),\ \mathbb{E}\left[\pi(A|X)(q(Z,A,X) - \frac{1}{f(A|U,X)})|A,U,X\right] = 0\ \ \mathrm{a.s.}\right\rbrace. \end{align}$$

Identification of GACE

With correctly specified bridge functions, the identification of GACE becomes possible. We define an integral operator $\mathcal{T}:L_2(W,A,X)\to L_2(W,X)$ as

\[(\mathcal{T}h)(w,x) = \int h(w,a,x)\pi(a|x)\mathrm{d}\mu(a).\]

Then analogous to the NUCA case, we define the corresponding estimators below:

$$\begin{align} \tilde{\phi}_\mathrm{IPW}(O;q_0) &= \pi(A|X)q_0(Z,A,X)Y,\\ \tilde{\phi}_\mathrm{REG}(O;h_0) &= (\mathcal{T}h_0)(W,X),\\ \tilde{\phi}_\mathrm{DR}(O;h_0,q_0) &= \pi(A|X)q_0(Z,A,X)(Y - h_0(W,A,X)) + (\mathcal{T}h_0)(W,X). \end{align}$$

Lemma 2. Suppose that Assumption 1 holds. For any $h_0\in\mathcal{H}_0$ and $q_0\in\mathcal{Q}_0,$

$$J = \mathbb{E}\tilde{\phi}_\mathrm{IPW}(O;q_0) = \mathbb{E}\tilde{\phi}_\mathrm{REG}(O;h_0) = \mathbb{E}\tilde{\phi}_\mathrm{DR}(O;h_0,q_0).$$

Proof. By assumption 1, we have the latent unconfoundedness $Z \perp(Y,W)\ \vert\ A,U,X.$

For the IPW estimator,

$$\begin{align} \mathbb{E}[\tilde{\phi}_\mathrm{IPW}(O;q_0)\,|\, A,U,X] &= \mathbb{E}[\pi(A|X)q_0(Z,A,X)\,|\, A,U,X]\,\mathbb{E}[Y\,|\, A,U,X]\\ &= \frac{\pi(A|X)}{f(A|U,X)}\mathbb{E}[Y\,|\, A,U,X],\\ \mathbb{E}[\tilde{\phi}_\mathrm{IPW}(O;q_0)\,|\,U,X] &= \int \frac{\pi(a|X)}{f(a|U,X)}\mathbb{E}\left[Y(a)\,|\, U,X\right]f(a|U,X)\mathrm{d}\mu(a)\\ &= \int \pi(a|X)\,\mathbb{E}\left[Y(a)\,|\, U,X\right]\mathrm{d}\mu(a)\\ &= \mathbb{E}\left[\left.\int Y(a)\pi(a|X)\mathrm{d}\mu(a)\right|\, U,X\right],\\ \mathbb{E}[\tilde{\phi}_\mathrm{IPW}(O;q_0)] &= \mathbb{E}\left[\mathbb{E}[\tilde{\phi}_\mathrm{IPW}(O;q_0)\,|\,U,X]\right]\\ &= \mathbb{E}\left[\mathbb{E}\left[\left.\int Y(a)\pi(a|X)\mathrm{d}\mu(a)\right|U,X\right]\right]\\ &= \mathbb{E}\left[\int Y(a)\pi(a|X)\mathrm{d}\mu(a)\right] = J. \end{align}$$

For the regression-based estimator,

$$\begin{align} \mathbb{E}[\tilde{\phi}_\mathrm{REG}(O;h_0)\,|\, U,X] &= \mathbb{E}\left[\left.\int h(W,a,X)\pi(a|X)\mathrm{d}\mu(a)\right| U,X\right]\\ &= \int k_0(a,U,X)\pi(a|X)\mathrm{d}\mu(a)\\ &= \mathbb{E}\left[\left.\int Y(a)\pi(a|X)\mathrm{d}\mu(a)\right|U,X\right],\\ \mathbb{E}[\tilde{\phi}_\mathrm{REG}(O;h_0)] &= \mathbb{E}\left[\mathbb{E}[\tilde{\phi}_\mathrm{REG}(O;h_0)\,|\, U,X]\right]\\ &= \mathbb{E}\left[\mathbb{E}\left[\left.\int Y(a)\pi(a|X)\mathrm{d}\mu(a)\right|U,X\right]\right]\\ &= \mathbb{E}\left[\int Y(a)\pi(a|X)\mathrm{d}\mu(a)\right] = J. \end{align}$$

For the doubly robust estimator, note that $Z\perp W\ \vert\ A,U,X,$ we have

$$\begin{align} \mathbb{E}[\tilde{\phi}_\mathrm{IPW}(O;q_0)\,|\, A,U,X] &= \frac{\pi(A|X)}{f(A,U,X)}\mathbb{E}[Y\,|\, A,U,X]\\ &= \mathbb{E}[\pi(A|X)q_0(Z,A,X)\,|\,A,U,X]\\ &\quad\; \times\mathbb{E}[h_0(W,A,X)\,|\,A,U,X]\\ &= \mathbb{E}[\pi(A|X)q_0(Z,A,X)h_0(W,A,X)\,|\,A,U,X], \end{align}$$

then

$$\mathbb{E}[\tilde{\phi}_\mathrm{IPW}(O;q_0)] - \mathbb{E}[\pi(A|X)q_0(Z,A,X)h_0(W,A,X)] = 0,$$

$$\mathbb{E}[\tilde{\phi}_\mathrm{DR}(O;h_0,q_0)] = \mathbb{E}[\tilde{\phi}_\mathrm{REG}(O;h_0)].$$

Therefore,

$$J = \mathbb{E}\tilde{\phi}_\mathrm{IPW}(O;q_0) = \mathbb{E}\tilde{\phi}_\mathrm{REG}(O;h_0) = \mathbb{E}\tilde{\phi}_\mathrm{DR}(O;h_0,q_0).$$

Remark. The bridge functions $h_0$ and $q_0$ only depends on observed variables. Once we find appropriate bridge functions, we can apply Lemma 2 to estimate the GACE.

Learning Bridge Functions from Observations

In the previous discussion, we find that the GACE is identifiable under the negative control framework even when there exists unobserved confounding. However, the definition of bridge functions $h_0\in\mathcal{H}_0$ and $q_0\in\mathcal{Q}_0$ is dependent on unobserved variables $U.$ To handle this challenge, we need to find some method to learn bridge functions from observable data $O=(Z,X,W,A,Y).$

Lemma 3. Under Assumption 1, any $h_0\in\mathcal{H}_0$ and $q_0\in\mathcal{Q}_0$ satisfy

$$\begin{align} \mathbb{E}[Y - h_0(W,A,X)|Z,A,X] = 0\ \ \mathrm{a.s.}, \\ \mathbb{E}\left[\pi(A|X)(q_0(Z,A,X) - \frac{1}{f(A|W,X)})|W,A,X\right]=0\ \ \mathrm{a.s.}. \end{align}$$

Proof. For the first equation, we have

$$\begin{align} & \mathbb{E}[Y - h_0(W,A,X)|Z,A,X]\\ &= \mathbb{E}[\mathbb{E}[Y - h_0(W,A,X)|Z,A,U,X]|Z,A,X]\\ &= \mathbb{E}[\mathbb{E}[Y - h_0(W,A,X)|A,U,X]|Z,A,X] = 0. \end{align}$$

The first equality holds by the law of total expectation, and the second by the conditional independence $Z \perp(Y,W)\ \vert\ A,U,X$ from latent unconfoundedness in Assumption 1. The third equality follows from the definition of $\mathcal{H}_0.$

For the second equation, we first deal with the generalized propensity score $f(W,X)$:

$$\begin{align} \mathbb{E}\left[\left.\frac{1}{f(A|U,X)}\right|W,A,X\right] &= \int\frac{f(u|W,A,X)}{f(A|u,X)}\mathrm{d}\mu(u)\\ &= \int \frac{f(u,W,A,X)}{f(A|u,X)f(W,A,X)}\mathrm{d}\mu(u)\\ &= \int \frac{f(A|u,W,X)f(u,W,X)}{f(A|u,X)f(W,A,X)}\mathrm{d}\mu(u)\\ &= \int \frac{f(u,W,X)}{f(W,A,X)}\mathrm{d}\mu(u)\\ &= \frac{1}{f(A|W,X)}, \end{align}$$

the fourth equality follows from $W\perp A\ \vert\ U,X.$ Then,

$$\begin{align} &\mathbb{E}\left[\pi(A|X)(q_0(Z,A,X) - \frac{1}{f(A|W,X)})|W,A,X\right] \\ &= \mathbb{E}\left[\pi(A|X)(q_0(Z,A,X) - \frac{1}{f(A|U,X)})|W,A,X\right] \\ &= \mathbb{E}\left[\left.\mathbb{E}\left[\pi(A|X)(q_0(Z,A,X) - \frac{1}{f(A|U,X)})|W,A,U,X\right]\right|W,A,X\right]\\ &= \mathbb{E}\left[\left.\mathbb{E}\left[\pi(A|X)(q_0(Z,A,X) - \frac{1}{f(A|U,X)})|A,U,X\right]\right|W,A,X\right] = 0. \end{align}$$

where the third equality follows from $W\perp Z\ \vert\ A,U,X.$

Remark. From Lemma 3, we can define classes of observed bridge functions as follows:

$$\begin{align} \mathcal{H}_0^{\text{obs}} &= \lbrace h\in L_2(W,A,X):\ \mathbb{E}[Y - h(W,A,X)|Z,A,X] = 0\ \ \mathrm{a.s.}\rbrace,\\ \mathcal{Q}_0^{\text{obs}} &= \left\lbrace q: \pi q\in L_2(Z,A,X),\ \mathbb{E}\left[\pi(A|X)(q(Z,A,X) - \frac{1}{f(A|W,X)})|W,A,X\right]=0\ \ \mathrm{a.s.}\right\rbrace. \end{align}$$

Since $\mathcal{H}_0^{\text{obs}},\mathcal{Q}_0^{\text{obs}}$ are independent from $U,$ they are learnable from observed data. Moreover, we have $\mathcal{H}_0\subseteq\mathcal{H}_0^{\text{obs}},\ \mathcal{Q}_0\subseteq\mathcal{Q}_0^{\text{obs}}.$

Completeness Condition

The construction of $\mathcal{H}_0^{\text{obs}},\mathcal{Q}_0^{\text{obs}}$ make it possible to learn bridge functions from observed data. However, since the true bridge function classes $\mathcal{H}_0, \mathcal{Q}_0$ are subsets of the observed bridge function classes, we cannot ensure that that functions learned from $\mathcal{H}_0^{\text{obs}},\mathcal{Q}_0^{\text{obs}}$ are valid bridge functions. Moreover, the existence of bridge functions, i.e. $\mathcal{H}_0\neq\emptyset,\mathcal{Q}_0\neq\emptyset,$ is not guaranteed. The completeness condition is proposed to deal with these problems.

Assumption 2 (Completeness). For any $g\in L_2(U,A,X),$ $\mathbb{E}[g(U,A,X)\vert Z,A,X] = 0$ almost surely if and only if $g(U,A,X)=0$ almost surely, and $\mathbb{E}[g(U,A,X)\vert W,A,X] = 0$ almost surely if and only if $g(U,A,X)=0$ almost surely.

The completeness condition mean that the negative controls $Z, W$ have sufficient variability relative to the variability of the unobserved confounders $U.$

Theorem 1. Under Assumptions 1 and 2, we have $\mathcal{H}_0=\mathcal{H}_0^{\text{obs}}, \mathcal{Q}_0=\mathcal{Q}_0^{\text{obs}}.$

Proof. In Lemma 3, we have proven that under Assumption 1, for any $h\in L_2(W,A,X)$ and $q$ with $\pi q\in L_2(Z,A,X)$

$$\begin{align} & \mathbb{E}[Y - h(W,A,X)|Z,A,X] = \mathbb{E}[\mathbb{E}[Y - h(W,A,X)|A,U,X]|Z,A,X],\\ & \mathbb{E}\left[\pi(A|X)(q(Z,A,X) - \frac{1}{f(A|W,X)})|W,A,X\right]\\ &= \mathbb{E}\left[\left.\mathbb{E}\left[\pi(A|X)(q(Z,A,X) - \frac{1}{f(A|U,X)})|A,U,X\right]\right|W,A,X\right]. \end{align}$$

By Assumption 2, we have

$$\begin{align} &\mathbb{E}[Y - h(W,A,X)|Z,A,X] = 0 \Leftrightarrow \mathbb{E}[Y - h(W,A,X)|A,U,X] = 0,\\ &\mathbb{E}\left[\pi(A|X)(q(Z,A,X) - \frac{1}{f(A|W,X)})|W,A,X\right] = 0\\ &\Leftrightarrow \mathbb{E}\left[\pi(A|X)(q(Z,A,X) - \frac{1}{f(A|U,X)})|A,U,X\right] = 0. \end{align}$$

Then by definition we have $\mathcal{H}_0=\mathcal{H}_0^{\text{obs}}, \mathcal{Q}_0=\mathcal{Q}_0^{\text{obs}}.$

Theorem 2. Under Assumptions 1, 2 and some regularity conditions, $\mathcal{H}_0\neq\emptyset,\mathcal{Q}_0\neq\emptyset.$ That is, the solutions of following equations exist:

$$\begin{align} \mathbb{E}[Y - h(W,A,X)|Z,A,X] &= 0,\\ \mathbb{E}\left[\pi(A|X)(q(Z,A,X) - \frac{1}{f(A|W,X)})|W,A,X\right] &= 0. \end{align}$$

Remark. If the completeness condition holds, we can ensure that the observed bridge functions learned from $\mathcal{H}_0^{\text{obs}},\mathcal{Q}_0^{\text{obs}}$ are valid. Moreover, the existence of bridge functions are guaranteed with some additional regularity conditions, hence the bridge functions are learnable from observed data.

References

  • Wang Miao, Xu Shi, and Eric Tchetgen Tchetgen. A confounding bridge approach for double negative control inference on causal effects. arXiv preprint arXiv:1808.04945, 2018.
  • Yifan Cui, Hongming Pu, Xu Shi, Wang Miao, and Eric Tchetgen Tchetgen. Semiparametric proximal causal inference. arXiv preprint arXiv:2011.08411, 2020.
  • Nathan Kallus, Xiaojie Mao, Masatoshi Uehara. Causal Inference Under Unmeasured Confounding With Negative Controls: A Minimax Learning Approach. arXiv preprint arXiv:2103.14029, 2022.
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