Alternative Measures of Direct and Indirect Effects

Consider the following example from [8], where the causal graph is

such that $X$ represents a drug treatment, $Z$ the presence of a certain enzyme in a patient’s blood, and $Y$ recovery. Moreover, we have that

Pearl imagines a scenario where someone proposes developing a cheaper drug that is equal to the existing one except for the lack of effect on enzyme production. To evaluate the new drug’s performance, he computes $TE(X,Y)=0.46$ and $NDE(X,Y)=0.32$, and concludes that the new drug will reduce the probability of recovery by $30\%$, i.e. $1-NDE(X,Y)/TE(X,Y)=0.3$. We can repeat the analysis using $DE(X,Y)$ instead of $NDE(X,Y)$:

which implies that $0.32\leq DE(X,Y)\leq 0.43$. An interval is returned because $p(X)$ is not given in the original example (it is not needed to compute $NDE(X,Y)$ or $NIE(X,Y)$). Therefore, we conclude that the new drug will reduce the probability of recovery by between $7\%$ and $30\%$, depending on $p(X)$.

The new drug development scenario described above corresponds to the following alternative causal graph:

where the edge $X\leftarrow\!\!\!\!\!\multimap Z$ means that there is an edge $X\leftarrow Z$ or $X\leftrightarrow Z$. The former represents that the presence of enzyme may have an effect on the patient taking the treatment, and the latter represents the potential existence of an unmeasured confounder between them. The drug performance in this alternative causal graph is simply $TE(X,Y)$, which can be computed by adjusting for $Z$, and thus it coincides with $DE(X,Y)$ in the original causal graph, since the two graphs are distribution equivalent. In other words, it is $DE(X,Y)$ rather than $NDE(X,Y)$ that should be used to answer the original question. Note that $DE(X,Y)=NDE(X,Y)=0.32$ if and only if $p(x)=0$, i.e. everyone is untreated. This is no coincidence because $NDE(X,Y)$ in the original causal graph coincides with the average effect of the treatment among the untreated in the alternative graph [9],²²2[9] prove the equivalence when $X\leftarrow Z$, but the proof also applies when $X\leftrightarrow Z$. rather than with $TE(X,Y)$ which is the correct answer to the original question.

Pearl also imagines a scenario where someone proposes developing a cheaper drug that is equal to the existing one except for the lack of direct effect on recovery, i.e. it just stimulates enzyme production as much as the existing drug. To evaluate the new drug’s performance, he computes $TE(X,Y)=0.46$ and $NIE(X,Y)=0.04$, and concludes that the new drug will reduce the probability of recovery by $91\%$, i.e. $1-NIE(X,Y)/TE(X,Y)=0.91$.³³3The small disagreements with the results in [8] are due to rounding. We can repeat the analysis using $IE(X,Y)$ instead of $NIE(X,Y)$:

which implies that $0.04\leq IE(X,Y)\leq 0.15$. Therefore, we conclude that the new drug will reduce the probability of recovery by between $67\%$ and $91\%$, depending on $p(X)$.

The latest new drug development scenario corresponds to the following alternative causal graph:

where the edge $X\leftrightarrow Y$ represents the potential existence of an unmeasured treatment-outcome confounder. The drug performance in this alternative causal graph is simply $TE(X,Y)$, which can be computed by the front-door criterion, and thus it coincides with $IE(X,Y)$ in the original causal graph, under our assumption that the direct path $X\rightarrow Y$ in the original graph is mediated by an unmeasured random variable (recall the previous section). In other words, it is $IE(X,Y)$ rather than $NIE(X,Y)$ that should be used to answer the original question. Note that $IE(X,Y)=NIE(X,Y)=0.04$ if and only if $p(x)=0$. Again, this is no coincidence because $NIE(X,Y)$ in the original causal graph coincides with the average effect of the treatment among the untreated in the alternative graph [1, 10], rather than with $TE(X,Y)$ which is the correct answer to the original question.

In this section, we study the following extension of the original causal graph with an unmeasured treatment-outcome confounder $V$.

Now, neither $NDE(X,Y)$ nor $NIE(X,Y)$ nor their total and controlled counterparts are identifiable from the observed data distribution $p(X,Z,Y)$ [1, 2]. However, $IE(X,Y)$ can be computed pretty much like before. First, we add the unmeasured mediator $U$. The edge $V\mathrel{\reflectbox{$\leftarrow\!\!\!\!\!\multimap$}}U$ means that there may be an edge $V\rightarrow U$ or $V\leftrightarrow U$.

Then, we group $U$ and $V$. Note that every probability distribution that is representable by the graph above is representable by the graph below, since all the independencies entailed by the latter hold in the former.

Finally, we apply the front-door criterion.

Like $NDE(X,Y)$ and its total counterpart, $DE(X,Y)$ is not identifiable from the observed data distribution $p(X,Z,Y)$ in the extended causal graph under consideration. However, it may be bounded if $V$ is binary and a binary proxy $W$ of $V$ is measured. The causal graph under consideration is then as follows.

In the literature, there are many cautionary tales about the bias that adjusting for the proxy of an unmeasured confounder introduces to the estimation of a causal effect [11, 12, 13]. For instance, [14] constructs an example where adjusting for the proxy is worse than not adjusting at all. However, there are conditions under which the opposite is true [15, 16, 17, 18]. We use some of these conditions here.

Recall that $DE(X,Y)$ is the part of $TE(X,Y)$ in the causal graph that remains after deactivating the path $X\rightarrow Z\rightarrow Y$. This is achieved by simply adjusting for $Z$:

where $TE(X,Y|z)$ is the average total effect of $X$ on $Y$ in the stratum $Z=z$. Let us define the observed or partially adjusted average total effect of $X$ on $Y$ in the stratum $Z=z$ as

Note that it can be computed from the observed data distribution $p(X,W,Z,Y)$. Rephrasing the results in [16, 17] to our scenario, if $E[Y|x^{\prime},z^{\prime},W]$ and $E[X|z^{\prime},W]$ are one nonincreasing and the other nondecreasing in $W$ for all $x^{\prime}\in\{x,\overline{x}\}$ and $z^{\prime}\in\{z,\overline{z}\}$, then $TE(X,Y|z^{\prime})\geq TE_{obs}(X,Y|z^{\prime})$ for all $z^{\prime}\in\{z,\overline{z}\}$. On the other hand, if $E[Y|x^{\prime},z^{\prime},W]$ and $E[X|z^{\prime},W]$ are both nonincreasing or both nondecreasing in $W$ for all $x^{\prime}\in\{x,\overline{x}\}$ and $z^{\prime}\in\{z,\overline{z}\}$, then $TE_{obs}(X,Y|z^{\prime})\geq TE(X,Y|z^{\prime})$ for all $z^{\prime}\in\{z,\overline{z}\}$. Note that the antecedents of these rules are testable from the observed data distribution $p(X,W,Z,Y)$. Not in all but in many cases, these rules enable us to bound $DE(X,Y)$ and even determine its sign. Specifically,

where $\boldsymbol{1}_{\neq}$ is 1 (respectively, 0) if $E[Y|x^{\prime},z^{\prime},W]$ and $E[X|z^{\prime},W]$ are one nonincreasing and the other nondecreasing (respectively, both nonincreasing or both nondecreasing) in $W$ for all $x^{\prime}\in\{x,\overline{x}\}$ and $z^{\prime}\in\{z,\overline{z}\}$.

This section addresses a problem of randomized controlled trials, namely long-time effect estimation from typically short-lived trials. Consider the following causal graph, where $X$ and $V$ are unmeasured.

We assume that the mediator $Z$ is a short-term effect of the treatment $X$, whereas $Y$ is a long-term effect of $X$. Randomized controlled trials are typically short-lived due to cost considerations and, thus, they are typically conducted to estimate short-term effects but not longer ones. Observational data, on the other hand, is much cheaper to obtain and, thus, they may include long-term outcome observations. Unfortunately, observational data is typically subject to unmeasured confounding, and mismeasurements due to self-reporting. Therefore, we assume that a randomized controlled trial was conducted to estimate $TE(X,Z)$, but not $TE(Z,Y)$ or $TE(X,Y)$. We also assume that the probability distribution $p(W,Z,Y)$ was estimated from observational data, where $W$ represents the self-reported treatment, which may differ from the actual unmeasured treatment $X$. Our goal is computing $TE(X,Y)$. Unfortunately, this cannot be done from the information available. However, the fact that $TE(X,Y)=IE(X,Y)$ implies, as we show below, that $TE(X,Y)$ can be bounded sometimes.

Our setup above is similar to the one by [19] with the difference that they assume no unmeasured confounding. Our setup is also close to the one by [20] with the difference that they consider linear and partial linear structural equation models, while we consider non-parametric models. Moreover, unlike us, these two works assume that the true treatment is available in the observational data.

Provided that $Z$ is binary, we have that

Let us define the observed or partially adjusted average total effect of $Z$ on $Y$ as

Note that it can be computed from the observed data distribution $p(W,Z,Y)$. If $E[Y|z^{\prime},W]$ and $E[Z|W]$ are one nonincreasing and the other nondecreasing in $W$ for all $z^{\prime}\in\{z,\overline{z}\}$, then $TE(Z,Y)\geq TE_{obs}(Z,Y)$ [16, 17]. On the other hand, if $E[Y|z^{\prime},W]$ and $E[Z|W]$ are both nonincreasing or both nondecreasing in $W$ for all $z^{\prime}\in\{z,\overline{z}\}$, then $TE_{obs}(Z,Y)\geq TE(Z,Y)$ [16, 17].⁴⁴4In the proofs of these results, $X$ is a parent of $V$. The results also hold when $X$ is a child of $V$, since (i) every probability distribution that is representable when $X$ is a child of $V$ is also representable when $X$ is a parent of $V$ and vice versa [2], and (ii) $TE(Z,Y)$ and $TE_{obs}(Z,Y)$ give each the same result in both cases. Note that the antecedents of these rules are testable from the observed data distribution $p(W,Z,Y)$. Not in all but in many cases, these rules together with the knowledge of $TE(X,Z)$ enable us to bound $IE(X,Y)$ and even determine its sign. Specifically,

where $\boldsymbol{1}_{\neq}$ is 1 (respectively, 0) if $E[Y|z^{\prime},W]$ and $E[Z|W]$ are one nonincreasing and the other nondecreasing (respectively, both nonincreasing or both nondecreasing) in $W$ for all $z^{\prime}\in\{z,\overline{z}\}$, and $\boldsymbol{1}_{\geq}$ is 1 if $TE(X,Z)\geq 0$ and 0 otherwise.

Finally, note that the last equation also holds if we add the edge $X\rightarrow Y$ to the causal graph under study. To see it, simply pre-process the graph with the three transformations at the beginning of Section 2.2.