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Suppose we're interested in the effect of $D$ on $Y$. Suppose that variables $D$ and $O$ are mutually dependent on a variable $C$, and that $Y$ is mutually dependent on variables $D$ and $O$. I find it quiet obvious that we could condition on $O$ to observe the desired effect. However, my professor and textbook say that equivalently, we could condition on "C". Personally, I do believe that this is incorrect for the following reason:

If we conditioned on $C$, then theoretically, there would be no variance in $D$, $O$ or $Y$, and so a causal relationship between $D$ and $Y$ could not be determined. But even if there were some variance in $D$ within each strata of $C$, there could very well also be some variance in $O$ within each strata of $C$. If this were the case, then even after having conditioned on $C$, variance in $Y$ could still be attributable to both variance in $D$ and variance in $O$. Hence, the direct causal effect of $D$ on $Y$ cannot be determined from conditioning on $C$.

Assuming that my professor and textbook are correct, where is my analysis lacking?

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First to help everyone see your case with a DAG the below:

enter image description here

To make this problem rigorous, replace "$Y$ mutually depends on $X$" with "$X$ causes $Y$". "Dependence" is a probabilistic phenomenon and is not directed, if cancer depends on smoking, smoking depends on cancer by Bayes' Rule. But smoking causes cancer, and not vice versa.

Also the direct effect (of $D$ on $Y$) is the effect without the presence of mediators. There is no mediation in this DAG, so modeling $P(Y|D)$ would be direct, albeit confounded. If--to correct you again--you mean you wish to control for confounding in the $P(Y | \text{do}(D))$ causal question, then the answer is: you're both right, and your argument is not about confounding, but rather about precision.

The trick to understanding this problem is understanding the backdoor criterion. This is defined in Pearl's Causality. The sets ${C}, {O}, and {C,O}$ all satisfy the backdoor criterion because they block every path from $D$ to $Y$ that has an arrow leading to $D$.

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  • $\begingroup$ So even though there's variance between O and Y, it doesn't change the fact that we can measure the variance between D and Y, and that's actually what we want? $\endgroup$ – David Dec 10 '18 at 18:16
  • $\begingroup$ @DavidS to summarize my answer: yes, conditioning on $C$ will work, and the estimate of covariance between $D, Y$ will be consistent with the covariance you would estimate adjusting for everything. (Think hard about the meaning of causal inference here) $\endgroup$ – AdamO Dec 10 '18 at 18:16
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I will just complement Adam's answer (+1) regarding precision.

As Adam has said, given the description of your problem (as in the DAG), adjusting either for $C$ or for $O$ is enough for identification of the causal effect of $D$ on $Y$, since both sets satisfy the backdoor criterion. What does that mean? It means that, given infinite samples, both adjusted estimates will converge to the causal effect of interest.

With finite samples, however, both sets are not equivalent, and they differ on the precision of your estimates. Among backdoor admissible sets of covariates, in order to get a more precise estimate of the causal effect of $D$ on $Y$, you want to look for two things: (i) you want to maximally reduce the variation of $Y$; (ii) you want to minimally reduce the variation of $D$.

Thus, in finite samples, considering your problem structure, adjusting for $O$ will give you more precise estimates than adjusting for $C$. Here is a simple simulation in R:

gen_data <- function(n = 1e2){
  c <- rnorm(n)
  d <- c + rnorm(n)/2
  o <- c + rnorm(n)
  y <- d + o + rnorm(n)
  data <- data.frame(c,d,o,y)
  data
}

set.seed(10)

summary(replicate(1000, coef(lm(y ~ d + c, gen_data()))["d"]))
#    Min. 1st Qu.  Median    Mean 3rd Qu.    Max. 
# 0.04868 0.82050 0.99792 0.99813 1.18057 2.11129 

summary(replicate(1000, coef(lm(y ~ d + o, gen_data()))["d"]))
#   Min. 1st Qu.  Median    Mean 3rd Qu.    Max. 
# 0.5899  0.9312  1.0066  1.0079  1.0867  1.3466 

In our example, although both estimates are unbiased, you can see the estimates adjusting for $O$ have substantially less variance.

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    $\begingroup$ I did not know about the difference is precision between condioning on O vs C. Great insight, Carlos! $\endgroup$ – ColorStatistics Dec 12 '18 at 21:37

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