When studying the effect of an exposure $T$ on an outcome $Y$, we can control for measured confounders $X$, but the treatment effect can still be biased by residual confounding due to unmeasured confounders $U$. One approach for detecting this kind of bias is to use a negative control outcome $N$ (e.g. here), which is not caused by $T$ but shares the same confounder $U$. If, after adjusting for confounders $X$ using a regression model (for example), there a non-zero effect of $T$ on $N$, then we know there is residual confounding.

This approach can detect confounding, but not necessarily estimate the magnitude of confounding bias, as explained in eAppendix 3 of this paper. The issue is that the effect of $U$ on $N$ is not necessarily equal to the effect of $U$ on $Y$.

However, if one views $N$ as a surrogate for unmeasured confounding, then why can we not simply add it as a covariate in our original regression model? That is, why can we not adjust our estimates of the effect of $T$ on $Y$ by adding $N$ to our measured confounders $X$?

I usually avoid adjusting for any post-treatment variable as it could lead to collider bias, but by assumption, $Y$ is not associated with $N$, so this is not an issue. That said, there could be M-bias. Is that the reason why this should be avoided?

  • $\begingroup$ Off-hand, I'd say that as $N$ isn't related to $T,$ it shouldn't (theoretically) change your regression coefficients much, if at all. $\endgroup$ Commented Dec 15, 2022 at 19:32
  • $\begingroup$ @AdrianKeister $N$ shares a common cause ($U$) with $T$, so even though it is not causally related, it can be strongly correlated because of the strength of the unmeasured confounder. $\endgroup$
    – user310374
    Commented Dec 15, 2022 at 19:35
  • $\begingroup$ But it won't shut off the backdoor path; you can always try simulating this experience with some known example. $\endgroup$ Commented Dec 15, 2022 at 19:37

1 Answer 1


That's an interesting thought, but naively, conditioning on $N$ doesn't closes the backdoor path $T \leftarrow U \rightarrow Y$. See the orange path in this DAG: enter image description here

It can be shown using a fairly simple simulation based on this DAG:

import numpy as np
import pandas as pd
import statsmodels.formula.api as smf

def generate_data(seed, M=10000):
    rng = np.random.default_rng(seed)

    x = rng.normal(size=M)
    u = rng.normal(size=M)

    t_logit = 1 + 0.5*x + 1*u + rng.normal(size=M)
    t_propensity = 1 / (1 + np.exp(-t_logit))
    t = rng.binomial(1, t_propensity)

    y = 1 + 2*t + 0.5*x + 1*u + rng.normal(size=M)

    n = 1 + 0.5*x + 1*u + rng.normal(size=M) # not affected by t

    data = pd.DataFrame({
        "x": x, "u": u, "t": t,
        "y": y, "n": n,
    # data.head()
    return data

Which running a 100 times:

res = []
for i in range(100):
    data = generate_data(i)
    correct = smf.ols("y ~ 1 + t + x + u", data=data).fit().params['t']
    misspecified = smf.ols("y ~ 1 + t + x", data=data).fit().params['t']
    nc_adjusted = smf.ols("y ~ 1 + t + x + n", data=data).fit().params['t']
    res.append((correct, misspecified, nc_adjusted))

res = pd.DataFrame(res, columns=["correct", "misspecified", "neg_control_adjusted"])

Shows the estimates adjusting for negative controls is still biased: enter image description here

However, if you play around with the correlation between $U$ and $N$ you are correct to assume that the stronger the correlation is the smaller the bias is (even bounded by the model with $U$ to the model without $U$) because $N$ becomes $U$ from information perspective: enter image description here

Which is in line with the line of work exploring the adjustment for proxies of confounders (which I recommend).

  • $\begingroup$ This is very helpful, @ehudk. Can you think of the potential risks of this approach? Adjusting for $U$ cannot lead to collider bias, because it is not downstream from $T$, but is there some other type of bias I would risk introducing by adjusting for $N$? $\endgroup$
    – user310374
    Commented Jan 9, 2023 at 2:49
  • $\begingroup$ You have to be very certain $N$ is not affected by $T$ because otherwise you might introduce post-treatment adjustment bias. However, my main concern with this approach (personally) is temporality issues. (Negative) outcomes occur during follow-up and are not available during baseline. This may hinder generalizability or usability of conclusions (think of a physician needing to enter patient's details into some risk calculator during meeting, they won't have access to future events [outcomes]) $\endgroup$
    – ehudk
    Commented Jan 14, 2023 at 13:02

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.