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Consider the following joint distribution for the random variables $A$ and $B$:

$$ \begin{array} {|r|r|}\hline & B=1 & B=2 \\ \hline A=1 & 49\% & 1\% \\ \hline A=2 & 49\% & 1\% \\ \hline \end{array}$$

Intuitively,

  • if I know A, I can predict very well B (98% accuracy!)
  • but I if know B, I can't say anything about A

Questions:

  • can we say that A causes B?
  • if yes, what is the mathematical way to conclude that A causes B?

thank you! (and apologies for the maybe "naive" question)

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    $\begingroup$ If you don't know A, you can predict B with 98% accuracy as well. It looks like the two are independent. $\endgroup$ – Cris Luengo Aug 24 '20 at 15:02
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    $\begingroup$ If this exercise attemps to illustrate the "correlation is not causation" mantra, it's a bad illustration, because the variables are NOT correlated here - actually they are independent. "if I know A, I can predict very well B (98% accuracy!)" But I you don't know A, you can predict it also, with the same accuracy. $\endgroup$ – leonbloy Aug 24 '20 at 15:09
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    $\begingroup$ @leonbloy This exercise is clearly designed to illustrate that the first two bullet points taken together are not enough to imply causation. Causation is tricky to define. If someone uses those two bullet points to explain what causation is, how quickly would the people around them realize that this is wrong? This exercise is trying to teach students to realize it very quickly. $\endgroup$ – Matt Aug 24 '20 at 16:47
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    $\begingroup$ But the first bullet point ("if I know A, I can predict very well B (98% accuracy!)" is wrong, Here, to know A does not help to predict B. $\endgroup$ – leonbloy Aug 24 '20 at 19:57
  • $\begingroup$ May I introduce you to tylervigen.com/spurious-correlations ? $\endgroup$ – John Doe Aug 24 '20 at 23:14
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can we say that A causes B?

No, this is (presumably) a simple observational study. To infer causation it is necessary (but not necessarily sufficient) to conduct an experiment or a controlled trial.

Just because you are able to make good predictions does not say anything about causality. If I observe the number of people who carry cigarette lighters, this will predict the number of people who have a cancer diagnosis, but it doesn't mean that carrying a lighter causes cancer.


Edit: To address one of the points in the comments:

But now I wonder: can there ever be causation without correlation?

Yes. This can happen in a number of ways. One of the easiest to demonstrate is where the causal relation is not linear. For example:

> X <- 1:20
> Y <- 21*X - X^2
> cor(X,Y)
[1] 0

Clearly Y is caused by X, yet the correlation is zero.

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    $\begingroup$ Thanks @Robert! Since you mentioned correlation: in this case the correlation is zero, because the probabilities of B don't vary when A varies. In other words, you don't even need to know A to predict B. So definitely there is no causation. But now I wonder: can there ever be causation without correlation? 🙂 $\endgroup$ – elemolotiv Aug 24 '20 at 8:50
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    $\begingroup$ Yes, there can be causation without correlation. It is possible that all the individual causal effects balance each other out and result in no average causal effect. Think about the exposure being beneficial in half the population and harmful in the other half. It is possible for these to cancel out and look like there is no correlation (although you could find correlations within the strata if you know those strata do exist) $\endgroup$ – pzivich Aug 24 '20 at 11:14
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    $\begingroup$ +1. The only thing that caught my eye is the line about the necessity of controlled trials/experiments. These probably are necessary in most fields, but they aren't sufficient to prove causation. You also need a theory that describes what is happening. $\endgroup$ – indigochild Aug 24 '20 at 17:57
  • $\begingroup$ @pzivich If causal effects cancel out, then there's no causation. $\endgroup$ – Acccumulation Aug 24 '20 at 18:00
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    $\begingroup$ @indigochild You don't need an explanation to have very strong evidence for causation. $\endgroup$ – Acccumulation Aug 24 '20 at 18:01
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Both of the previous answers are good, but I want to dive into the weeds on this question a little more. So we know that correlation is not causation, but correlation is also not not causation. So when do we get to say that correlation is causation. Unfortunately, the data itself can never tell us this, we can only arrive at this by imposing assumptions on the data.

Simple Example: I am going to use directed acyclic graphs (DAGs) since they graphically encode the assumptions. Let's focus on three variables: $A$, $B$, and $U$ (you can extend this to more, but the basic concepts remain the same). $U$ is some variable we did not have the opportunity to collect. Each arrow in the DAG indicates a causal relationship, with the direction of the arrow indicating what causes what. For three variables (and the ordering restriction), following are some possible DAGs that will result in a correlation between $A$ and $B$:

Examples of directed acyclic graphs consistent with correlation between A and B

Correlation is causation in only DAGs numbered 1, 2, and 3; which requires appealing to outside knowledge (although 3 is tricky since $U$ being a common cause of both $A$ and $B$ can flip the relationship from the true causal direction, e.g. $A$ is protective from $B$ in reality but $U$ makes it look harmful).

One way to determine whether correlation is consistent with causation is if we conducted a randomized experiment. If we did not randomize based on $U$ and $B$ was measured after $A$ was randomized, then we know that an arrow from $U$ to $A$ and $B$ to $A$ are implausible. Therefore, we can say that the correlation is causation. Alternatively, maybe we have some subject matter knowledge on the topic of $A$ and $B$ that says there are no common causes (unlikely in reality but this is only an example), similarly we can say that correlation is causation.

The important part is that the assumptions used to claim correlation is causation are supported by outside knowledge. How and exactly what outside knowledge is needed is an important issue.

Conclusion: There are a variety of frameworks and formal assumptions that can be used to make the claim that a certain correlation is causation. The key part is that the data alone cannot tell you whether a correlation is or isn't causation. Some outside assumptions or procedures must be applied in order to distinguish non-causal correlations from causal correlations.

Aside: As to my example of a scenario with causation but no correlation, DAGs are assumed to be faithful. This basically means that there are no perfect cancellations that occur (all the individual causal effects don't cancel out perfectly to result in no average causal effect). Because of this, it is a little trickier to claim that no correlation means no causation.

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    $\begingroup$ +1 Excellent response. @pzivich. One thing that I'll add is that the OP may want to examine the topics on CV of causal inference: stats.stackexchange.com/search?q=Causal+Inference. $\endgroup$ – StatsStudent Aug 24 '20 at 13:14
  • $\begingroup$ "but correlation is also not not causation" Correlation is correlated with causation. $\endgroup$ – Acccumulation Aug 24 '20 at 18:12
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    $\begingroup$ All answers are repeating the mantra "correlation is not causation". But this question is not about that, here we have ZERO correlation. $\endgroup$ – leonbloy Aug 24 '20 at 19:55
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No, you cannot say A causes B. The table you have only describes associations between A and B. Even if you know A accurately predicted B a large percentage of the time , that does not imply that A causes B. It may in fact, be that A causes some other, confounding variable C to occur that is highly correlated with B.

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    $\begingroup$ " The table you have only describes associations between A and B." No, the table shows A and B has ZERO correlation. $\endgroup$ – leonbloy Aug 24 '20 at 19:56
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    $\begingroup$ The only statement was in comparison to causation. $\endgroup$ – StatsStudent Aug 24 '20 at 21:43
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  1. Prediction means that entropy is reduced. That is, if A predicts B, then the entropy of the distribution of B is greater than the entropy of distribution B conditioned on A.

  2. Prediction is symmetric. If A predicts B, then B predicts A (barring degenerate cases).

  3. Causation is not symmetric. Causation refers to an asymmetric relationship between two events. So it follows that prediction does not mean causation.

  4. In the case that you present, A and B do not predict each other. While the entropy of B given A is low, it's just as low without knowing A.

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