# Does causation imply correlation?

Correlation does not imply causation, as there could be many explanations for the correlation. But does causation imply correlation? Intuitively, I would think that the presence of causation means there is necessarily some correlation. But my intuition has not always served me well in statistics. Does causation imply correlation?

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Problem is, if you look up "imply" in a dictionary you'll see both "suggest" and "necessitate." –  rolando2 Feb 1 at 23:54

As many of the answers above have stated, causation does not imply linear correlation. Since a lot of the correlation concepts come from fields that rely heavily on linear statistics, usually correlation is seen as equal to linear correlation. The wikipedia article is an alright source for this, I really like this image:

Look at some of the figures in the bottom row, for instance the parabola-ish shape in the 3rd example. This is kind of what happens in @StasK answer (with a little bit of noise added). Y can be fulled caused by X but if the numeric relationship is not linear and symmetric, you will still have a correlation of 0.

The word you are looking for is mutual information: this is sort of the general non-linear version of correlation. In that case, your statement would be true: causation implies high mutual information.

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It's usually but not always true that high mutual information accompanies causation. See @gung's answer where "if the cause is perfectly correlated with another causal variable with exactly the opposite effect." –  Neil G Apr 12 '12 at 21:02
The argument of two causes with opposite effects that always cancels each other doesn't make much sense to me as a cause. I can always assume there are unicorns causing something, and gremlins cancelling their efforts perfectly; I avoid this since it's silly. But maybe I am misunderstanding your point. –  Artem Kaznatcheev Apr 12 '12 at 22:41
His example is more extreme than it needs to be. It's possible for you to have Boolean variables $A, B$, and $C$ such that $A$ and $B$ are causes of $C$, and $C = A + B$ (mod 2). Then, absent of knowledge of $B$, $A$ and $C$ have no mutual information. $B$ is an undiscovered confounder — what you are calling "gremlins" even though it is something very common. –  Neil G Apr 12 '12 at 23:15
@NielG I agree with your first sentence, but not the second. Just because A & B causes C, doesn't mean that A causes C and B causes C. I don't see why cause has to be distributive over &. –  Artem Kaznatcheev Apr 13 '12 at 21:45
The reason that A is nevertheless a cause of C is because changing A will still change C. So, C is dependent on A even when we don't observe B. –  Neil G Apr 13 '12 at 21:51

The strict answer is "no, causation does not necessarily imply correlation".

Consider $X\sim N(0,1)$ and $Y=X^2\sim\chi^2_1$. Causation does not get any stronger: $X$ determines $Y$. Yet, correlation between $X$ and $Y$ is 0. Proof: The (joint) moments of these variables are: $E[X]=0$; $E[Y]=E[X^2]=1$; $${\rm Cov}[X,Y]=E[ (X-0)(Y-1) ] = E[XY]-E[X]1 = E[X^3]-E[X]=0$$ using the property of the standard normal distribution that its odd moments are all equal to zero (can be easily derived from its moment-generating-function, say). Hence, the correlation is equal to zero.

To address some of the comments: the only reason this argument works is because the distribution of $X$ is centered at zero, and is symmetric around 0. In fact, any other distribution with these properties that would have sufficient number of moments would have worked in place of $N(0,1)$, e.g., uniform on $(-10,10)$ or Laplace $\sim \exp(-|x|)$. An oversimplified argument is that for every positive value of $X$, there is an equally likely negative value of $X$ of the same magnitude, so when you square the $X$, you can't say that greater values of $X$ are associated with greater or smaller values of $Y$. However, if you take say $X\sim N(3,1)$, then $E[X]=3$, $E[Y]=E[X^2]=10$, $E[X^3]=36$, and ${\rm Cov}[X,Y]=E[XY]-E[X]E[Y]=36-30=6\neq0$. This makes perfect sense: for each value of $X$ below zero, there is a far more likely value of $-X$ which is above zero, so larger values of $X$ are associated with larger values of $Y$. (The latter has a non-central $\chi^2$ distribution; you can pull the variance from the Wikipedia page and compute the correlation if you are interested.)

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@DQdlM: The standard random variable has vanishing odd central moments, due to the evenness of the density. Matthew: The answer is no, as StasK has demonstrated, because correlation is not the only type of dependence. –  Emre Apr 12 '12 at 1:15
@DQdlM: see the bottom middle graph in the first image on the Wikipedia correlation page. That's StasK's case. It only works when x is equally distributed about the origin (ie, if $X\sim N(3,1)$, correlation will be fairly high) –  naught101 Apr 12 '12 at 1:17
It should also be noted that this answer only applies to linear correlation/covariance. There are other measures of correlation that are non-linear, such as distance correlation, and you can see from that page, that StatK's answer won't apply (covariance is non-zero for the $Y=X^2$ case). –  naught101 Apr 12 '12 at 1:28
Won't any distribution with first and third moments of zero work (no symmetry needed)? –  cardinal Apr 12 '12 at 20:28
@cardinal: yeah, I guess we all learned these kinds of simple counterexamples in grad school... and yes, from the derivation of the covariance, you only need the first and the third moments to be zero. If you have a non-trivial example of an asymmetric distribution that has a zero third moment (finely tuned probability masses over five or six points do not count), I'd be very curious to see it though. –  StasK Apr 12 '12 at 20:42

Essentially, yes.

Correlation does not imply causation because there could be other explanations for a correlation beyond cause. But in order for A to be a cause of B they must be associated in some way. Meaning there is a correlation between them - though that correlation does not necessarily need to be linear.

As some of the commenters have suggested, it's likely more appropriate to use a term like 'dependence' or 'association' rather than correlation. Though as I've mentioned in the comments, I've seen "correlation does not mean causation" in response to analysis far beyond simple linear correlation, and so for the purposes of the saying, I've essentially extended "correlation" to any association between A and B.

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I tend to reserve the word correlation for linear correlation, and use dependence for nonlinear relations that may or may not have linear correlation. –  Memming Apr 11 '12 at 20:34
@Memming I would too, save for the fact that people trot out "Correlation does not imply causation" re: fairly complex non-linear association. –  EpiGrad Apr 11 '12 at 20:36
Memming is right. You need to define correlation if you don't mean Pearson correlation. –  Neil G Apr 11 '12 at 20:52
@NeilG Or for that matter, one may be able to get a linear Pearson correlation by transforming one variable or the other. The problem is the adage itself is over-simplified. –  EpiGrad Apr 11 '12 at 20:59
@EpiGrad: Both good points. In common parlance, correlation is just more of A coincides with more B. I think your answer would benefit from making your use of a broad definition of correlation clear. –  Neil G Apr 11 '12 at 21:01

Adding to @EpiGrad 's answer. I think, for a lot of people, "correlation" will imply "linear correlation". And the concept of nonlinear correlation might not be intuitive.

So, I would say "no they don't have to be correlated but they do have to be related". We are agreeing on the substance, but disagreeing on the best way to get the substance across.

One example of such a causation (at least people think it's causal) is that between the likelihood of answering your phone and income. It is known that people at both ends of the income spectrum are less likely to answer their phones than people in the middle. It is thought that the causal pattern is different for the poor (e.g. avoid bill collectors) and rich (e.g. avoid people asking for donations).

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