I know that correlation does not imply causality but does an absence of correlation imply absence of causality?

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    $\begingroup$ To quote Andrew Gelman, "Correlation does not even imply correlation." $\endgroup$
    – user78229
    Commented Jul 3, 2016 at 14:55
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    $\begingroup$ No. A can be the cause of B, but only affect it nonlinearly. $\endgroup$
    – Neil G
    Commented Jul 3, 2016 at 15:03
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    $\begingroup$ "Correlation correlates with causation. (Just not very much.)" $\endgroup$
    – Adrian
    Commented Jul 3, 2016 at 15:04
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    $\begingroup$ Please look at this page for the contrapositive. If causality does not imply correlation, then no correlation does not imply no causality. $\endgroup$
    – EdM
    Commented Jul 3, 2016 at 15:27
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    $\begingroup$ While it's a good start to flag that correlation doesn't imply causation, and then discuss details, I've long thought why single out correlation? I put it down to assonance, and the idea attractive to teachers (me too) that students with some effort can remember a slogan and use it in their thinking. But truth is, not much in statistics implies causation. Otherwise put, this warning often comes in the correlation chapter or the correlation lecture, but it belongs everywhere. $\endgroup$
    – Nick Cox
    Commented Jul 6, 2016 at 13:11

6 Answers 6


does an absence of correlation imply absence of causality?

No. Any controlled system is a counterexample.

Without causal relationships control is clearly impossible, but successful control means - roughly speaking - that some quantity is being maintained constant, which implies it won't be correlated with anything, including whatever things are causing it to be constant.

So in this situation, concluding no causal relationship from lack of correlation would be a mistake.

Here's a somewhat topical example.

  • $\begingroup$ An intuitive way to think about it $\endgroup$
    – Repmat
    Commented Jul 3, 2016 at 17:26
  • $\begingroup$ +1, interesting take. However, it seems to imply that causation could be present while correlation of any kind is absent. That can't be true. If some event causes another there'll be some" kind of correlation present, tht _constant which you mentioned will be in the form of nonlinear correlation $\endgroup$
    – Aksakal
    Commented Jul 5, 2016 at 13:16
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    $\begingroup$ +1 Bravo! When I saw the question title in the side bar, I was all "This needs answering from a systems perspective." You nailed it. $\endgroup$
    – Alexis
    Commented Jul 5, 2016 at 17:32
  • $\begingroup$ If from an absense of correlation one removes the causality will the remaining functioning be candidate to label "casuality"? $\endgroup$
    – ttnphns
    Commented Jul 7, 2016 at 1:55
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    $\begingroup$ Not sure I'm understanding @ttnphns's question, but I think the answer is: if you snap the brake cable (or detach the accelerator pedal) then hills will indeed start to show their causal impact on the speed of a car. $\endgroup$ Commented Jul 7, 2016 at 23:58

No. Mainly because by correlation you most likely mean linear correlation. Two variables can be correlated nonlinearly, and may show no linear correlation. It's easy to construct an example like that, but I'll give you an example which is closer to your (narrower) question.

Let's look at the random variable $x$, and the non random function $f(x)=x^2$, with which we create a random variable $y=f(x)$. The latter is clearly caused by the former variable, not just correlated. Let's draw a scatter plot:

enter image description here

Nice, clear nonlinear correlation picture, but in this case it's also direct causality. However, the linear correlation coefficient is non significant, i.e. there's no linear correlation despite obvious nonlinear correlation, and even causality:

>> x=randn(100,1);
>> y=x.^2;
>> scatter(x,y)
>> [rho,pval]=corr(x,y)

rho =


pval =


UPDATE: @Kodiologist is right in the comment. It can be shown mathematically that linear correlation coefficient for these two variables is zero indeed. In my example $x$ is the standard normal variable, so we have the following: $$E[x]=0$$ $$E[x^2]=1$$ $$E[x\cdot x^2]=E[x^3]=0$$ Hence, the covariance (and subsequently the correlation) is zero: $$Cov[x,x^2]=E[x \cdot x^2]-E[x]E[x^2]=0$$

We'd get the same result for any symmetrical distribution, such as uniform $U[-1,1]$.

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    $\begingroup$ Non-significance does not imply truth of the null hypothesis. What's important in your example is that the population correlation coefficient is 0. $\endgroup$ Commented Jul 3, 2016 at 15:28
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    $\begingroup$ Why do you believe the OP means linear correlation? $\endgroup$ Commented Jul 4, 2016 at 23:21
  • $\begingroup$ @immibis, because causation must result in some kind of nonlinear correlation. $\endgroup$
    – Aksakal
    Commented Jul 4, 2016 at 23:26
  • $\begingroup$ Why is the correlation zero? The covariance is $E[X^3] - E[X^2]E[X]$, and in general for a random variable $X$ then $E[X^3] \neq E[X^2]E[X]$.. It holds for $X$ standard normal though $\endgroup$
    – Ant
    Commented Jul 6, 2016 at 7:53
  • $\begingroup$ @Ant, I'm using standard normal for $x$ in the MATLAB example. I updated my post to make it clear. Thanks for pointing it out. $\endgroup$
    – Aksakal
    Commented Jul 6, 2016 at 13:00

No. In particular, random variables can be dependent but uncorrelated.

Here's an example. Suppose I have a machine that takes a single input $x ∈ [-1, 1]$ and produces a random number $Y$, which is equal to either $x$ or $-x$ with equal probability. Clearly $x$ causes $Y$. Now let $X$ be a random variable uniformly distributed on $[-1, 1]$ and select $Y$ with $x = X$, inducing a joint distribution on $(X, Y)$. $X$ and $Y$ are dependent, since

$$ P(X < -\tfrac{1}{2})P(|Y| < \tfrac{1}{2}) = \tfrac{1}{4} \cdot \tfrac{1}{2} = \tfrac{1}{8} ≠ 0 = P(X < -\tfrac{1}{2},\; |Y| < \tfrac{1}{2}). $$

However, the correlation of $X$ and $Y$ is 0, because

$$ \operatorname{Corr}(X, Y) = \frac{\operatorname{Cov}(X, Y)}{σ_Xσ_Y} = \frac{E[XY] - E[X]E[Y]}{σ_Xσ_Y} = \frac{0 - 0\cdot0}{σ_Xσ_Y} = 0. $$

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    $\begingroup$ Actually, this is a bad example in my opinion. X doesn't cause Y. A binary variable absent of the model PresenceOfX is the actual cause with a correlation of 1. What you prove is actually that the value of X doesn't influence Y. $\endgroup$ Commented Jul 3, 2016 at 15:21
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    $\begingroup$ I am really at a loss for how you could feel that the choice of $x$ does not cause $Y$. Perhaps you should specify what you mean by "cause". $\endgroup$ Commented Jul 3, 2016 at 15:24
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    $\begingroup$ @user2088176 Here's a quick proof that the choice of $x$ causes $Y$. Let's use a counterfactual model of causation, in which $x$ is an index into a set of possible distributions for $Y$. If $x = \frac{1}{2}$, then $Y$ is $\frac{1}{2}$ or $-\frac{1}{2}$ with equal probability. If $x = \frac{3}{4}$, then $Y$ is $\frac{3}{4}$ or $-\frac{3}{4}$ with equal probability. Since counterfactuals distinguished by the value of $x$ imply distinct distributions for $Y$, the choice of $x$ causes $Y$. $\endgroup$ Commented Jul 3, 2016 at 15:44
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    $\begingroup$ This example would perhaps be simpler (and still work) if we limit $x$ to $[0,1]$. $\endgroup$
    – JiK
    Commented Jul 4, 2016 at 7:49
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    $\begingroup$ What about the simple and standard example: $X \sim \mathcal N(0,1)$ and $X^2$. They are uncorrelated but the $\chi^2(1)$-distributed $X^2$ is perfectly dependent on $X$. $\endgroup$
    – Therkel
    Commented Jul 4, 2016 at 8:17

Maybe looking at it from a computational perspective will help.

As a concrete example, take a pseudorandom number generator.

Is there a causal relationship between the seed you set and the $k^\text{th}$ output from the generator?

Is there any measurable correlation?


The better answer to the question is that correlation is a statistical, mathematical, and/or physical relationship while causation is a metaphysical relationship. You can't LOGICALLY get from correlation (or non-correlation) to causation, without a (large) set of assumptions binding the metaphysics to the physics. (One example is that what two people might agree to be "a rational observer" is to a large degree arbitrary and probably ambiguous). If A pays B to do C which results in D, what is D's cause? There's simply no rational reason to choose C or B or A (or any of A's precursor events). Control theory deals with systems in realms where they are under control. One way to get a dependent variable under control is to reduce the response of that variable to the possible range of (controlled) variation of the independent variable to statistical noise. For instance, we know air pressure correlates to health (just try breathing vacuum), but if we control air pressure to 1 +/-0.001 atm, how likely is ANY variation of air pressure to effect health?

  • $\begingroup$ The distinction you're after is 'observed in a sample' (correlation) vs dependency that exists whether or not it is observed in a sample (physics). There's no role for metaphysics in this explanation (though some for physical assumption). Springs have elastic limits whether or not they ever reach them. Or in a more homely example: a sugar cube is soluble - a clearly causal concept implying, roughly, that if you drop it in tea it will dissolve. But this causal property is purely due to its physical structure. Sugar cubes would be soluble even if we never thought to dissolve any of them. $\endgroup$ Commented Jul 20, 2016 at 13:41
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    $\begingroup$ You are correct, of course, that without causal assumptions into an argument, you don't get causal conclusions out of it. But there's really nothing very metaphysical about that! $\endgroup$ Commented Jul 20, 2016 at 13:42
  • $\begingroup$ fwiw the counterfactual theory of causality (e.g. Pearl or Woodward) is exactly designed to make sense of "If A pays B to do C which results in D, what is D's cause? There's simply no rational reason to choose C or B or A". The only old fashioned notion and unhelpful notion that these theories put to rest is that we can always make sene of the idea that there is the cause of something. Of course there isn't. $\endgroup$ Commented Jul 20, 2016 at 13:45

Yes, contrary to previous replies. I'm going to take the question as nontechnical, particularly the definition of "correlation". Maybe I'm using it too broadly, but see my second bullet. I hope it will be considered appropriate to discuss other answers here, because they illuminate different portions of the question. I'm drawing on Pearl's approach to causation, and in particular my take on it in some papers with Kevin Korb. Woodward probably has the clearest nontechnical account.

  • @conjugateprior says "any controlled system is a counterexample". Yes, to the stronger claim that noncorrelation observed in your experiment implies no causation. I'm going to assume the question is more general. Certainly one experiment might have failed to control for masking causes, or inappropriately controlled for common effects, and hidden the correlation. But if $x$ causes $y$, there will be a controlled experiment where that relationship is revealed. Almost all definitions or accounts of causation treat it as a difference that makes a difference. Therefore no causation without (some kind of) correlation. If there is a direct link $x \rightarrow y$ in a causal Bayesian network, it does not mean that $x$ always makes a difference to $y$, only that there is some experiment fixing all other causes of $y$ where wiggling $x$ wiggles $y$.

  • @aksakal has a great example why linear causation is insufficient. Agreed, but I want to be broad and nontechnical. If $y=x^2$, it's incomplete to tell a client that $y$ is uncorrelated with $x$. So I'll use correlation very broadly to mean a difference in $x$ that is reliably associated with a difference in $y$. It can be as nonlinear or nonparametric as you like. Threshold effects are fine ($x$ makes a difference to $y$, but only over a finite range, or only by being larger or smaller than a particular value, like voltage in digital circuits).

  • @Kodiologist creates an example where $y = \mathrm{Unif}({x,-x})$, so $|y| = |x|$ but no linear correlation. But clearly there is a discoverable relationship, so correlated in the broad sense.

  • @Szabolcs uses random number generators to show an output stream constructed to appear uncorrelated. Like the digits of $\pi$, the stream appears random but is deterministic. I agree you're unlikely to find the relationship if given only the data, but it's there.

  • @Li Zhi notes you can't logically jump from correlation to causation. Yes, no causes in, no causes out. But the question begins from causation: does it imply correlation? In the air pressure example, we have a threshold effect. There is a range where air pressure is uncorrelated with health. Indeed plausibly where it has no causal effect on health. But there is a range where it does. That is sufficient. But probably better to note ranges where there is and is not an effect. If $A \rightarrow B \rightarrow C \rightarrow D$, then there is correlation all along the chain, because there is causation. Repeated observation (or experiment) can show that $A$ does not directly cause $D$ but the correlation is there because there is a causal story.

I do not know what @user2088176 had in mind, but I think if we take the question very generally, then the answer is yes. At least I think that's the answer required of the causal discovery literature and the interventionist account of causation. Causes are differences that make a difference. And that difference will be revealed, in some experiment, as persistent association.

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    $\begingroup$ I was hoping to approach this from a simpler perspective and non-technical, as you have. What does "cause" mean? Presumably it involves change in something leading to a change in something else. I cannot fathom causality without some kind of correlation. $\endgroup$
    – Behacad
    Commented Jul 13, 2016 at 0:17
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    $\begingroup$ @Behacad I think the contrast is between some kind of correlation (the sort of thing you can observe) and some kind of dependency (which may never be triggered). There are untriggered dependencies but there are no unobserved correlations. This is why causation has a counterfactual element to its definition, whereas correlation does not. $\endgroup$ Commented Jul 15, 2016 at 23:51

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