Two random variables A and B are statistically independent. That means that in the DAG of the process: $(A {\perp\!\!\!\perp} B)$ and of course $P(A|B)=P(A)$. But does that also mean that there's no front-door from B to A?

Because then we should get $P(A|do(B))=P(A)$. So if that's the case, does statistical independence automatically mean lack of causation?


3 Answers 3


So if that's the case, does statistical independence automatically mean lack of causation?

No, and here's a simple counter example with a multivariate normal,

n <- 1e6
a <- 0.2
b <- 0.1
c <- 0.5
z <- rnorm(n)
x <- a*z + sqrt(1-a^2)*rnorm(n)
y <- b*x - c*z + sqrt(1- b^2 - c^2 +2*a*b*c)*rnorm(n)
cor(x, y)

With corresponding graph,

enter image description here

Here we have that $x$ and $y$ are marginally independent (in the multivariate normal case, zero correlation implies independence). This happens because the backdoor path via $z$ exactly cancels out the direct path from $x$ to $y$, that is, $cov(x,y) = b - a*c = 0.1 - 0.1 = 0$. Thus $E[Y|X =x] =E[Y] =0$. Yet, $x$ directly causes $y$, and we have that $E[Y|do(X= x)] = bx$, which is different from $E[Y]=0$.

Associations, interventions and counterfactuals

I think it's important to make some clarifications here regarding associations, interventions and counterfactuals.

Causal models entail statements about the behavior of the system: (i) under passive observations, (ii) under interventions, as well as (iii) counterfactuals. And independence on one level does not necessarily translate to the other.

As the example above shows, we can have no association between $X$ and $Y$, that is, $P(Y|X) = P(Y)$, and still be the case that manipulations on $X$ changes the distribution of $Y$, that is, $P(Y|do(x)) \neq P(Y)$.

Now, we can go one step further. We can have causal models where intervening on $X$ does not change the population distribution of $Y$, but that does not mean lack of counterfactual causation! That is, even though $P(Y|do(x)) = P(Y)$, for every individual their outcome $Y$ would have been different had you changed his $X$. This is precisely the case described by user20160, as well as in my previous answer here.

These three levels make a hierarchy of causal inference tasks, in terms of the information needed to answer queries on each of them.

  • 1
    $\begingroup$ Thank you, that's exactly what I was looking for. So I guess my confusion was caused (no pun intended) from thinking that statistical independence also means D-separation between the two variables. But it only works the other way around, correct? $\endgroup$ Commented Jul 16, 2018 at 7:40
  • $\begingroup$ @user1834069 that's right, d-separation implies independence, but independence does not imply d-separation. These two are examples where the distribution is unfaithful to the graph, and you can see it depends on the choice of parameterization. If we change the parameters, then dependence shows up again. $\endgroup$ Commented Jul 16, 2018 at 16:35
  • $\begingroup$ Nice example. If I remember correctly, this is one of the non-testable assumptions of mining causal data mining from observational data. For linear models in SEM, Pearl's book also mentions that the set of coefficients that result in an unfaithful distribution is of measure 0. $\endgroup$
    – Vimal
    Commented Jul 17, 2018 at 21:05
  • $\begingroup$ For the top answer, it shows a case where cov(X,Y)=0, but there is clearly a causation from X to Y. However, uncorrelated does NOT mean independent. We can only say X and Y is uncorrelated, but we can never say X and Y is independent in that case. Thus, I think independence means lack of causation. $\endgroup$
    – Xingyu Su
    Commented Sep 26, 2023 at 17:22

Suppose we have a lightbulb controlled by two switches. Let $S_1$ and $S_2$ denote the state of the switches, which can be either 0 or 1. Let $L$ denote the state of the lighbulb, which can be either 0 (off) or 1 (on). We set up the circuit such that the lighbulb is on when the two switches are in different states, and off when they're in the same state. So, the circuit implements the exclusive or function: $L = \text{XOR}(S_1, S_2)$.

By construction, $L$ is causally related to $S_1$ and $S_2$. Given any configuration of the system, if we flip one switch, the state of the lightbulb will change.

Now, suppose both switches are actuated independently according to a Bernoulli process, where the probability of being in state 1 is 0.5. So, $p(S_1=1) = p(S_2=1) = 0.5$, and $S_1$ and $S_2$ are independent. In this case, we know from the design of the circuit that $P(L=1) = 0.5$ and, furthermore, $p(L \mid S_1) = p(L \mid S_2) = p(L)$. That is, knowing the state of one switch doesn't tell us anything about whether the lighbulb will be on or off. So $L$ and $S_1$ are independent, as are $L$ and $S_2$.

But, as above, $L$ is causally related to $S_1$ and $S_2$. So, statistical independence does not imply lack of causation.

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    $\begingroup$ user, you are right that this example has causation with lack of dependence, as I explain here stats.stackexchange.com/questions/26300/…, however in this example we also have that $P(L|do(S_1)) = P(L)$, so it does not answer the OP's question directly. $\endgroup$ Commented Jul 15, 2018 at 19:23
  • $\begingroup$ user, question please: what about $p(L|S_1, S_2)$? I.e. is it equal to $p(L)$ as well? I personally think, for any $(v_L, v_1, v_2) \in \{0,1\}^3$, $p(L=v_L|S_1=v_1) = p(L=v_L|S_2=v_2) = 0.5$, but $p(L=v_L|S_1=v_1, S_2=v_2) \in \{0, 1\}$. Am I right? (I see it's not really related, but I want to double check my understanding) $\endgroup$
    – caveman
    Commented May 24, 2019 at 22:40

Based on your question, you can think like this:

$ P(A B) = P(A) P(B)$ when $A$ and $B$ are independent. You can similarly imply

$P(AB)/P(A) = P(B|A) = P(B)$. Also,

$P(AB)/P(B) = P(A|B) = P(A)$.

In this regards, I believe that independence means a lack of causation. However, dependence doesn't necessarily imply causation.

  • 2
    $\begingroup$ I'm asking if $P(AB)=P(A)P(B) $ means that $P(A|do(B))=P(A)$ ? (using the Pearl Do-calculus notation) $\endgroup$ Commented Jul 15, 2018 at 15:12

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