Why does collider adjustment in a shielded triplet tend to cause independence

causalitydagmutual informationrstructural-equation-modeling

I created a causal model in which $X$ causes $Y$ and $Z$, and $Y$ causes $Z$ in the following way:

set.seed(2021)
N <- 10000

X <- purrr::rbernoulli(N)
Y <- X + purrr::rbernoulli(N)
Z <- 2*X + 3*Y + purrr::rbernoulli(N)

I created it this way so that the variables are discrete. That's it. The equivalent DAG would be the one below:

People working with causal inference are probably more used to unshielded triplets, in which case we would have no edge between $X$ and $Y$ and therefore a v-structure. In this hypothetical situation, $X$ and $Y$ are independent but become [spuriously] dependent when conditioning on $Z$, a collider.

However, going back to the diagram I showed, there is direct association between $X$ and $Y$ and if we adjust on $Z$, we open a blocked path, in the sense that we add some spurious dependence between $X$ and $Y$ through $Z$. What's driving me nuts is that the mutual information between $X$ and $Y$ not only is larger than $X$ and $Y$ conditioned by $Z$, but the latter is $0$! That is,

$I(X;Y) > I(X;Y|Z) = 0$. In R:

infotheo::condinformation(X,Y)
infotheo::condinformation(X,Y,Z)

I tried changing the equations for $Y$ and $Z$ and yet, the zero is always there (almost always, check the end of the question) for $I(X;Y|Z)$, whenever it's a shielded triplet. Even if I do it with continuous variables and normal distributions for the noise, I still find the same thing.

set.seed(2021)
N <- 10000

X <- rnorm(N, mean=10, sd=2)
Y <- X + rnorm(N, mean=10, sd=2)
Z <- X + Y + rnorm(N, mean=10, sd=2)

miic::discretizeMutual(X,Y, plot=FALSE)$info
miic::discretizeMutual(X,Y, matrix_u=matrix(Z), plot=FALSE)$info

But then, if I change a bit the structural equation for $Z$, I get something different from zero.

set.seed(2021)
N <- 10000

X <- rnorm(N, mean=10, sd=2)
Y <- X + rnorm(N, mean=10, sd=2)
Z <- 2*X + 3*Y + rnorm(N, mean=10, sd=2)

miic::discretizeMutual(X,Y, plot=FALSE)$info
miic::discretizeMutual(X,Y, matrix_u=matrix(Z), plot=FALSE)$info

I also get a value different from zero if I make the distribution of the noise in $Z$ explicitly different from the noise in $X$ and $Y$.

set.seed(2021)
N <- 10000

X <- rnorm(N, mean=10, sd=2)
Y <- X + rnorm(N, mean=10, sd=2)
Z <- X + Y + rnorm(N, mean=100, sd=10)

miic::discretizeMutual(X,Y, plot=FALSE)$info
miic::discretizeMutual(X,Y, matrix_u=matrix(Z), plot=FALSE)$info

I don't understand what's happening here. I tried to draw a few diagrams, see what paths would be blocked or opened, what would happen with correlation between the noises, but I can't think of a way that adjusting for the collider $Z$ would make $X$ and $Y$ independent. A hypothesis would be some sort of cancelling of effects between the two paths, but I changed the equations in ways that I didn't expect it to happen and still… The $0$ is there.

Could you please explain to me what's happening here? Both analytically, if possible, and intuitively. By intuitively, I'm referring to the intuitive explanations for the unshielded triplets. Adjusting for $B$ in $A \rightarrow B \leftarrow C$ makes A and C dependent. Adjusting for $B$ in $A \rightarrow B \rightarrow C$ or $A \leftarrow B \rightarrow C$ make $A$ and $B$ independent. Something along these lines 🙂

Best Answer

Here you can find a formal example of linear causal model that share your DAG. Which OLS assumptions are colliders violating?

I consider (there and here) the case where all noises are independent each others. Moreover now I consider here the particular case where noises are standard Normal. Moreover, like in your example, I start considering all three causal parameters equal to $1$.

As showed in my example (in the link) the causal coefficient/effect of $X$ on $Y$ can be consistently estimated from the regression of $Y$ on $X$. No controls are needed, indeed if we add a collider ($Z$) as control the regression coefficient of $X$ is no more a consistent estimator of the causal coefficient of $X$ on $Y$; worse, it do not represent any causal parameter of SCM. It is a useless regression coefficient.

Actually under the just suggested parametrization, the useless regression coefficient converge precisely to $0$; from normality assumption we can said that $X$ and $Y$ are independent conditioned on $Z$.

Now you said that

I tried changing the equations for $Y$ and $Z$ and yet, the zero is always there (almost always, check the end of the question) for $I(X;Y|Z)$, whenever it's a shielded triplet. Even if I do it with continuous variables and normal distributions for the noise, I still find the same thing. ... But then, if I change a bit the structural equation for $Z$, I get something different from zero. ... A hypothesis would be some sort of cancelling of effects between the two paths, but I changed the equations in ways that I didn't expect it to happen and still... The 0 is there.

This do not seems me completely clear and true. Indeed from my example if the causal parameter in the structural equation for $Y$ is different from $1$ the useless regression coefficient become different from zero. For example if the causal parameter in that structural equation is $2$ the conditional independence claimed do not hold. Indeed the useless regression coefficient converrge to $0,5$

Moreover following the alternative parameterization that you suggest ($2$ and $3$ as causal parameters in structural equation for $Z$ and retain $1$ in the equation for $Y$) the useless regression coefficient converge to $-0,5$.

In general, different values of causal parameters can produce null or positive or negative value for the useless regression coefficient. Moreover even the distribution of structural errors matters.

Finally

Why does collider adjustment in a shielded triplet tend to cause independence?

the independence you claimed can happen in particular cases (parameters combinations) but in general it do not hold. Indeed the general message is that control for collider is a bad idea.

Related Solutions

Solved – Does statistical independence mean lack of causation

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,

set.seed(100)
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,

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.

Independence of Cause and Mechanism – Causality

I would say this is a matter of terminology. What people in causality mean when they talk about mechanisms (also known as Independent Causal Mechanisms, causal Markov kernels, causal conditionals, etc.) are the "true" causal conditional distributions. In your case, you have only one mechanism $P(B \mid A)$.

As you correctly pointed out, the independence of cause and mechanism is just a way of saying that if you intervene on a variable, you don't change the conditional of downstream variables.

I don't know exactly what do you mean by showing this is true, this is a bit of an "axiom" of causality. But if I was forced to do so, then I would start with the causal Markov condition, which states that a causal graph can be decomposed into the "true" conditional distributions.

Best Answer

Related Solutions

Solved – Does statistical independence mean lack of causation

Independence of Cause and Mechanism – Causality

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