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I'm just starting to learn about causal inference methods, focused on Pearl's do-calculus. So the point between Pearl's causal graphs and rules for manipulating causal graphs appears to be to turn a causal graph into a statistical model (e.g. a linear regression).

So you might have a causal graph such as $Z \rightarrow X, X \rightarrow M, M \rightarrow Y, Z \rightarrow Y$ (Z is a confounder of X and Y, but X also partially causes Y through M).

enter image description here

If my aim is to figure out the causal effect of X on Y and I just did the naive thing and setup a linear regression with $Y = aX + e$ and tried to estimate a regression coefficient, then of course I would get a biased estimate due to the presence of the confounder Z. On the other hand, if I do $Y = aX + bM + cZ + e$, then I will block the effect of X by conditioning on the mediator M. So again, knowing the causal graph will tell me to condition on Z but not on M, i.e. $Y = aX + cZ + e$ is the correct statistical model that allows me to estimate a causal effect.

But is $Y = aX + cZ + e$ (as a regression model, not a math equation) also a causal model (albeit a "wrong" causal model)? If I manipulate $X$ it tells me what happens to $Y$. Doesn't it correspond to the causal graph $X \rightarrow Y, Z \rightarrow Y$ ? If so, then is the Pearl method just finding a transform of a causal graph into another causal graph that is easier to work with or represent as a regression?

edit: I think my causal graph analysis in this simple example is wrong, but hopefully the broader point is still clear

edit #2:

If I write a model $Y + aX + bZ + e$ in a programming language, I could do so as a function, e.g. in Python

 def model(a,X,c,Z):
     return a*X + b*Z + np.random.randn()

So if I change the input $X$, it will cause the output of $model(...)$ to change, but I cannot do the opposite. Isn't that a causal model?

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    $\begingroup$ Nope. It is pretty easy to find real-world examples where no set of models on any arrangement of observed variables give unbiased causal estimates. $\endgroup$
    – Alexis
    Nov 19, 2021 at 2:40
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    $\begingroup$ Right- I get that, there are issues of identifiability. But my question is that there doesn’t seem to be a mathematical difference between a causal model and a statistical model- it’s just a matter of which one you assume to actually represent the world. It seems like we start with an assumed model that we think represents the real world, the causal model, and then we see if we can derive another model from it using only observable variables that is in some way in bijection with the assumed causal model. And sometimes it may not be possible to identify that bijective model. $\endgroup$ Nov 19, 2021 at 3:10
  • $\begingroup$ I very much agree with the close of your comment. $\endgroup$
    – Alexis
    Nov 19, 2021 at 4:20
  • $\begingroup$ You may be interested in Econometrics. In Econometrics statistical models are built in a way that reflect causal relations. That starts on how data are collected. $\endgroup$
    – Pere
    Nov 19, 2021 at 13:26
  • $\begingroup$ In the model "$Y=aX+cZ+e$" there is no concept of "manipulating" the values of the explanatory variable, not even an implied one. To suppose there is such a meaning would be tantamount to asserting association is equivalent to causation. $\endgroup$
    – whuber
    Nov 19, 2021 at 14:09

3 Answers 3

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But is $Y = aX + cZ + e$ (as a regression model, not a math equation) also a causal model (albeit a "wrong" causal model)? If I manipulate $X$ it tells me what happens to $Y$. Doesn't it correspond to the causal graph $X \rightarrow Y, Z \rightarrow Y$ ?

It may correspond to the causal graph $X \rightarrow Y, Z \rightarrow Y$...

... But it can just as well be $Y \rightarrow X, Y \rightarrow Z$.

Statistical models are present wherever there are causal models, but those statistical models are not equivalent/identical to the causal models themselves.

A statistical model only describes the correlation, and it does not (need to) describe the causation. You can describe and fit statistical models without a description of an underlying causal model.

Or at least, certainly the statistical model alone does not tell you anything about the causation (except that there is some underlying causal mechanism, but we do not know which). In this sense it is not equivalent to a causal model.

You could see a statistical model as the shadow of a causal model.


Edit

If I write a model $Y + aX + bZ + e$ in a programming language, I could do so as a function, e.g. in Python

 def model(a,X,c,Z):
     return a*X + b*Z + np.random.randn()```

In this case you are programming a causal model and not a statistical model. You are specifically defining the random noise as being added to the a*X+b*Z. But this does not need to be the case in order to get that there is a relation:

$$ Y|X,Z \sim N(aX + bZ, \sigma^2)$$

Consider the data below

symmetry

The statistical model is that X and Y follow a bivariate distribution. But can you tell the causal model from it? Do we have $X = aY + \text{noise}$ or do we have $Y = aX +\text{noise}$ ? They can result in the same statistical distribution, but the causal models are different.

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  • $\begingroup$ Made update to post $\endgroup$ Nov 19, 2021 at 15:37
  • $\begingroup$ @Sextus Empiricus: I liked your explanation on causal versus statistical. But, in the example that you gave where you showed how the causal graph could be written in two different ways, what about the case where one happened to know that $X$ and $Z$ occurred BEFORE $Y$ in time ( a time series framework ). In that case, could one say that a statistical model is also a causal model ? Thanks. $\endgroup$
    – mlofton
    Nov 19, 2021 at 16:13
  • $\begingroup$ @mlofton no, that would be a post hoc ergo proper hoc fallacy. Example: every year we have increased sales of ice cream (on the northern hemisphere) in June, July, August and following that we have lots of leaves falling from trees in September, Oktober, November. But, ice cream sales do not cause leaves falling from trees, even though they happend before in time. $\endgroup$ Nov 19, 2021 at 16:19
  • $\begingroup$ Thanks for the explanation. In my causal model using the programming language, I can use it to do regression and get regression coefficients. Wouldn’t they be the same as a statistical model regression? $\endgroup$ Nov 19, 2021 at 19:10
  • $\begingroup$ @BrandonBrown that's why I called a statistical model the shadow of a causal model. Your programming model casts a shadow that is a statistical model, but you can not reason backward that this particular shadow is equal to the shape that created it. And many different shapes/models can create the same shadow. $\endgroup$ Nov 19, 2021 at 20:24
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edit: I think my causal graph analysis in this simple example is wrong, but hopefully the broader point is still clear

To the extent that you correctly identified that M is a mediator and Z is a confounder, your analysis is correct. If this is the right causal model (based either on background knowledge or causal discovery), there is only one thing you didn't mention explicitly: X is not a direct cause of Y. You can measure some total effect due to the effect mediated through M, but if you control M, you can make X and Y independent (Markov property).

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Are all statistical models also causal models?

No, they aren't.

So the point between Pearl's causal graphs and rules for manipulating causal graphs appears to be to turn a causal graph into a statistical model (e.g. a linear regression).

This statement, at best, is confusing. The main point of Pearl works is to separate clearly: statistical concepts vs causal concepts. Your question itself reveal that you do not grasped this, not yet. In fact statements like the follow

But is $Y=aX+cZ+e$ (as a regression model, not a math equation) also a causal model (albeit a "wrong" causal model)? If I manipulate $X$ it tells me what happens to $Y$. Doesn't it correspond to the causal graph $X→Y,Z→Y$ ? If so, then is the Pearl method just finding a transform of a causal graph into another causal graph that is easier to work with or represent as a regression?

confirm my impression, definitely.

Apart that a regression equation remain a respectable math equation (too much general concept) causal graph do not stay for represent some regression. Worse, the (causal) concept of manipulation cannot apply to regression. Indeed causal graphs represent structural equations and them must be clearly separated from regression equations. Indeed Pearl underscore repeatedly the opportunity to use different notations.

If I write a model Y+aX+bZ+e in a programming language, I could do so as a function, e.g. in Python

It looks like a causal model. It is so because the definitions demanded from programming language imply a kind of equal sign that is definitional ($:=$), it is different from the standard equal sign ($=$). The standard equal sign is logically symmetric, the other is not and precisely for this reason it is what a structural equation (causal concept) need.

Said that, the main link between causal and statistical model is about identification (read here:Why do we need identification in causal inference?). For identification purpose Pearl suggest do-calculus (read here:What's the purpose of do-calculus?)

For more details read here:

Under which assumptions a regression can be interpreted causally?

Criticism of Pearl's theory of causality

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  • $\begingroup$ I understand that when we observe some data, all we get is essentially some joint distribution. And I guess a "statistical model" purely deals in terms of defining probability distributions, and probability distributions alone never encode causal relationships. But to my naive eyes it seems when we actually do a regression in practice, we do it using Y := aX + b and not just Y = aX + B, so in actuality we are encoding a structural relationship. However, we cannot determine if that is the correct causal model, so the data alone can rarely identify the correct causal model. Maybe just semantics $\endgroup$ Nov 23, 2021 at 17:39
  • $\begingroup$ Your first part of the comment is ok. Your interpretation of regression is prone to any kind of misunderstanding. Read here: stats.stackexchange.com/questions/484914/regression-and-the-cef/… and links suggested above, mainly stats.stackexchange.com/questions/493211/…. $\endgroup$
    – markowitz
    Nov 23, 2021 at 17:50
  • $\begingroup$ Let me add that this misunderstanding is not only your. It is common in Econometrics literature, read here stats.stackexchange.com/questions/477705/… $\endgroup$
    – markowitz
    Nov 23, 2021 at 17:57

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