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I'm dealing with the following graph, taken from Causality by Pearl (2009):

enter image description here

The book says that in order to identify $\beta$, the coefficient of regression $X = \theta_1 Z + r_1$ is good, and for $\alpha$, the first coefficient ($\theta_2$) of the regression $Y = \theta_2 X + \theta_3 Z + r_2$ is what we need (we must adjust for $Z$).

Moreover, it is said that the total causal effect of $Z$ on $Y$ cannot be directly identified from any regression, but we can compute it indirectly as $\alpha \beta = \theta_1\theta_2$.

First question:
If I'm understanding correctly, the last problem come from the bidirected arc (unmeasured common causes). So if we delete it, then we can directly compute the total effect from $Y = \theta_4 Z + r_3$ and $\theta_4 = \alpha \beta $. Is this correct?

Now, I know that any graph can be represented as a system of structural equations (linear SEM), but I'm having some trouble. If so, then it seems to me that the following is correct:

$$Y = \alpha X + \epsilon_Y$$

$$X = \beta Z + \epsilon_X$$

$$Z = \epsilon_Z$$

And so we can write

$$Y = \alpha (\beta Z + \epsilon_X) + \epsilon_Y = \alpha\beta Z + \alpha \epsilon_X + \epsilon_Y$$

And $\text{cov}(\epsilon_Y, \epsilon_X) = 0$ and $\text{cov}(\epsilon_X, \epsilon_Z) = 0$; but $\text{cov}(\epsilon_Y, \epsilon_Z) \neq 0$ because we have bidirected arc.

Now I see from the last covariance that $E[\epsilon_Y \mid Z] = 0$ does not hold, and so the regression $Y = \theta_4 Z + r_3$ does not identify $\alpha \beta$. But I do not see why $E[\epsilon_Y \mid X] = 0$ does not hold, so $Y = \theta_5 X + r_5$ (not adjusted for $Z$) seems to me enough for identification of $\alpha$, but it should not be ($Z$ is needed). What am I missing in the system?

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  • $\begingroup$ What does "good for $\alpha$ ..." mean at the end? $\endgroup$ – The Pointer May 3 at 15:38
  • $\begingroup$ "good" for identification $\endgroup$ – markowitz May 3 at 15:40
  • $\begingroup$ Perhaps you can add some detail instead of just having "...", because it isn't clear to me what you're referring to there. $\endgroup$ – The Pointer May 3 at 15:40
  • $\begingroup$ I would to help you but I don't know how. I tried to use the notation clearly as possible, what part is unclear? $\endgroup$ – markowitz May 3 at 16:03
  • $\begingroup$ This isn't for my benefit, but for the benefit of others who might be in a position to help you. Why is there $\dots$ after $\alpha$? $\endgroup$ – The Pointer May 3 at 16:04
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The answer to your first question is yes. With the bidirected arc, the correlation between $Z$ and $Y$ is composed of two components, the correlation through the bidirected arc and the causal path from $Z$ to $X$ to $Y$. A simple regression of $Y$ on $Z$ gives you the whole thing, which cannot separate the two components to give you only the causal component. Without the bidirected arc, the correlation between $Z$ and $Y$ only has the causal component $\alpha\beta$, so the regression of $Y$ on $Z$ gives you the causal effect of $Z$ on $Y$.

For your other question: $E[\epsilon_Y|X]=0$ does not hold. As you mentioned, $E[\epsilon_Y|Z]=0$ does not hold. That's due to a connected path $\epsilon_Y\leftrightarrow\epsilon_Z\rightarrow Z$. Note that $X$ is an "extension" on that path, so $X$ is connected to $\epsilon_Y$ through the path $\epsilon_Y\leftrightarrow\epsilon_Z\rightarrow Z\rightarrow X$. This path is not causal (because of the bidirected arc). So again, regressing $Y$ on $X$ will give you the whole thing (causal + correlation) instead of just the causal path ($\alpha$).

A general and quick method you can refer to for checking those is the Wright's path tracing rules. Wright's rules allow you to equate the covariance between two variables to the sum of paths between them, where each path is the multiplication of the causal coefficients on that path. In your example, denote the causal coefficient on the bidirected arc as $\epsilon_{ZY}$, then $cov_{ZY}=\epsilon_{ZY}+\alpha\beta$, $cov_{ZX}=\beta$, $cov_{XY}=\alpha+\epsilon_{ZY}\beta$. It can be seen that only the regression of $Z$ on $X$ gives the causal effect $\beta$, while the other two both contain a correlation component. For more details, check out https://en.wikipedia.org/wiki/Path_analysis_(statistics)#Path_tracing_rules

P.S.: In this answer, I abused the term "correlation" for ease of explanation, and the Wright's rule equations are for standardized models, which need to be slightly modified if the model is unstandardized. For unstandardized models, correlation $\rho_{YX}$, covariance $cov_{YX}$, and regression coefficient $\beta_{YX}$ are not exactly equal. But the answers to your questions do not change. For more details, check out https://ftp.cs.ucla.edu/pub/stat_ser/r459-reprint-errata.pdf

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  • $\begingroup$ Welcome to CV, nice first answer! $\endgroup$ – Noah May 4 at 2:46
  • $\begingroup$ @Noah Thank you!! $\endgroup$ – Chiii Zhang May 4 at 4:58
  • $\begingroup$ First of all thanks for answer and the article (+1). Then, I conclude that if $cov(\epsilon_Z, \epsilon_Y)=0$ (no bidirected arc) we have $E[\epsilon_Y|X] = 0 $, so regressing $Y$ on $X$ permit us to identify $\alpha$. Moreover regressing $Y$ on $X$ and $Z$ permit us to identify $\alpha$ as well but it is not needed. It is correct? $\endgroup$ – markowitz May 4 at 8:55
  • $\begingroup$ Nice to see you here, Chi! $\endgroup$ – Carlos Cinelli May 4 at 21:13
  • $\begingroup$ @markowitz Yes that’s correct. $\endgroup$ – Chiii Zhang May 5 at 0:29

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