In the multiple linear regression analysis if regressors are non-stochastic the causal interpretation of parameters is automatically permitted? I think so, because it seems me that the model can be interpreted as "true causal model" but I'm not sure. Expecially I'm not sure about the role of something like that control variables.

  • $\begingroup$ Dear Tim, I don't write "true deterministic model" but "true causal model". I mean "true model" as used in several econometrics textbook (as data generating process) and I intend it as explicitly causal. If I estimate a sample counterpart of it surely I achieve causal parameters. If I estimate under-specified model probably not. My doubt is about the link with experimental language when I have fixed design matrix in repeated sample. This strategy, at least in certain sense, is like to built a true causal model or not? $\endgroup$ – markowitz Nov 10 '18 at 15:06
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    $\begingroup$ I ask this because I currently focused on causal interpretation in observational study but some author use fixed (non stochastic) regressors too. See for example here: stats.stackexchange.com/questions/374959/… $\endgroup$ – markowitz Nov 10 '18 at 15:07
  • $\begingroup$ What do you mean by "non-stochastic regressors"? $\endgroup$ – Tim Nov 10 '18 at 16:58
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    $\begingroup$ Shortly, as I write before, fixed in repeated sample. Actually in the texts that I read the meaning of "non-stochastic regressors" is vague. In any case the only meaning that seems me possible is something like "treatment variables" in experimental sense. $\endgroup$ – markowitz Nov 11 '18 at 11:52
  • $\begingroup$ If this was true, then every experimental data would enable us to draw direct causual conclusions and this is simply not the case. $\endgroup$ – Tim Nov 11 '18 at 12:11

In the multiple linear regression analysis if regressors are non-stochastic the causal interpretation of parameters is automatically permitted?

No, the fact that regressors are non-stochastic makes no difference for whether the estimates identify a causal effect. For causal interpretation of parameters you need a causal model.

For simplicity, assume everything has mean zero and unit variance. Let $X$ and $Z$ be two fixed vectors of size $n$ with $(n-1)^{-1}\sum_{i} X_{i}Z_{i} = \sigma_{xz} \neq 0$. Assume you do not observe $Z$.

Now let the structural equation for $Y$ be:

$$ Y = \gamma Z + U_{y} $$

Where $U_{y}$ is a zero mean, normally distributed random variable. Thus, there is no causal effect of $X$ on $Y$.

However, the regression of $X$ on $Y$ is given by,

$$ \begin{align} \hat{\beta} &= (n-1)^{-1}\sum_{i=1}^{n} X_{i}Y_{i} \\ &=(n-1)^{-1}\left(\gamma \sum_{i=1}^{n} X_{i}Z_{i} + \sum_{i=1}^{n} X_{i}U_{yi}\right)\\ &= \gamma\sigma_{xz} + (n-1)^{-1}\sum_{i=1}^{n} X_{i}U_{yi} \end{align} $$

Where the only "random" part is $U_{y}$, Thus, taking the expectation gives us:

$$ \begin{align} E[\hat{\beta}] &= \gamma\sigma_{xz} + (n-1)^{-1}\sum_{i=1}^{n} X_{i}E[U_{yi}]\\ &= \gamma\sigma_{xz} \end{align} $$

Which is different from zero and clearly does not have a causal meaning. We can also do asymptotics by letting the size of $X$ and $Z$ grow and keeping $\sigma_{xz}$ fixed. The very book you mention in the comments and in your other question, Brooks (2014), mentions how omitted variables can bias the estimate.

Bear in mind this is just an example of omitted variable bias, there are several other things that can go wrong, and they have nothing to do with whether you treat $X$ as stochastic or not. The bottom line here is that confounding, missing data, selection bias --- and many other problems---can still be present, whether you treat the regressors as random or "non-stochastic". You need to make assumptions about the presence or absence of those things, which are causal concepts, and for that you need a causal --- not a regression --- model.

  • $\begingroup$ Hi Carlos thanks you for your answer, but I disagree. Your demonstration is essentially the same of observational study. The text of my question is probably too short but I specify (see comment with Tim) that “In any case the only meaning that seems me possible [for fixed/non-stichastic regressors] is something like "treatment variables" in experimental sense”. Later … random sampling, in repeated sample scheme, and random assignment of the treatments are assumed as minimal conditions. Probably these are thing related with manipulations. $\endgroup$ – markowitz Nov 13 '18 at 16:43
  • $\begingroup$ . Shortly, for me, in fixed non-stochastic regressors scheme the dependence between two regressors, or among many, is not acceptable case. Ideally, among non random variables the correlation, or any kind of dependence, do not exist. Numerically is possible that correlation, or other dependence, exist but at least in large sample it must disappear. The same is true between regressor and errors (that remain random). Persistent correlation that you propose is not acceptable in my opinion. $\endgroup$ – markowitz Nov 13 '18 at 16:44
  • $\begingroup$ In any case researcher can always manipulate the treatment for achieve the substantially (numerical) independence among regressors (treatments). The independence among regressors seems me, at least, the base case in experiment. The same between regressors and errors. Moreover the omitted variable problem don’t play any role in the case that I have in mind. Researcher known, indeed builds, the “true” regressors; is a nonsense to use another set of variable as regressors. $\endgroup$ – markowitz Nov 13 '18 at 16:44
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    $\begingroup$ @markowitz among non random variables the correlation, or any kind of dependence, do not exist. ---> this sentence is completely incorrect. Just imagine a deterministic process where I experimentally choose to set $X = (1, 2, 3, ...)$ and $Z = (1, 2,3 ...)$. The correlation of $X$ and $Z$ is one, and it will always be one. $\endgroup$ – Carlos Cinelli Nov 13 '18 at 17:08
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    $\begingroup$ @markowitz this claim is clearly false, otherwise by your logic even the very relationship between $X$ and $Y$ would always need to be absent, since $X$ is not random and $Y$ is random. $\endgroup$ – Carlos Cinelli Nov 13 '18 at 17:31

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