Let’s say we have a certain DGP $y=c+ax+bz+ error$. I take this to mean that conditional on the values of $x$ and $z$ in a certain instance (and no additional information )the expected value of $y$ can be inferred by plugging the values into the above equation. Now let’s say $x$ and $z$ are correlated. A different DGP may now be $y=d+mx+error$ where m may be higher(or lower) than $a$ due to the correlation of $x$ and $z$. Now wouldn’t $y=d+mx+error$ also be a valid DGP given that the expected value of $y$ conditional on $x$ (and no other information) is given correctly?

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    $\begingroup$ the DGP refers to the true process which generates the data. It is different from a model of the data. So it doesn't really make sense to talk about validity in terms of the DGP since the DGP is the truth. Once you have the DGP then you can think about models that try to describe that DGP and whether they are useful or not. $\endgroup$ – Andrew M Dec 29 '19 at 3:13
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    $\begingroup$ It does not make sense to compare two different DGP's since they are referring to two completely different realities. On the other hand for a given DGP you can consider many different models and discuss whether they are valid or not so if y=c+ax+bz+e is the DGP then a model could be y=d+mx+error. Note then with your model you will not be able to recover the "true" DGP parameters of c and a unless b=0 or using other methods (i.e. instrumental variables etc). $\endgroup$ – Andrew M Dec 29 '19 at 3:22
  • $\begingroup$ @AndrewM Why do you use scare quotes around 'true'? Do you think that the data generating process is a correct description of the process that generates the data or do you not? Scare quotes can only fudge the issue. $\endgroup$ – user17900 Jan 2 '20 at 13:12
  • $\begingroup$ @afreelunch I used the quotes because it was meant that the truth depends on how we define it. What we say is the true process in one setting may not be the true process in another. I was trying to highlight that fact. Was not trying to fudge anything. $\endgroup$ – Andrew M Jan 22 '20 at 4:37

There seem to be (at least) two ways of interpreting the concept of a 'data generating process':

  1. The data generating process is a correct description of the causal process that generates the values of the dependent variable $Y$.
  2. The data generating process specifies the expected value of the dependent variable $Y$ given the values taken the independent variables.

Now the answer to your question seems to depend on which interpretation we take:

  1. If we follow interpretation 1, then these cannot be both data generating processes: either variable $Z$ causally determines $Y$ or it does not.
  2. If we follow interpretation 2, then these can both be valid data generating processes. As you point out, there is no reason to suppose that $\mathbb{E}[Y|X = x]$ is the same as $\mathbb{E}[Y|X = x, Z = z]$.
  • $\begingroup$ @user17900: Nice differentiation but 1) is how I've always thought of the DGP. I think 2) is the model. $\endgroup$ – mark leeds Jan 26 at 17:13

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