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Suppose I have three variables. $A$ and $U$ are continuous variables but $U$ is unobserved. $Y$ is the binary outcome. $A$ and $U$ are independent.

Let the true model be from the typical probit or logit setup, $$Y = \mathbf{1}\{\beta_0 + \beta_1A + \beta_2U + \epsilon = Y^* > 0 \},$$ where $\epsilon$ is normal or logistic noise.

I want to say the DAG simply features two edges, $A \rightarrow Y$ and $U \rightarrow Y$. Then, there is no backdoor path featuring $A$ so it would seem that $\beta_1$ is identified without controlling for $U$.

However, I know this is true of OLS (supposing we observed the latent $Y^*$), and not true of logistic or probit models. (previously covered here)

Does this mean the DAG should be written differently for binary $Y$? We could say $A\rightarrow Y^*$, $U \rightarrow Y^*$, and $Y^* \rightarrow Y$? But I don't see how that shows there is an identification problem, and is it even necessary?

Is there a way to write the DAG that highlights why there is OVB in the binary $Y$ case but not for continuous $Y^*$?

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You are confusing confounding and noncollapsibility. In logistic and probit models, the difference between the focal coefficient in a covariate-adjusted vs. unadjusted model is not due just to confounding, but rather also to the fact that in nonlinear models, marginal relationships are not equal to conditional relationships. It's only the case in a few models, the linear model being one of them, that they are equal. For example, in Cox regression models, the covariate-adjusted conditional hazard ratio is not equal to the marginal hazard ratio even in the absence of confounding. See the discussion in Greenland, Pearl, and Robins (1999), for example. Noncollapsibility is not a very intuitive concept. Fcold's answer uses a similar analysis to that presented in Mood (2010).

I think you have some confusion regarding the term "omitted variable bias", which is a non-causal term used in econometrics, though often used to refer to confounding. Omitted variable bias simply refers to the bias in a coefficient that results from omitting a variable from the model. This bias can take many forms, one of which is confounding, but there are others, including noncollapsibility. You are equating omitted variable bias with confounding by assuming that the DAG (which represents causal, not parametric, relationships) needs to be adjusted to capture this bias. But again, omitted variable bias is not a causal concept; it may be due to confounding but may also be due to noncollapsibility. Therefore, there is no reason to adjust the DAG.

Note that if your model describes a structural (i.e., causal) model, then in the absence of confounding, while the coefficient $\beta^*$ on $A$ in the model omitting $U$ may not be equal to the coefficient $\beta$ on $A$ in the data-generating model, it may still have a valid causal interpretation (i.e., as the marginal effect of $A$ in the population). Basically, $\beta$ is not the only causal parameter associated with the data-generating model; $\beta^*$ is another, and there are still many others. In the absence of confounding, many of these can be estimated without bias even if $U$ is not observed. It just so happens that the specific causal parameter $\beta$ may not be estimable without bias.

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I would argue that there is no Omitted variable bias here. What you have is a bias on the coefficients produced by a different scale on the errors. Because probit model imposes a constrain on the variance (=1), the "bias" you see is more related to the new unobserved scale in the errors. However, what you are interested is not the coefficient on the latent variable, but the effect on p(y=1|X).

If you concentrate on that parameter, there shouldn't be an OVB

One way to see this is the following

Consider 2 variables x1 and x2, and the error u. All are IID N(0,1)

your latent $y^*=x_1-x_2 + u$ and your dummy simply $y=(y^*>0)$

Now run the following regressions:

$ p(y|x1,x2) = F(a_0+a_1 x1 + a_2 x_2)$

$ p(y|x1) = F(b_0+b_1 x1 )$

$ p(y|x2) = F(c_0+ c_2 x_2)$

What you will see is that $b_1$ will be smaller than $a_1$ because now the latent error is now larger (var=2), but probit model standardizes the variance to be =1. So all coefficients have to be adjusted by the square root of 2.

However, if you estimate the marginal effects. You will find that they are still the same in all models.

F

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  • $\begingroup$ What calculation are you suggesting? I think $\Pr(Y=1|A)$ will depend on the distribution of $U$. Suppose $U$ has, compared to $A$, very extreme values in our sample so that it basically determines $Y$. Then we might see almost no effect from $A$. But if instead $U$ is low variance, then we will get a substantial effect from $A$. In either case $U$ and $A$ are drawn independently. I've simulated this with $A$ a Bernouli(0.5) rv, $U$ being normal with high/low variance, then checking the cross tabs for $A$ and $Y$. Similar results with $A$ normal. $\endgroup$
    – Pburg
    Oct 15, 2021 at 19:26
  • $\begingroup$ Absolutely, the effect of A on P(y) will change depending on how large is Var(U). Reduction to the absurd. IF Var(U) goes to infinity, A has no effect. However, you can do the following. 1) y* = x1 + x2 + u. with u~N(0,1). then dy = y*>0 . Here you can compare the results modeling P(y=1|X) = F(x1) and F(x1,x2). Coefficients will change, but marginal effects will be almost the same. $\endgroup$
    – Fcold
    Oct 16, 2021 at 0:52

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