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I have a Poisson model where the true relationship is: $$E(y\mid x,z)=\exp(b_1+b_2\times x+b_3\times z)$$ but z is not observable and so it is omitted from the estimated regression.

I read here that when z is independent from x, the estimate of $b_2$ should not be biased. Is it possible to derive the sign or magnitude of the bias in the case when x and z are not independent? (The paper discusses this case briefly but I was not able to follow the argument there.)

The specific case to which I would like to apply the answer is where the omitted variable is z=x*u with some unobservable $u \sim N(0,1)$.

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2 Answers 2

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It is indeed possible in some circumstances to say something analytic.

$$E[Y|X]=E\left[\exp(b_1+b_2\times X +b_3Z)\mid X\right]=E\left[\exp(b_1+b_2\times X)\mid X] \times E[\exp(b_3Z)\mid X\right]$$

Now $$E\left[\exp(b_3Z)\mid X\right]=e^{b_3}\,E[\exp Z|X]$$

so$$\log E\left[\exp(b_3Z)\mid X\right]=b_3\log\,E[\exp Z|X]$$

So the bias in $b_2x$ will be $b_3\log\,E[\exp Z|X=x]$, which gives an indication of the direction and magnitude.

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A direct way to approach this would be to simulate data under your assumptions about the nature of $z$ and its association with outcome ($b_3$) and repeat modeling with and without the $b_3 z$ term. If you already have data that you have modeled, seeing what happens when you include that term (under reasonable assumptions, based on your understanding of the subject matter) is a good way to evaluate the sensitivity of your outcome estimates to the potential omitted-variable bias.

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  • $\begingroup$ Thank you for the suggestion. I have run simulations and these are indeed informative, but I am wondering if it is possible say something more definite in relatively simple cases. Thank you again, nevertheless. $\endgroup$
    – lippi
    Commented Oct 14 at 18:12

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