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From |
Phil Schumm <pschumm@uchicago.edu> |

To |
statalist@hsphsun2.harvard.edu |

Subject |
Re: st: Poisson model with interaction term |

Date |
Wed, 27 Feb 2008 09:46:31 -0600 |

On Feb 26, 2008, at 4:49 AM, Maarten buis wrote:

The problem I have is this: In the case of non-linear models you would expect the effect of x1 to change when x2 changes even if we do not enter the interaction term.

<snip>

I think (but I am not sure) that the method by Norton et al. gives the combined change in the effect of x1, i.e. the change in effect of x1 that would have occured anyhow and the change in effect due to the interaction term together. I think that in many case this would be reasonable, but I can also imagine situations where you just want to know the effect of the interaction term net of the change in effect that would occur anyhow.

Just to make sure I understand the question, if you define the interaction effect as Norton et al. do (i.e., as the mixed partial derivative of the mean of y), then, as they note, this may be non- zero even if the coefficient on the interaction term itself (i.e., the interaction term in the linear predictor) is zero. I believe that this is what you are referring to in the first quote above.

Now, if you fit a model without the interaction term (i.e., constraining its coefficient to be zero), you can compute the interaction effect (i.e., the mixed partial derivative); call this A. You can then refit the model with the interaction term included (let's assume the estimated coefficient is non-zero), and recompute the interaction effect (call this B). What you are asking, I think, is if there is a way to decompose B into A plus another part that is due to the addition of the interaction term to the model. Is that correct?

If so, I don't think it can be done. This is because, when you add the interaction term to the model, the coefficients on the first order terms will change. IOW, A and B are computed with different values for the same quantities, and thus you can't decompose B in terms of A. You could, I suppose, simply compare A and B numerically, however, if the model without the interaction term doesn't fit the data well, interpretation of A is problematic.

Don't know if that helps at all...

-- Phil

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**References**:**Re: st: Poisson model with interaction term***From:*Maarten buis <maartenbuis@yahoo.co.uk>

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