Unscaled Deviance is independent

jmoore13jmoore13
edited May 2024 in GLM.ModelRefinement

I'm confused by why exactly Goldburd uses the term "independent" when talking about the dispersion parameter in unscaled deviance. Unscaled Deviance literally is reliant on the value of the dispersion parameter in the formula, so by definition it isn't independent and depends on what the dispersion parameter is.

I get that they're saying without the dispersion parameter, we couldn't compare distributions since Scaled Deviance doesn't account for difference in distributions andthe dispersion paraemter brings that in, but is there a reason they're using "independent" that I'm not understanding?

Comments

  • adminadmin

    I agree with you, this feels like a poor choice of language in the source. Unscaled deviance is a function of the dispersion parameter so is definitely not independent of the dispersion parameter.

    What they're trying to get at is unscaled deviance allows you to compare any two models irrespective of their dispersion parameters. If you try comparing scaled deviance then you only have a valid comparison if the two models being compared have the same dispersion parameter.

  • jmoore13jmoore13

    Thanks! That makes sense.

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