# Re: [R] paradox about the degree of freedom in a logistic regression model

From: Peter Dalgaard <P.Dalgaard_at_biostat.ku.dk>
Date: Fri, 07 Dec 2007 15:14:52 +0100

Bin Yue wrote:
> Dear all:
> "predict.glm" provides an example to perform logistic regression when the
> response variable is a tow-columned matrix. I find some paradox about the
> degree of freedom .
> > summary(budworm.lg)
>
> Call:
> glm(formula = SF ~ sex * ldose, family = binomial)
>
> Deviance Residuals:
> Min 1Q Median 3Q Max
> -1.39849 -0.32094 -0.07592 0.38220 1.10375
>
> Coefficients:
> Estimate Std. Error z value Pr(>|z|)
> (Intercept) -2.9935 0.5527 -5.416 6.09e-08 ***
> sexM 0.1750 0.7783 0.225 0.822
> ldose 0.9060 0.1671 5.422 5.89e-08 ***
> sexM:ldose 0.3529 0.2700 1.307 0.191
> ---
> Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1
>
> (Dispersion parameter for binomial family taken to be 1)
>
> Null deviance: 124.8756 on 11 degrees of freedom
> Residual deviance: 4.9937 on 8 degrees of freedom
> AIC: 43.104
>
> Number of Fisher Scoring iterations: 4
>
> This is the data set used in regression:
> 1 1 19 M 0
> 2 4 16 M 1
> 3 9 11 M 2
> 4 13 7 M 3
> 5 18 2 M 4
> 6 20 0 M 5
> 7 0 20 F 0
> 8 2 18 F 1
> 9 6 14 F 2
> 10 10 10 F 3
> 11 12 8 F 4
> 12 16 4 F 5
>
> The degree of freedom is 8. Each row in the example is thought to be
> one observation. If I extend it to be a three column data.frame, the first
> denoting the whether the individual is alive , the secode denoting the sex,
> and the third "ldose",there will be 12*20=240 observations.
> Since my data set is one of the second type , I wish to know whether
> the form of data set affects the result of regression ,such as the degree of
> freedom.
> Regards,
> Bin Yue
>
>
Yes. Never use the deviance in binary logistic regression. Only use differences in deviance between models, each of which satisfy requirements for asymptotic theory (in your case, you could compare your model with that described by sex*factor(ldose)). Another striking example is this

y <- rbinom(1000, prob=.5, size=1)
summary(glm(y~-1,binomial))

now try it with different data

y <- rbinom(1000, prob=.01, size=1)
summary(glm(y~-1,binomial))

and think about it. Then consider the same thing with y~1.

As Brian keeps telling me, there IS a sense in which the residual deviances make sense in such cases, but it is not as a means of testing the model adequacy.

> -----
> Best regards,
> Bin Yue
>
> *************
> student for a Master program in South Botanical Garden , CAS
>
>

```--
O__  ---- Peter Dalgaard             Øster Farimagsgade 5, Entr.B
c/ /'_ --- Dept. of Biostatistics     PO Box 2099, 1014 Cph. K
(*) \(*) -- University of Copenhagen   Denmark          Ph:  (+45) 35327918
~~~~~~~~~~ - (p.dalgaard_at_biostat.ku.dk)                  FAX: (+45) 35327907

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