# [R] What is se.fit in a glm predict list?

peter dalgaard pdalgd at gmail.com
Thu Mar 2 09:50:52 CET 2017

```It comes from linearizing log(x) around the fitted value, derivative of log is 1/x, so
aa\$se.fit/aa\$fit == bb\$se.fit

For diverged estimates, this gets dubious as the linearization only works in a neighborhood and for x not too close to the singularity of log() at zero.

-pd

> On 02 Mar 2017, at 09:22 , Patrick Connolly <p_connolly at slingshot.co.nz> wrote:
>
>
> I'm trying  to calculate a CI for predictions from a Poisson GLM object egg.glm.
>
> Browse[2]> aa <- as.data.frame(predict(egg.glm, newdat, type = "response", se.fit = TRUE)[-3])
> Browse[2]> bb <- as.data.frame(predict(egg.glm, newdat, se.fit = TRUE)[-3])
> Browse[2]> aa
>           fit       se.fit
> 1 6.144212e-07 0.0005114257
> 2 2.452632e+01 5.4657657443
> 3 1.440000e+01 2.5817126393
> 4 4.389796e+01 4.5533997800
> 5 3.820455e+01 4.4827326393
> 6 6.226667e+01 5.6589154967
> Browse[2]> bb
>         fit       se.fit
> 1 -14.302585 832.36979026
> 2   3.199747   0.22285311
> 3   2.667228   0.17928560
> 4   3.781868   0.10372691
> 5   3.642954   0.11733506
> 6   4.131426   0.09088194
> Browse[2]>
>
> bb\$fit is clearly log of aa\$fit but just what is se.fit?  How do I use
> it to get a CI which is calculated on the log scale?  The first one is
> a bit messy since it is entirely from zeros.  Should I remove those or
> would that be unnecessary?
>
> TIA
>
> --
> ~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.
>   ___    Patrick Connolly
> {~._.~}                   Great minds discuss ideas
> _( Y )_  	         Average minds discuss events
> (:_~*~_:)                  Small minds discuss people
> (_)-(_)  	                      ..... Eleanor Roosevelt
>
> ~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.
>
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--
Peter Dalgaard, Professor,
Center for Statistics, Copenhagen Business School
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Email: pd.mes at cbs.dk  Priv: PDalgd at gmail.com

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