unexpected GAM result - at least for me!
On 3/31/2008 9:01 AM, Monica Pisica wrote:
Thanks Duncan. Yes i do have variation in the lidar metrics (be, ch, crr, and home) although i have a quite high correlation between ch and home. But even if i eliminate one metric (either ch or home) i end up with a deviation of 99.99. The species has values of 0 and 1 since i try to predict presence / absence. Do you think it is still a valid result?
I repeat: look at the data. Compare the observed and predicted. That's the only way to know whether this is reasonable or not. If you're getting reasonable predictions, then it's a valid fit. (The tests and approximations used in the reported p-values may not be at all valid. I don't know what the requirements are for those in a GAM, but if you're getting a perfect fit, then they probably aren't being met.) Duncan Murdoch
Thanks again, Monica
> Date: Mon, 31 Mar 2008 08:47:48 -0400 > From: murdoch at stats.uwo.ca > To: pisicandru at hotmail.com > CC: r-help at r-project.org > Subject: Re: [R] unexpected GAM result - at least for me! > > On 3/31/2008 8:34 AM, Monica Pisica wrote:
> > > > Hi > > > > > > I am afraid i am not understanding something very fundamental....
and does not matter how much i am looking into the book "Generalized Additive Models" of S. Wood i still don't understand my result.
> > > > I am trying to model presence / absence (presence = 1, absence = 0)
of a species using some lidar metrics (i have 4 of these). I am using different models and such .... and when i used gam i got this very weird (for me) result which i thought it is not possible - or i have no idea how to interpret it.
> >
> >> can3.gam <- gam(can>0~s(be)+s(crr)+s(ch)+s(home), family = 'binomial') > >> summary(can3.gam)
> > Family: binomial > > Link function: logit > > Formula: > > can> 0 ~ s(be) + s(crr) + s(ch) + s(home) > > Parametric coefficients: > > Estimate Std. Error z value Pr(>|z|) > > (Intercept) 85.39 162.88 0.524 0.6 > > Approximate significance of smooth terms: > > edf Est.rank Chi.sq p-value > > s(be) 1.000 1 0.100 0.751 > > s(crr) 3.929 8 0.380 1.000 > > s(ch) 6.820 9 0.396 1.000 > > s(home) 1.000 1 0.314 0.575 > > R-sq.(adj) = 1 Deviance explained = 100% > > UBRE score = -0.81413 Scale est. = 1 n = 148 > > > > Is this a perfect fit with no statistical significance, an
over-estimating or what???? It seems that the significance of the smooths terms is "null". Of course with such a model i predict perfectly presence / absence of species.
> > > > Again, i hope you don't mind i'm asking you this. Any explanation
will be very much appreciated.
> > Look at the data. You can get a perfect fit to a logistic regression > model fairly easily, and it looks as though you've got one. (In fact, > the huge intercept suggests that all predictions will be 1. Do you > actually have any variation in the data?) > > Duncan Murdoch
In a rush? Get real-time answers with Windows Live Messenger. <http://www.windowslive.com/messenger/overview.html?ocid=TXT_TAGLM_WL_Refresh_realtime_042008>