On 10-Sep-03 Christoph Lehmann wrote: > I have the follwoing data > V1 V2 > 1 -5.8000000 0 > 2 -4.8000000 0 > 3 -2.8666667 0 > 4 -0.8666667 0 > 5 -0.7333333 0 > 6 -1.6666667 0 > 7 -0.1333333 1 > 8 1.2000000 1 > 9 1.3333333 1 > > and I want to know, whether V1 can predict V2: of course it can, since > there is a perfect separation between cases 1..6 and 7..9 > > How can I test, whether this conclusion (being able to assign an > observation i to class j, only knowing its value on Variable V1) holds > also for the population, our data were drawn from? > > Means, which inference procedure is recommended? Logistic regression, > for obvious reasons makes no sense.
This is not so much an R question, nor really a "which procedure" question, since standard procedures are not usually equipped to deal with such situations (beyond telling you in some way that the situation is "out of bounds"). However, you can certainly investigate it by writing little R programs to look at it from various points of view. Let 'm' denote the location parameter for the CDF which models the probability, and 's' the scale parameter (e.g. a logistic function). For a start, clearly the maximum of the likelihood is 1, achieved when s=0 and m is any value between -0.7333.. and -0.1333.. You can investigate the variation of the likelihood as m and s vary by evaluating expressions like m<-(-.07);s<-1.0;L<-plogis((V1-m)/s);2*sum(V2*log(L)+(1-V2)*log(1-L)) For instance, for any value of s>0, find the value of m which maximises this. Then you can get an indication about your question by looking for the value of s such that this maximised value (with sign changed) is just on (say) the 5% point of a chisq[df=1]; my gropings suggest that s=0.8, m=(-0.1) (approx). This gives you a pair (m,s) which is just consistent with your data by this criterion. What, for instance, is the probability for any value of V1 that V2=1/0? E.g. for m=-0.1,s=0.8, consider the range -2 <= x <=2 (step=0.1): m<-(-0.10);s<-0.8;x<-0.1*(-20:20);L<-plogis((x-m)/s);L [1] 0.08509905 0.09534946 0.10669059 0.11920292 0.13296424 0.14804720 [7] 0.16451646 0.18242552 0.20181322 0.22270014 0.24508501 0.26894142 [13] 0.29421497 0.32082130 0.34864514 0.37754067 0.40733340 0.43782350 [19] 0.46879063 0.50000000 0.53120937 0.56217650 0.59266660 0.62245933 [25] 0.65135486 0.67917870 0.70578503 0.73105858 0.75491499 0.77729986 [31] 0.79818678 0.81757448 0.83548354 0.85195280 0.86703576 0.88079708 [37] 0.89330941 0.90465054 0.91490095 0.92414182 0.93245331 so that P(V2=1) can be substantial (>0.1) for V1 as low as -1.8, and P(V2=0) likewise for V2 as high as +1.6; yet this (m,s) is not "rejected" on likelihood grounds. So, in answer to your substantive question, it would seem that your data do not support the generalisation you are asking about. And so on; you can plot things out, etc. You can do a simulation study: for a given (m,s), say the pair above, and a set of V1 values like those which you have, what is the probability that you get a set of results (V2) which show "perfect separation"?:-- find what proportion of simulations satisfy max(which(V1[V2==0])) < min(which(V1[V2==1])) Explore a grid of (m,s) values and estimate this proportion; smooth the estimates and plot a contour diagram ... and so on! Use R as a tool for questions like this, and do not necessarily expect to find a procedure which is tailor-made for (e.g.) this particular question! Best wishes, Ted. -------------------------------------------------------------------- E-Mail: (Ted Harding) <[EMAIL PROTECTED]> Fax-to-email: +44 (0)870 167 1972 Date: 10-Sep-03 Time: 20:24:06 ------------------------------ XFMail ------------------------------ ______________________________________________ [EMAIL PROTECTED] mailing list https://www.stat.math.ethz.ch/mailman/listinfo/r-help