Hi Jarrod,
This is great. I have one final question: using the above equations, I am
finding it impossible for Pr(A) to ever be the greatest (in fact, Pr(C ) is
always largest using my data set). Even if I put in tiny values for eta_1
and eta_635 (0.0001 and 0.0001), all values just become essentially equal.
See results from some made-up combinations below:
eta_1 eta_635 Pr (A) Pr (B) Pr (C ) 0.0001 0.0001 0.333311111
0.333344444 0.333344444 0.0001 0.5 0.274061108 0.274088515 0.451850377 0.5
0.0001 0.274061108 0.451850377 0.274088515
Why would this be the case? I know from the data that A is in fact the most
common outcome.
Best,
Justine
On Tue, Jul 29, 2014 at 1:38 AM, Jarrod Hadfield <j.hadfield at ed.ac.uk>
wrote:
Hi Justine,
If you get the predictions on the link scale, and denote these as eta_1
and and eta_635 for the first observation, then
Pr(A) = 1/(1+exp(eta_1)+exp(eta_635))
Pr(B) = exp(eta_1)/(1+exp(eta_1)+exp(eta_635))
Pr(C) = exp(eta_635)/(1+exp(eta_1)+exp(eta_635))
There is some code for doing this (and marginalising any random effects)
in the CourseNotes (p97 after Eq. 5.7).
Cheers,
Jarrod
Quoting Justine <jsmith5 at ucsc.edu> on Mon, 28 Jul 2014 23:45:09 +0000
(UTC):
Jarrod Hadfield <j.hadfield at ...> writes:
Hi Justine,
The first 634 predictions are for B vs A, and the second 634 are for C
vs A. If you want the predicted probabilities of falling in category
A, B or C you'll have to do it by hand I'm afraid.
Cheers,
Jarrod
Hi Jarrod,
Thanks so much for clearing that up. Just to make sure I'm absolutely
clear,
if column 1 is 0.228, and column 635 is 0.092, than for data point #1 the
probability option B is more likely than option A is 0.228 and C more
likely
than A is 0.092? Does this indicate that A (the reference value) is the
most
likely? Can I calculate its relative probability by subtracting the other
two values from 1? I'm happy to assign the categories by hand, but I want
to
make sure I am interpreting the output correctly.
Thanks again,
Justine