Right, makes sense. So, by normalize, I need to replace the counts in the matrix with probabilities. So, I would divide everything by the sum of all the counts in the matrix?
On Thu, Jun 20, 2013 at 12:16 PM, Sean Owen <sro...@gmail.com> wrote: > I think the quickest answer is: the formula computes the test > statistic as a difference of log values, rather than log of ratio of > values. By not normalizing, the entropy is multiplied by a factor (sum > of the counts) vs normalized. So you do end up with a statistic N > times larger when counts are N times larger. > > On Thu, Jun 20, 2013 at 9:52 AM, Dan Filimon > <dangeorge.fili...@gmail.com> wrote: > > My understanding: > > > > Yes, the log-likelihood ratio (-2 log lambda) follows a chi-squared > > distribution with 1 degree of freedom in the 2x2 table case. > > A ~A > > B > > ~B > > > > We're testing to see if p(A | B) = p(A | ~B). That's the null > hypothesis. I > > compute the LLR. The larger that is, the more unlikely the null > hypothesis > > is to be true. > > I can then look at a table with df=1. And I'd get p, the probability of > > seeing that result or something worse (the upper tail). > > So, the probability of them being similar is 1 - p (which is exactly the > > CDF for that value of X). > > > > Now, my question is: in the contingency table case, why would I > normalize? > > It's a ratio already, isn't it? > > > > > > On Thu, Jun 20, 2013 at 11:03 AM, Sean Owen <sro...@gmail.com> wrote: > > > >> someone can check my facts here, but the log-likelihood ratio follows > >> a chi-square distribution. You can figure an actual probability from > >> that in the usual way, from its CDF. You would need to tweak the code > >> you see in the project to compute an actual LLR by normalizing the > >> input. > >> > >> You could use 1-p then as a similarity metric. > >> > >> This also isn't how the test statistic is turned into a similarity > >> metric in the project now. But 1-p sounds nicer. Maybe the historical > >> reason was speed, or, ignorance. > >> > >> On Thu, Jun 20, 2013 at 8:53 AM, Dan Filimon > >> <dangeorge.fili...@gmail.com> wrote: > >> > When computing item-item similarity using the log-likelihood > similarity > >> > [1], can I simply apply a sigmoid do the resulting values to get the > >> > probability that two items are similar? > >> > > >> > Is there any other processing I need to do? > >> > > >> > Thanks! > >> > > >> > [1] http://tdunning.blogspot.ro/2008/03/surprise-and-coincidence.html > >> >
