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From Sean Owen <sro...@gmail.com>
Subject Re: Log-likelihood ratio test as a probability
Date Thu, 20 Jun 2013 09:16:50 GMT
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.filimon@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 <srowen@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.filimon@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
>>

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