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From David Hall <>
Subject Re: Probability from log likelihood in LDA output
Date Tue, 07 Dec 2010 04:46:04 GMT

The scores aren't (log) normalized until they're loaded in the map
phase. Take a look at LDAState. The array

private final double[] logTotals; // log \sum p(w|t) for topic=1..nTopics

in LDAState has normalization constants.  The method
logProbWordGivenTopic is intended for access...  LDADriver#createState
is a round about way of creating an LDA State.

-- David

On Mon, Dec 6, 2010 at 12:06 PM, Quiroz Hernandez, Andres
<> wrote:
> Thanks for your quick reply, Ted. It looks like either the probabilities are not normalized
or the function being used is not a simple sum of log probabilities, because exp does not
always return a value between 0 and 1. I will take a look at the code to see if I can find
exactly how the value is calculated (but if anyone knows the function used, and if I can directly
invert it to find P(w|t) please let me know).
> Thanks again,
> Andres
> -----Original Message-----
> From: Ted Dunning []
> Sent: Monday, December 06, 2010 11:57 AM
> To:
> Subject: Re: Probability from log likelihood in LDA output
> Yes.  I should be possible to use exp to get the actual probability.  The
> fact that it is a sum
> of log probabilities just means that the probability is a product of
> probabilities.
> It is possible that the probabilities are not normalized, but that would be
> a bit surprising for
> this kind of algorithm.
> On Mon, Dec 6, 2010 at 8:02 AM, Quiroz Hernandez, Andres <
>> wrote:
>> Hello,
>> As I understand it, the output for LDA is a log likelihood value for
>> each word/topic pair, which is a function of log(P(w|t)). Is it possible
>> to invert that function to obtain P(w|t)? I have a feeling it is not,
>> since it looks like the final value is obtained as a sum of log
>> probabilities, but I just wanted to check, since an output as a
>> probability is more readable than the likelihood value given.
>> Thanks,
>> Andres

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