Ngram models and the Sparsity problem - PowerPoint PPT Presentation

ngram models and the sparsity problem n.
Skip this Video
Loading SlideShow in 5 Seconds..
Ngram models and the Sparsity problem PowerPoint Presentation
Download Presentation
Ngram models and the Sparsity problem

play fullscreen
1 / 36
Download Presentation
Ngram models and the Sparsity problem
Download Presentation

Ngram models and the Sparsity problem

- - - - - - - - - - - - - - - - - - - - - - - - - - - E N D - - - - - - - - - - - - - - - - - - - - - - - - - - -
Presentation Transcript

  1. Ngram models and the Sparsity problem John Goldsmith

  2. The task • Find a probability distribution for the current word in a text (utterance, etc.), given what the last n words have been. (n = 0,1,2,3) • Why this is reasonable • What the problems are

  3. Why this is important • Probability is the only common currency that can be used to relate information from several sources: • Language model (“prior”) plus right-now information from sound or writing • Probability (joint event: the current input + its analysis) = probability (analysis) * prob(current input|analysis) Bayesian analysis

  4. If you take logs • Log prob ( your joint analysis ) = Sum of Log prob ( linguistic analysis ) + Log prob ( likelihood of that data, given this linguistic analysis ) Find the analysis that maximizes this sum.

  5. Why an n-gram model is reasonable The last few words tells us a lot about the next word: • collocations • prediction of current category: the is followed by nouns or adjectives • semantic domain

  6. Reminder about applications • Speech recognition • Handwriting recognition • POS tagging

  7. Problem of sparsity • Words are very rare events (even if we’re not aware of that), so • What feel like perfectly common sequences of words may be too rare to actually have in our training corpus

  8. What’s the next word? in a ____ with a ____ the last ____ shot a _____ open the ____ over my ____ President Bill ____ keep tabs ____

  9. Example: • Corpus: five Jane Austen novels • N = 617,091 words • V = 14,585 unique words • Task: predict the next word of the trigram “inferior to ________” • from test data, Persuasion: “[In person, she was] inferior to both [sisters.]” borrowed from Henke, based on Manning and Schütze

  10. Instances in the Training Corpus:“inferior to ________” borrowed from Henke, based on Manning and Schütze

  11. Maximum Likelihood Estimate: borrowed from Henke, based on Manning and Schütze

  12. Maximum Likelihood Distribution = DML • probability is assigned exactly based on the n-gram count in the training corpus. • Anything not found in the training corpus gets probability 0.

  13. Actual Probability Distribution: borrowed from Henke, based on Manning and Schütze

  14. Conundrum • Do we stick very tight to the “Maximum Likelihood” model, assigning zero probability to sequences not seen in the training corpus? • Answer: we simply cannot; the results are just too bad.

  15. Smoothing • We need, therefore, some “smoothing” procedure • which adds some of the probability mass to unseen n-grams • and must therefore take away some of the probability mass from observed n-grams

  16. And linguistics? • The theory of syntax can be viewed as a contribution to the back-off conundrum: syntactic categories are the first back-off route, and linear distance may be less good than syntactic closeness for the conditioning words.

  17. Discounting, back-off, and deleted interpolation • These words all go with “smoothing”. • “Smoothing” describes the general problem we face: getting probability mass to the great unseen. • “Discounting” describes who we take probability mass away from, and how much….

  18. “Back-off” and “deleted interpolation” (a special case of linear interpolation) are the two standard ways of redistributing the probability mass taken away by discounting.

  19. Back-off and deleted interpolationfor a given context: What is probability of words {wi}i in the context: following “in the__” (e.g., pocket) ? Words that were found in this context get a probability a bit less thanand with backoff, the held-back probability mass is distributed over words in the context “the __”. And how?

  20. Probability mass is distributed over “the WORD” pretty much in proportion to how often each word appears in the context “the___”. But even there, we hold some of the probability mass, and assign it to all words independent of context.

  21. Deleted Interpolation • Is linear: for any word in context (e.g., pocket after in the), we choose three ls and take its probability to be the weighted average of the trigram, bigram, and unigram models:l1P(pocket|in the) + l2P(pocket|the) + l3P(pocket) • If we fixed the ls, we would only need to insist that they sum to 1.0. But…

  22. We don’t fix them: we allow them to vary, depending on the context (“in the”); we need to do some fancier calculations then (Expectation-Maximization).

  23. General ideas about discounting Three closely related ideas that are widely used.

  24. “Sum of counts” method of creating a distribution You can always get a distribution from a set of counts by dividing each count by the total count of the set. “bins”: name for the different preceding n-grams that we keep track of. Each bin gets a probability, and they must sum to 1.0

  25. Zero knowledge Suppose we give a count of 1 to every possible bin in our model. If our model is a bigram model, we give a count of 1 to the V2 conceivable bigrams. (V if unigram, V3 if trigram, etc.) Admittedly, this model assumes zero knowledge of the language…. We get a distribution for each bin by assigning probability 1/V2 to each bin. Call this distribution DN.

  26. Too much knowledge • Give each bin exactly the number of counts that it earns from the training corpus. • If we are making a bigram model, then there are V2 bins, and those bigrams that do not appear in the training corpus get a count of 0. • We get the Maximum Likelihood distribution by dividing by the total count = N.

  27. Laplace (“Adding one”) Add the bin counts from the Zero-knowledge case (1 for each bin, V2 of them in bigram case) and the bin counts from the Too-much knowledge (score in training corpus) • Divide by total number of counts = V2 + N • Formula: each bin gets probability (Count in corpus + 1) / (V2 + N)

  28. Lidstone’s Law Choose a number l, between 0 and 1, for the count in the NoKnowledge distribution. Then the count in each bin is Count in corpus + l And we assign probability to it (where the number of bins is V2, because we’re considering a bigram model: If l = 1 this is Laplace; If l = 0.5, this is Jeffrey-Perks Law If l = 0, this is Maximum Likelihood

  29. Another way to say this… • We can also think of Laplace as a weighted average of two distributions, the No Knowledge distribution and the MaximumLikelihood distribution…

  30. 2. Averaging distributions Remember this: If you take weighted averages of distributions of this form: l * distribution D1 + (1- l) * distribution D2 the result is a distribution: all the numbers sum to 1.0 This means that you split the probability mass between the two distributions (in proportion l/1- l) then divide up those smaller portions exactly according to D1 and D2.

  31. “Adding 1” (Laplace) Is it clear that

  32. this is a special case of l DN + (1- l )DML where l = V2/(V2+N). How big is this? if V= 50,000, then V2 = 2,500,000,000. This means that if our corpus is 2 and a half billion words, we are still reserving half of our probability mass for zero knowledge – that’s too much. l = V2/(V2+N) = 2,500,000,000/5,000,000,000 = 0.5

  33. Good-Turing discounting • The central problem is assigning probability mass to unseen examples, especially unseen bigrams (or trigrams), based on known vocabulary. • Good-Turing estimation says that a good estimate for the total probability of unseen n-grams is the total number of 1-grams seen = N1/N.

  34. So we take the probability mass assigned empirically to n-grams seen once, and assign it to all the unseen n-grams (we know how many there are: if the vocabulary is of size V, then there are Vn n-grams: if we have seen T distinct n-grams, then each unseen n-gram gets probability:

  35. So unseen n-grams got all of the probability mass that had been earned by the n-grams seen once. So the n-grams seen once will grab all of the probability mass earned by n-grams seen twice, then (uniformly) distributed:

  36. So n-grams seen twice will take all the probability mass earned by n-grams seen three times…and we stop this foolishness around the time when observed frequencies are reliable, around 10 times. all unseen ngrams Counts seen 1x seen 2x 3x 4x 5x pred 1x pred 2x 3x 4x 5x MODEL: assigns probabilities