When learning Add-1 smoothing, I found that somehow we're adding 1 to each word in our vocabulary but not considering start-of-sentence and end-of-sentence as two words in the vocabulary. Let me throw an example to explain.


Assume we've a corpus of three sentences: "John read Moby Dick", "Mary read a different book", and "She read a book by Cher". After training our bi-gram model on this corpus of three sentences, we need to evaluate the probability of a sentence "John read a book", i.e. to find $P(John\; read\; a\; book)$

To differentiate John appearing anywhere in a sentence from it's appearance at the beginning, and likewise for book appearing at the end, we rather try to find $P(<s>John\; read\; a\; book<\backslash s>)$ after introducing two more words $<s>$ and $<\backslash s>$, indicating start of a sentence, and end of a sentence respectively.

Finally, we arrive at the

$P(<s>John\; read\; a\; book<\backslash s>)$ as $P(John|<s>)P(read|John)P(a|read)P(book|a)P(<\backslash s>|book)=\frac{1}{3}\frac{1}{1}\frac{2}{3}\frac{1}{2}\frac{1}{2}$

My Question: Now, to find $P(Cher\; read\; a\; book)$, using Add-1 smoothing (Laplace smoothing) shouldn't we add the word 'Cher' that appears first in a sentence? And to that, we must add $<s>$ and $<\backslash s>$ in our vocabulary. With this, our calculation becomes

$P(Cher|<s>)P(read|Cher)P(a|read)P(book|a)P(<\backslash s>|book)=\frac{0+1}{3+13}\frac{0+1}{1+13}\frac{2+1}{3+13}\frac{1+1}{2+13}\frac{1+1}{2+13}$

The 13 added to each numerator is due to the unique word count of the vocabulary which has 11 English words from our 3-sentence corpus plus 2 tokens - start and end of a sentence. In few places, I see 11 is added instead of 13 to the numerator, wondering what I'm missing here!


It depends on the definition of vocabulary (V).

Most teaching examples only include words in the vocabulary for simplicity. Start and stop of sentences tags can also included in the vocabulary.

Vocabulary can also include punctuation, or stop words can be removed from the vocabulary.

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    $\begingroup$ AFAIK, what happens underneath during a Laplace smoothing is that, we add 1 for each cell in a $N\times N$ matrix (since, our model is 2-gram) - each word maps to a row and a column. Assuming N=11 and after adding 1 word to each 11x11 cells, P(a|read), for example, is then $\frac{2+1}{3+11}$. For the row corresponding to the word 'a' has 11 columns and each is collecting 1 new word - this explains the increment by 11 in the denominator. The numerator increases by one, because one column out of the 11 for the row 'a', corresponds to column 'read'. [contd..] $\endgroup$ – KGhatak Oct 9 at 18:47
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    $\begingroup$ However, without any row for <s> how can we have a cell that can corresponds to numerator of P(Cher|<s>). Those in favor of not not including <s> and <\s> in the dictionary, how would they explain this! $\endgroup$ – KGhatak Oct 9 at 18:47

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