# In smoothing of n-gram model in NLP, why don't we consider start and end of sentence tokens?

When learning Add-1 smoothing, I found that somehow we are 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 give an example to explain.

Example:

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

Finally, we arrive at the

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

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 $$$$ and $$<\backslash s>$$ in our vocabulary. With this, our calculation becomes:

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

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 am missing here.

• 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..] Oct 9 '20 at 18:47