Natural Language Processing · TEE · formulas

Unigram estimation and complete worked examples

N-grams & probability

Formulas explained through full worked examples

Maximum likelihood estimation (MLE) chooses probabilities equal to observed relative frequencies. In a unigram model, every word is predicted independently of its surrounding words. If a word occurs c times in N training tokens, its estimate is c/N. To score m specified words, multiply their m unigram probabilities.

P_{\mathrm{MLE}}(w)=\frac{C(w)}{N}\qquad P(w_1,\ldots,w_m)\approx\prod_{i=1}^{m}P(w_i)

Choose a tokenization convention before counting. Lowercasing, punctuation handling and boundary markers can change the vocabulary and counts. Never count words with one convention and form the denominator with another. A unigram model gives the same product to rearrangements of the same word multiset; that is one of its limitations.

Question. Train an unpadded, case-insensitive unigram model on “I like tea” and “I like coffee”. Find the probability of “I like tea”.

Given

Training sentences: I like tea; I like coffee. Use lowercase words; exclude punctuation and sentence boundaries.

To find

The vocabulary, word probabilities and test-sentence probability.

Formula and why it applies

P(w)=C(w)/N,\qquad P(I\ like\ tea)=P(I)P(like)P(tea)

The unigram approximation ignores order and uses total token count for every denominator.

Solution

Step 1. Count every training token

Each sentence contains three words, so N = 3+3 = 6.

Step 2. Count types and frequencies

Word	Count	MLE probability
i	2	2/6 = 1/3
like	2	2/6 = 1/3
tea	1	1/6
coffee	1	1/6

Vocabulary size V = 4.

Step 3. Write one factor per test word

P=\frac{2}{6}\times\frac{2}{6}\times\frac{1}{6}

Step 4. Multiply the fractions

P=\frac{4}{216}=\frac{1}{54}\approx0.018519

Final answer. The unigram sentence probability is approximately 0.018519.

Question. A corpus contains counts a: 4, b: 3 and c: 1. Find the unigram probability of “a b a”.

Given

C(a)=4, C(b)=3, C(c)=1. No boundary markers.

To find

P(a b a).

Formula and why it applies

P(a\,b\,a)=P(a)P(b)P(a)

A repeated test word contributes a probability factor at every occurrence; do not include it just once.

Solution

Step 1. Find the total

N=4+3+1=8

Step 2. Estimate the required probabilities

P(a)=4/8=1/2,\qquad P(b)=3/8

Step 3. Use both occurrences of a

P=\frac12\times\frac38\times\frac12

Step 4. Simplify

P=\frac3{32}=0.09375

Final answer. P(a b a) = 0.09375.

Yaad rakho: Repeated word ka probability bhi repeat multiply hoga.

Question. An unsmoothed unigram model is trained on “red blue red”. Find the probability of “red green”.

Given

N=3. Counts: red=2, blue=1, green=0. The test contains red and green.

To find

The unsmoothed test probability and the reason for its value.

Formula and why it applies

P(red\ green)=P(red)P(green)

MLE assigns zero to an unseen outcome. Smoothing is needed if an unseen in-vocabulary item must receive probability.

Solution

Step 1. Estimate red

P(red)=2/3

Step 2. Estimate the unseen word

P(green)=0/3=0

Step 3. Multiply

(2/3)\times0=0

Final answer. Unsmoothed probability = 0. This reflects missing training evidence, not proof that the phrase is impossible.