Natural Language Processing · M2 · mocks

Q5 (3 marks)

M2 Mock C · mixed numerical practice

M2 · confirmed 2 + 2 + any two of 3 + 3 + 3 · worked practice

Q5

In a WSD training set, financial bank and river bank each have two labelled contexts. Token totals are N(F)=6 and N(R)=6. Vocabulary size is 8. Counts are C(loan,F)=2, C(loan,R)=0, C(water,F)=0, C(water,R)=3. Use add-one Naive Bayes to classify context {loan,water}. (3 marks)

Given

P(F)=P(R)=2/4=0.5; |V|=8; both class token totals=6.

To find

Smoothed scores and winning sense.

Formula and why it applies

P(w\mid c)=\frac{C(w,c)+1}{N_c+|V|}LaTeX: P(w\mid c)=\frac{C(w,c)+1}{N_c+|V|}
s(c)=P(c)\prod_wP(w\mid c)LaTeX: s(c)=P(c)\prod_wP(w\mid c)

Class priors use labelled-context counts; word likelihood denominators use class token totals. Do not exchange those two totals.

Solution

Step 1. Find the denominator

6+8=14LaTeX: 6+8=14

Step 2. Financial likelihoods

P(loan\mid F)=\frac3{14},\quad P(water\mid F)=\frac1{14}LaTeX: P(loan\mid F)=\frac3{14},\quad P(water\mid F)=\frac1{14}

Step 3. River likelihoods

P(loan\mid R)=\frac1{14},\quad P(water\mid R)=\frac4{14}LaTeX: P(loan\mid R)=\frac1{14},\quad P(water\mid R)=\frac4{14}

Step 4. Compute financial score

s(F)=\frac12\times\frac3{14}\times\frac1{14}=\frac3{392}\approx0.007653061LaTeX: s(F)=\frac12\times\frac3{14}\times\frac1{14}=\frac3{392}\approx0.007653061

Step 5. Compute river score

s(R)=\frac12\times\frac1{14}\times\frac4{14}=\frac4{392}\approx0.010204082LaTeX: s(R)=\frac12\times\frac1{14}\times\frac4{14}=\frac4{392}\approx0.010204082

Step 6. Choose

4/392 > 3/392; the water evidence outweighs the loan evidence under these counts.

River sense wins. If normalised, P(R|context)=4/7≈0.571429.