Unit V · Lecture 36 of 40 · B.Sc. Data Science, 1st Year, Semester II
Two attributes A and B are said to be independent if the possession (or non-possession) of one attribute does not affect the probability of possessing the other. Mathematically, A and B are independent if and only if:
(AB) = (A) × (B) / N
That is, the observed joint frequency equals the expected frequency under independence.
If 60% of a town is literate and 50% is employed, then if literacy and employment are independent, we'd expect 60% × 50% = 30% of the town to be both literate AND employed. If the actual percentage is close to 30%, the attributes are independent — knowing someone is literate doesn't change the chance they're employed. If the actual percentage is very different from 30%, the attributes are associated — there's a connection.
If A and B are independent, the expected (theoretical) frequency of each cross class is:
E(AB) = (A)(B) / N
E(Aβ) = (A)(β) / N
E(αB) = (α)(B) / N
E(αβ) = (α)(β) / N
These are the frequencies we'd expect if knowing A tells us nothing about B.
A and B are independent if any one of these holds:
(AB) = (A)(B)/N
(Aβ) = (A)(β)/N
(αB) = (α)(B)/N
(αβ) = (α)(β)/N
(If one holds, all four hold automatically.)
Two attributes A and B are associated if they are not independent — that is, if the observed joint frequency (AB) differs from the expected frequency (A)(B)/N. Association means that knowing whether a person has attribute A gives information about whether they have attribute B.
(AB) > (A)(B)/N
The attributes occur together more often than expected by chance. Having A makes it more likely to have B.
Example: Smoking and lung cancer — smokers are more likely to get lung cancer than non-smokers.
Also means: (αβ) > (α)(β)/N — the absences also go together.
(AB) < (A)(B)/N
The attributes occur together less often than expected. Having A makes it less likely to have B.
Example: Exercise and obesity — exercisers are less likely to be obese.
Also means: (Aβ) > (A)(β)/N — presence of A goes with absence of B.
Independent: (AB) = (A)(B)/N → Observed = Expected
Positive association: (AB) > (A)(B)/N → Observed > Expected
Negative association: (AB) < (A)(B)/N → Observed < Expected
In a survey of 400 people: (A) = 200, (B) = 160, (AB) = 100. Are A and B independent?
Expected frequency under independence:
E(AB) = (A)(B)/N = 200 × 160 / 400 = 80
Observed frequency: (AB) = 100
Since (AB) = 100 > 80 = E(AB), the attributes A and B are positively associated (not independent).
They occur together more often than expected by chance.
| (B) | (β) | Total | |
|---|---|---|---|
| (A) | 100 (exp: 80) | 100 (exp: 120) | 200 |
| (α) | 60 (exp: 80) | 140 (exp: 120) | 200 |
| Total | 160 | 240 | 400 |
In a group of 500: (A) = 250, (B) = 200, (AB) = 100. Are A and B independent?
E(AB) = (A)(B)/N = 250 × 200 / 500 = 100
Observed (AB) = 100
Since (AB) = E(AB) = 100, the attributes A and B are independent.
Knowing whether someone has A gives no information about whether they have B.
| (B) | (β) | Total | |
|---|---|---|---|
| (A) | 100 | 150 | 250 |
| (α) | 100 | 150 | 250 |
| Total | 200 | 300 | 500 |
In a study of 300 patients: (A) = 150 (received treatment), (B) = 100 (recovered), (AB) = 20 (received treatment AND recovered). Determine the type of relationship.
E(AB) = (A)(B)/N = 150 × 100 / 300 = 50
Observed (AB) = 20
Since (AB) = 20 < 50 = E(AB), the attributes are negatively associated.
Patients who received the treatment recovered less often than expected by chance. (This would be a very concerning result for the treatment!)
Three attributes A, B, C are mutually independent if:
(ABC) = (A)(B)(C) / N²
And all pairwise expected frequencies also match:
(AB) = (A)(B)/N, (AC) = (A)(C)/N, (BC) = (B)(C)/N
A, B, C are pairwise independent if each pair is independent:
(AB) = (A)(B)/N
(AC) = (A)(C)/N
(BC) = (B)(C)/N
Note: Pairwise independence does NOT guarantee mutual independence. All three conditions plus the triple condition must hold.
1. Complete independence: All pairs and the triple are independent.
2. Pairwise association: Some or all pairs are associated, but the triple may or may not be.
3. Conditional independence: A and B are independent when C is held fixed, but may appear associated overall (Simpson's paradox).
Q1: A and B are independent if:
Q2: If (AB) > (A)(B)/N, the attributes are:
Q3: N=200, (A)=100, (B)=80. If A and B are independent, (AB) should be:
Q4: Positive association means:
Q5: For mutual independence of three attributes, we need:
1. N=500, (A)=250, (B)=200, (AB)=120. Test for independence. If not independent, what type of association?
2. N=400, (A)=200, (B)=180, (AB)=90. Test for independence.
3. N=600, (A)=300, (B)=240, (AB)=80. Test for independence and interpret.
4. If A and B are independent and (A)=150, (B)=120, N=300, find all four cross-class frequencies.
5. Explain why pairwise independence of three attributes does not guarantee mutual independence. Give a conceptual example.
Measures of Association & Contingency Tables — Quantifying the strength of association