- 2070 Asar · 4 marks
Correlation coefficient lies between −1 and +1
Short answer
The Pearson correlation coefficient measures linear association: r = Cov(X,Y)/(σx σy). Proof: let u = (x−x̄)/σx, v = (y−ȳ)/σy; Σu²/n=1, Σv²/n=1. Then Σ(u±v)² ≥ 0 gives 2 ± 2r ≥ 0, hence −1 ≤ r ≤ +1.
What do you mean by correlation coefficient? Show that correlation coefficient lies between -1 and +1.
Correlation coefficient
The (Karl Pearson) correlation coefficient measures the strength and direction of the linear relationship between two variables X and Y:
r = Cov(X, Y)/(σx σy) = Σ(x − x̄)(y − ȳ) / √[Σ(x − x̄)² Σ(y − ȳ)²]
r > 0 means variables move in the same direction, r < 0 in opposite directions, r = 0 means no linear relationship.
Proof that −1 ≤ r ≤ +1
Let u = (x − x̄)/σx and v = (y − ȳ)/σy. Then Σu²/n = 1, Σv²/n = 1 and Σuv/n = r.
Since a sum of squares is never negative,
Σ(u ± v)² ≥ 0
Σu² + Σv² ± 2Σuv ≥ 0
Dividing by n:
1 + 1 ± 2r ≥ 0
2 ± 2r ≥ 0, i.e. 1 ± r ≥ 0
Taking the upper sign: 1 + r ≥ 0, so r ≥ −1. Taking the lower sign: 1 − r ≥ 0, so r ≤ +1.
Hence −1 ≤ r ≤ +1.
Equality: r = +1 when Σ(u − v)² = 0, i.e. u = v for every point (perfect positive linear relation); r = −1 when Σ(u + v)² = 0, i.e. u = −v (perfect negative linear relation).
(Alternatively, by the Cauchy–Schwarz inequality, [Σ(x − x̄)(y − ȳ)]² ≤ Σ(x − x̄)² Σ(y − ȳ)², so r² ≤ 1.)
✓ Answer checked by Claude Sonnet 5.5
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