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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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