prove that coefficient of correlation always lies between -1 to +1

coefficient of correlation = r = cov(x,y)σxσy
σx2= (x-x¯)2n

σy2= (y-y¯)2n

cov(x,y) = (x-x¯)(y-y¯)n

we have to prove -1r1
let us consider two terms x-x¯σx and y-y¯σy

take sum of the square of these two variables
x-x¯σx±y-y¯σy2
since it is a square term then it is surely a positive number
that is x-x¯σx±y-y¯σy20
opening its square we get
x-x¯σx2+y-y¯σy2+ 2x-x¯σxy-y¯σy0
on further simplifying it, we get
‚Äč(x-x¯)2σ2x+(y-y¯)2σ2y±2(x-x¯)(y-y)¯σxσy0
divide the entire expression by n
(x-x¯)2nσ2x+(y-y¯)2nσ2y±2(x-x¯)(y-y)¯nσxσy0
using the formula of variance of x, variance of y and covariance of x and y, we get
σ2xσ2x+σ2yσ2y±2cov(x,y)σxσy0

1+1±2r0

2+2r02-2r1-rr-1or2-2r022r1rtherefore we conclude that-1r1always.

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