Properties of coefficient of correlation

2. Properties of coefficient of correlation:-
1) The coefficient of correlation always lies between -1 and
+1 i.e, −1 ≤ 𝑟 ≤ +1
2) The correlation coefficient is symmetrical with respect to
X and Y i.e 𝑟𝑥𝑦 = 𝑟𝑦𝑥
3) The coefficient of correlation is the geomatric mean of the
two regression coefficient.
r = √𝑏 × 𝑑 Or r = √ 𝑏 𝑦𝑥 × 𝑏 𝑥𝑦
4) It does not depend upon the units employed
5) It is independent of orgin and unit of measurement
6) The coefficient of cerrelation is unaffected by change of
origin and scale i.e 𝑟𝑥𝑦 = 𝑟𝑢𝑣
7) The coefficient of cerrelation is a pure number.

3. Example-7:
i) Calculate regression co-efficient by𝑥 and 𝑏𝑥𝑦 and
calculate correlation with the help of regression
coefficients for the following pairs of observations.
ii) Calculate Karl Pearson’s coefficient of correlation and
then verify that.
X 1 2 3 4 5 6 7 8
Y 12 14 16 18 20 22 24 26
Solution:
We know that the correlation coefficient is the geometric mean of the
two regression coefficients.
𝑟 = √𝑏𝑦𝑥 × 𝑏𝑥𝑦
𝑏𝑦𝑥 =
𝑛∑𝑥𝑦−∑𝑥∑𝑦
𝑛∑𝑥2−(∑𝑥)2
𝑏𝑥𝑦 =
𝑛∑𝑥𝑦−∑𝑥∑𝑦
𝑛∑𝑦2−(∑𝑦)2
The necessary calculations for regression coefficients are given
below.
𝒙 𝒚 𝒙𝒚 𝒙 𝟐
𝒚 𝟐
1 12 12 1 144
2 14 28 4 196
3 16 48 9 256
4 18 72 16 324
5 20 100 25 400
6 22 132 36 484
7 24 168 49 576
8 26 208 64 676
∑x=36 ∑y=152 ∑xy=768 ∑x2
=204 ∑y2
=3056

4. 𝑏𝑦𝑥 =
𝑛∑𝑥𝑦−∑𝑥∑𝑦
𝑛∑𝑥2−(∑𝑥)2
𝑏𝑦𝑥 =
8(768)−(36)(152)
8(204)−(36)2
𝑏𝑦𝑥 =
6144−5472
1632−1296
𝑏𝑦𝑥 =
672
336
𝑏𝑦𝑥 = 2
𝑏𝑥𝑦 =
𝑛∑𝑥𝑦−∑𝑥∑𝑦
𝑛∑𝑦2−(∑𝑦)2
𝑏𝑥𝑦 =
8(768)−(36)(152)
8(3056)−(152)2
𝑏𝑥𝑦 =
6144−5472
24448−23104
𝑏𝑥𝑦 =
672
1344
𝑏𝑥𝑦 = 0.5
We know that correlation coefficient is the geometric mean of the two
regression coefficients i.e.
𝑟 = √𝑏𝑦𝑥 × 𝑏𝑥𝑦
𝑟 = √2 × 0.5
𝒓 = 𝟏

5. (ii) Karl Pearson’s co-efficient of correlation.
𝑟 =
𝑛∑𝑥𝑦−∑𝑥∑𝑦
√(𝑛∑𝑥2−(∑𝑥)2)(𝑛∑𝑦2−(∑𝑦)2)
𝑟 =
8(768)−(36)(152)
√(8(204)−(36)2)(8(3056)−(15)2)
𝑟 =
6144−5472
√(1632−1296)(24448−23104)
𝑟 =
672
√(336)(1344)
𝑟 =
672
√451584
𝑟 =
672
672
𝒓 = 𝟏
Hence Proved
𝒓 = √𝒃𝒚𝒙 × 𝒃𝒙𝒚

6. Example-8:
If 𝑏 𝑦𝑥 = 51.9 and 𝑏 𝑥𝑦 = 0.019
Find coefficient of determination
Solution:
𝑟𝑥𝑦
2
= 𝑏 𝑦𝑥 × 𝑏 𝑥𝑦 × 100
𝑟𝑥𝑦
2
= (51.9)(0.019) × 100
𝒓 𝒙𝒚
𝟐
= 𝟗𝟖. 𝟔𝟏%
It means that 98.61% of the variation in the 𝑦-variable is explained or
accounted for 𝑏𝑦 variation in the 𝑥-variable.
Example-9:
For the following two sets, the regression lines for each set are
respectively.
i) 𝑦 = 1.94𝑥 + 10.83 (𝑦 𝑜𝑛 𝑥) and
𝑥 = 0.15𝑦 + 6.18 (𝑥 𝑜𝑛 𝑦)
ii) 𝑦 = −1.96𝑥 + 15 (𝑦 𝑜𝑛 𝑥) and
𝑥 = −0.45𝑦 + 7.16 (𝑥 𝑜𝑛 𝑦)
Find coefficient of correlation in each case.
Solution:
i) Regression coefficient 𝑦 on 𝑥 (𝑏𝑦𝑥) = 1.94
Regression coefficient 𝑥 on 𝑦 (𝑏𝑥𝑦) = 0.15
𝑟 = √𝑏𝑦𝑥 × 𝑏𝑥𝑦
𝑟 = √(1.94)(0.15)
𝒓 = 𝟎. 𝟓𝟒

7. ii) Regression coefficient 𝑦 on 𝑥 (𝑏𝑦𝑥) = −1.96
Regression coefficient 𝑥 on 𝑦 (𝑏𝑥𝑦) = −0.45
𝑟 = √𝑏𝑦𝑥 × 𝑏𝑥𝑦
𝑟 = √(1.96)(0.45)
𝒓 = −𝟎. 𝟗𝟒
It is to be noted when both regression coefficients are negative then
“𝑟” is also negative.
Example-10:
If 𝑛 = 18, ∑𝑥 = 638, ∑𝑦 = 41, ∑𝑥𝑦 = 1569.5, ∑𝑥2
= 25814,
∑𝑦2
= 101.45
i) Find simple coefficient of correlation.
ii) If 𝑢 =
𝑥−3
5
and 𝑣 =
𝑦
20
then what would be the coefficient of
correlation between 𝑢 and 𝑣.
Solution:
𝑟 =
𝑛∑𝑥𝑦 − (∑𝑥)(∑𝑦)
√{𝑛(∑𝑥2) − (∑𝑥)2}{𝑛(∑𝑦2) − (∑𝑦)2}
𝑟 =
18(1569.5) − (638)(41)
√{18(25814) − (638)2}{18(101.45) − (41)2}
𝑟 =
28251 − 26158
√(464652 − 407044)(18261 − 1681)

8. 𝑟 =
2093
√(57608)(145.1)
𝑟 =
2093
√8358920.8
𝑟 =
2093
2891.18
𝒓 = 𝟎. 𝟕𝟐
(ii)Correlation is unaffected by the change of origin and scale.
i.e.
𝑟𝑢𝑣 = 𝑟𝑥𝑦
𝒓 𝒖𝒗 = 𝟎. 𝟕𝟐