# Relationship between different types of correlation coefficients

Let,

$$r_{1(2.34...p)}$$ = Correlation between $$x_1$$ and $$x_{2.34...p}$$. The latter being the residuals after regressing $$x_2$$ on $$x_3 , x_4 ....x_p$$.

$$r_{1.234..p}$$ = Multiple correlation coefficient of regressing $$x_1$$ on $$x_2 , x_3, x_4....x_p$$

Prove that -

$${r_{1.23...p}}^2 = {r_{1p}}^2 + {r_{1(p-1.p)}}^2 + ...... + {r_{1(2.34...p)}}^2$$

I tried writing the $$correlation^2$$ coefficients first in terms of $$covariance^2$$ by variance*variance. Variance of $$x_1$$ will cancel out from both the sides. Then I tried substituting all the residuals/fitted values in the covariances with linear combinations of $${x_i}'s$$, but to no avail. How to prove this equality?

$${r_{1.23...p}}^2 = {r_{1p}}^2 + {r_{1(p-1.p)}}^2 + ...... + {r_{1(2.34...p)}}^2$$

$$r_{12.34...p}$$ = Partial Correlation between 1 and 2 removing the effects of 3,4,...p.

$$x_{1.34...p}$$ = Residuals of 1 after regressing on 3,4,...p

$$s_{11.34....p}$$ = Variance of residuals of 2 after regressing on 3,4...p

$$r_{12.34...p}^2$$ = $$\left({Cov(x_{1.34...p},x_{2.34...p})}\over {\sqrt(s_{11.34....p}) \sqrt(s_{22.34...p})}\right)^2$$

$$r_{12.34...p}^2$$ = $$\left({Cov(x_{1},x_{2.34...p})}\over {\sqrt(s_{11.34....p}) \sqrt(s_{22.34...p})}\right)^2$$

Because, normal equations, residuals of $$x_1$$ with $$x_3,x_4...x_p$$ will give zero when multiplied with $$x_3,x_4....x_p$$ Multiplying dividing with $$(\sqrt(s_{11}))^2$$

$$r_{12.34...p}^2$$ = $$(r_{1(2.34...p)})^2 \times s_{11} \over s_{11.34....p}$$

Now,

$$x_{1.23....p}$$ = Values of $$x_1$$ regressed on $$x_2, x_3....x_p$$

We look at $$\sum_{i} ((x_{1.23....p})_i)^2 = \sum_{i} ((x_1)_i)\times((x_{1.23....p})_i)$$

= $$\sum_{i} ((x_{1.34...p})_i)\times ((x_{1.23....p})_i)$$

Writing $$((x_{1.23....p})_i) = ((x_1)_i) - \sum_{j=2}^{p}b_j \times ((x_j)_i)$$

= $$\sum_{i} ((x_{1.34...p})_i)\times((x_1)_i) - \sum_{j=2}^{p}((x_{1.34...p})_i)\times b_j \times ((x_j)_i)$$

= $$\sum_{i} ((x_{1.34...p})_i)\times((x_1)_i) - ((x_{1.34...p})_i)\times b_2 \times ((x_2)_i)$$

We know, $$b_2 = b_{12.34...p}$$ ( Coefficient of $$x_2$$ when $$x_1$$ is regressed on $$x_2,x_3...x_p$$ is same as partial relation coefficient between residuals of $$x_1$$ and $$x_2$$ after removing the effects of $$x_3,x_4...x_p$$

= $$\sum_{i} ((x_{1.34...p})_i)\times((x_1)_i) - ((x_{1.34...p})_i)\times b_{12.34...p} \times ((x_2)_i)$$

= $$\sum_{i} ((x_{1.34...p})_i)\times((x_1)_i) - ((x_{1.34...p})_i)\times b_{12.34...p} \times ((x_{2.34...p})_i)$$

= $$\sum_{i} (x_{1.34...p})_i) \times (((x_1)_i) - b_{12.34...p} \times (((x_{2.34...p})_i)$$

= $$\sum_{i} ((x_{1.34...p})_i))^2 - ((x_{1})_i)\times b_{12.34...p} \times ((x_{2.34...p})_i)$$

So,

$$s_{11.23...p} = s_{11.34...p} - b_{12,34,,,p} \times \sum_{i} ((x_1)_i) \times ((x_{2.34...p})_i)$$

Using

$$b_{12.34...p}$$ = $$r_{12.34...p} \sqrt{s_{11.34...p}} \over \sqrt{s_{22.34...p}}$$

1 - $${r_{12.34...p}}^2$$ = $$s_{11.23...p} \over s_{11.34..p}$$

and

1 - $${r_{1.23...p}}^2$$ = $$s_{11.23...p} \over s_{11}$$

in the two equations derived above, cancelling and manipulating, we will get the desired result.