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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.