0
$\begingroup$

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?

$\endgroup$
0
$\begingroup$

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

$r_{12.34...p}$ = 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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.