Is the correlation coefficient of the residuals related to the correlation coefficient of the original distirbution?

Question

I have 10 different 2D datasets which each have 200 datapoints. Each data point has a 'quality' property ($Q$) which somewhat depends on the area ($A$) of that datapoint. There is a (relatively) strong linear relationship between these two properties for most of the datasets (row 1 of the figure). Call the correlation coefficients here $r_1$.

I then calculate the residuals ($R = Q - \hat Q$) of the linear relationship between the quality and the area (row 2). There is still a relatively strong relationship between the residual and the quality (row 3). Call the correlation coefficients here $r_2$. [Subquestion: what does this mean about the relationship between quality and area?]

If I take the Pearson correlation coefficient of these two linear relationships, these very strongly follow the pattern $r_1^2 + r_2^2 = 1$. My main question: Is this mathematically expected?

I know $ r_1 = \frac{cov(Q, A)}{\sigma_Q \sigma_A}$ and $ r_2 = \frac{cov(Q, R)}{\sigma_Q \sigma_R}$ but I am not able to make any significant inroads as to a proof of the above equality. I have also tried experimenting with some random data, and it seems as though this relationship does not hold in general. So my other question is, what features of the data set will result in this property holding?

Answer 1

I ended up finding the proof using properties of the covariance.

Claim: Given two variables $A$ and $B$ and the linear regression between them $B = mA + c + R$ (where $R$ are the residuals and thus $cov(A, R) = 0$), $\rho_{A,B}^2 + \rho_{B,R}^2 = 1$.

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Proof: By definition, $\displaystyle \rho_{X,Y} = \frac{cov(X, Y)}{\sigma_X \sigma_Y}$.

Then,

\begin{eqnarray*} \rho_{A,B}^2 + \rho_{B,R}^2 &=& \frac{cov^2(A, B)}{\sigma^2_A \sigma^2_B} + \frac{cov^2(B, R)}{\sigma^2_B \sigma^2_R} \\ &=& \frac{cov^2(A, mA + c + R)}{\sigma^2_A \sigma^2_B} + \frac{cov^2(R, mA + c + R)}{\sigma^2_R \sigma^2_B} \\ &=& \frac{[mcov(A,A) + cov(A,R)]^2}{\sigma^2_A \sigma^2_B} + \frac{[mcov(A,R) + cov(R,R)]^2}{\sigma^2_R \sigma^2_B} \\ &=& \frac{[m \sigma^2_A]^2}{\sigma^2_A \sigma^2_B} + \frac{[\sigma^2_R]^2}{\sigma^2_R \sigma^2_B} \\ &=& \frac{m^2 \sigma^2_A}{\sigma^2_B} + \frac{\sigma^2_R}{\sigma^2_B} \\ &=& \frac{m^2 \sigma^2_A + \sigma^2_R}{cov(mA+c+R, mA+c+R)}\\ &=& \frac{m^2 \sigma^2_A + \sigma^2_R}{m^2 cov(A,A) + 2mcov(A,R) + cov(R,R)}\\ &=& \frac{m^2 \sigma^2_A + \sigma^2_R}{m^2 \sigma^2_A + \sigma^2_R}\\ &=& 1 \end{eqnarray*}