Correlation between three variables

Question

Here is my question. If A and B has a correlation of 0.8, B and C has a correlation of 0.6. What is the range of correlation between A and C?

Answer 1

Let $A=(a_1,\ldots,a_n)$.

Let $\bar a=(a_1+\cdots+a_n)/n$.

Let $A_\text{corrected} = (a_1,\ldots,a_n) - (\bar a,\ldots,\bar a)$.

Let $\displaystyle \operatorname{ss}_A^2 = \langle A_\text{corrected},A_\text{corrected}\rangle$ where $\langle\cdot, \cdot\rangle$ is the dot product, the usual inner product and $\operatorname{ss}$ stands for "sum of squares". And let $\operatorname{ss}_A = \sqrt{\operatorname{ss}_A}$$.

Then $\displaystyle \left\langle \frac{A_\text{corrected}}{\operatorname{ss}_A}, \frac{A_\text{corrected}}{\operatorname{ss}_A} \right\rangle = 1$.

And similarly for $B$ and $C$.

Then the correlation is $\displaystyle \operatorname{corr}(A,B) = \left\langle \frac{A_\text{corrected}}{\operatorname{ss}_A}, \frac{B_\text{corrected}}{\operatorname{ss}_B} \right\rangle = \cos\theta_{AB}$, where $\theta_{AB}$ is the angle between these two unit vectors.

Then $\theta_{AB} = \arccos 0.8$ and $\theta_{BC}= \arccos0.6$.

The correlation $\cos\theta_{AC}$ must be between $\cos(\theta_{BC} \pm \theta_{AB})$. Notice that $\theta_{AB}+\theta_{BC} = 90^\circ$ because of the Pythagorean theorem.

This puts the correlation between $0$ and $0.96$ --- not a very narrow range, but it rules out a negative correlation between $A$ and $C$.

Here I've dealt only with a finite sample in which I've assigned each point a weight of $1/n$, but the whole thing can also work for discrete distributions on infinite set and for continuous distributions and other distributions. One would use integrals rather than sums for continuous distributions.