corr(x,y) = a, corr(y,z) = b, corr(x,z) = ? from $cos(\alpha+\beta)$ why it is a range.

Question

Question: $$corr(x,y) = a, corr(y,z) = b, corr(x,z) = ?$$ Here $corr$ is the correlation.

I know the answer

$$cos(\alpha + \beta) \leq corr(x,z) \leq cos(\alpha - \beta), \textrm{where } cos(\alpha) = a, cos(\beta) = b.$$

And the idea is making the correlation matrix positive semi-definite: $$ \left|\begin{pmatrix} 1 & a & x \\ a & 1& b \\ x & b& 1 \\ \end{pmatrix}\right| \geq 0$$ But from the point view of sample vectors $\textbf{x}, \textbf{y}, \textbf{z},$ without loss of generality , we assume they are all mean 0. Then the correlation is the $cos$ angle of two vectors. Thus $$corr(x,z) = cos(\alpha - \beta)\textrm{ or } cos(\alpha + \beta).$$

Why will it be a range? Can anyone point my mistake?

Seems you use corr(x,y)=a and b at same time. Intentionally? — coffeemath, Oct 07 '20 at 14:26
@coffeemath sorry, it's a typo, have corrected it, I don't understand from the point view of geometry, how can it be a range but not two numbers. — user6703592, Oct 07 '20 at 16:47
@user6703592, Do you have a proof why this is the case. I don't understand why it is cosine — YoYO Man, Nov 16 '20 at 04:00

J.G. · Accepted Answer · 2020-10-07T17:40:50.780

3

While random variables comprise a vector space (I've discussed this perspective in detail before), it's not $2$-dimensional, so the third angle needn't be $\alpha\pm\beta$. If you want to visualize this, it's best to consider the in-face angles meeting at the apex of a tetrahedron, whose other vertices' positions relative to that apex are the three vectors $x,\,y,\,z$.

edited Oct 07 '20 at 17:40

answered Oct 07 '20 at 17:28

J.G.

118,053

exactly! I am restricted within the 2-dim space. Thanks. – user6703592 Oct 07 '20 at 17:37

corr(x,y) = a, corr(y,z) = b, corr(x,z) = ? from $cos(\alpha+\beta)$ why it is a range.

1 Answers1