Let $A, B, C$ be three random variables. Suppose their Pearson correlation coefficients are $\rho(A, B) = \frac{1}{2}$ and $\rho(B, C) = \frac{1}{2}$, what is the possible range for $\rho(A, C)$?
I thought of two ways to compute this range.
Approach 1 (using cosine):
$\rho(A, B) = \frac{1}{2}$ means that the angle between $A$ and $B$ is $arccos(\frac{1}{2}) = 30$ degrees. Similarly, the angle between $B$ and $C$ is 30 degrees. Therefore, the angle between $A$ and $C$ can range from 0 degree to 60 degrees. Applying the cosine function to both bounds, we know that $\frac{1}{2} = cos(60^o) \leq \rho(A, C) \leq cos(0^o) = 1$.
In summary, $\frac{1}{2} \leq \rho(A, C) \leq 1$ is the desired range.
Approach 2 (using the positive semi-definite property of the correlation matrix):
Recall that correlation matrices are positive semi-definite. The Sylvester's Criterion says that a matrix is positive semi-definite if and only if all of its principal minors are non-negative.
In our case, our correlation matrix is \begin{align*} \begin{pmatrix} 1 & \frac{1}{2} & \rho(A, C) \newline \frac{1}{2} & 1 & \frac{1}{2} \newline \rho(A, C) & \frac{1}{2} & 1 \end{pmatrix} \end{align*} assuming we order the rows/columns by A, B, and then C.
Therefore, we need \begin{align*} \begin{vmatrix} 1 & \rho(A, C) \newline \rho(A, C) & 1 \end{vmatrix} \geq 0 \text{ and also } \begin{vmatrix} 1 & \frac{1}{2} & \rho(A, C) \newline \frac{1}{2} & 1 & \frac{1}{2} \newline \rho(A, C) & \frac{1}{2} & 1 \end{vmatrix} \geq 0 \end{align*}
Solving for $\rho(A, C)$, I get that $-\frac{1}{2} \leq \rho(A, C) \leq 1$, which is slightly different from what I get in the cosine approach. The lower bound is $-\frac{1}{2}$ instead of $\frac{1}{2}$
I am confused. Which one is correct? Why are they different?
