If I know the correlation between two pairs of random variables (A,B) and (B,C), can I determine the correlation of the pair (A,C)? If not, can I at least constrain it to some range?
I'm interesting in generating a covariance matrix where certain pairwise correlation values are determined, and the rest are as small (i.e. close to 0) as possible.
Best Answer
Unfortunately no, we cannot determine $\rho_{AC}$ given just $\rho_{AB}$ and $\rho_{BC}$. You can derive the theoretical bounds \begin{align*} \rho_{AC} \ge \max\{2(\rho_{AB} + \rho_{BC}) - 3, 2\rho_{AB}\rho_{BC} - 1\} \end{align*} Proof. Some notation. I let $\sigma_{AB} = \text{Cov}(A,B)$ and $\sigma_A^2 = \text{Var}(A)$.
Let's first prove $\rho_{AC} \ge 2(\rho_{AB} + \rho_{BC}) - 3$. Recall the identity \begin{align*} 2 E[X^2] + 2E[Y^2] = E[(X+Y)^2] + E[(X-Y)^2] \end{align*} hence $2E[Y^2] \le E[(X+Y)^2] + E[(X-Y)^2]$. Set \begin{align*} X = \widetilde{B} - (\widetilde{A} + \widetilde{C})/2 \quad \text{and} \quad Y = (\widetilde{A} - \widetilde{C})/2 \end{align*} where $\widetilde{C} = (C - E[C])/\sigma_C$, the normalized random variable, and similarly for $\widetilde{A}, \widetilde{B}$. Upon substitution and simplification, we get \begin{align*} \frac{1}{2}(2 - 2\rho_{AC}) \le (2 - 2\rho_{AB}) + (2 - 2\rho_{BC}) \iff \rho_{AC} \ge 2(\rho_{AB} + \rho_{BC}) - 3 \end{align*} To prove $\rho_{AC} \ge 2\rho_{AB}\rho_{BC} - 1$, consider the random variable \begin{align*} W = 2 \frac{\sigma_{AB}}{\sigma_B^2}B - A \end{align*} We can verify $\sigma_W^2 = \sigma_A^2$, and hence $\sigma_{WC} \le \sigma_{W}\sigma_{C} = \sigma_ A \sigma_C$ by the Cauchy-Schwarz inequality. On the other hand, you may compute \begin{align*} \sigma_{WC} = 2 \frac{\sigma_{AB}}{\sigma_B^2}\sigma_{BC} - \sigma_{AC} \end{align*} Reorganizing all this, we prove $\rho_{AC} \ge 2\rho_{AB}\rho_{BC} - 1$.