Inferring covariance cov[X,Z] from cov[X,Y] and cov[Y,Z] of known distributions

Question

Suppose X, Y and Z are real random variables of known distributions. If one knows the covariance $COV(X,Y)$ and $COV(Y,Z)$, is it possible to infer $COV(X,Z)$?

Answer 1

In general, no. For instance, if the known distributions are "$\psi_{a,b}$ nonzero only on an interval $[a, b]$, with value $(b-a)^{-1}$", then letting $$ X = \psi_{-1, 1} \\ Y = \psi_{0,2},\\ Z = \psi_{1, 3} $$ we have that the correlations of $Y$ with $X$ and $Z$ are equal, but $cov(X, Z)= 0$. On the other hand, if we let $Z = X$, we have the same equality of the original covariances, but $cov(X, Z) = 1$.

Answer 2

No. $COV(A,B)=E[AB]-E[A]\cdot E[B]$. Say you know $E[X]$, $E[Y]$, and $E[Z]$ since they are known distributions. Also you know $E[XY]$ and $E[YZ]$ because you know $COV(X,Y)$ and $COV(Y,Z)$. But to infer $COV(X,Z)$ we still need to know $E[XZ]$ which we can not get from the previous information.

Answer 3

You can't, but you can bound it. First, you bound $\text{cor}(X, Z)$ as a function of the two other correlations. The reasoning behind the bound is that the correlation matrix of the vector $(X, Y, Z)$ must be positive semidefinite. Without repeating the algebra, you write down the determinant of the $3\times 3$ correlation matrix and impose that it be nonnegative. This is a quadratic inequality in $\rho_{XZ}$. It is is satisfied when is $$ |\rho_{XZ} - \rho_{XY}\rho_{YZ}| \le \sqrt{(1- \rho_{XY}^2)(1- \rho_{YZ}^2)} $$ The inequality below is the one used most often. Now, onto the original question. You mentioned that the marginal distributions of the three variables are known, hence their standard deviations are (as well as the correlations $\rho_{XY}, \rho_{YZ}$). Hence you can rewrite the bound as $$ \left| \frac{\text{cov}(X,Z)}{\sigma_X\sigma_Z} - \frac{\text{cov}(X,Y)\text{cov}(Y,Z)}{\sigma_X\sigma_Y^2\sigma_Z} \right| \le \sqrt{(1- \rho_{XY}^2)(1- \rho_{YZ}^2)} $$ or $$ \left|\text{cov}(X,Z) - \text{cov}(X,Y)\text{cov}(Y,Z)\frac{1}{\sigma_Y^2} \right| \le \sigma_X\sigma_Z \sqrt{(1- \rho_{XY}^2)(1- \rho_{YZ}^2)} $$

Answer 4

Following gappy's idea, although we cannot characterize $\text{cov}(X,Z)$, it must depend on $\text{cov}(X,Y)$ and $\text{cov}(Y,Z)$ because the covariance matrix of $(X,Y,Z)$ must be semidefinite positive. Hence, using the Schur complement of that matrix (or equivalently, the partial correlations) we can derive the following condition for a quadratic polynomial of $\text{cov}(X,Z)$: \begin{equation} a\ \text{cov}(X,Z)^2 - 2b\ \text{cov}(X,Z) - c \leq 0 \end{equation} where $a=\sigma_y^2$, $b=\text{cov}(X,Y)\text{cov}(Y,Z)$, and $c=\sigma_x^2 \sigma_y^2 \sigma_z^2 - \sigma_x^2 \text{cov}(Y,Z)^2 - \sigma_z^2\text{cov}(X,Y)^2$. This condition should be equivalent to the bound shown by gappy. Equality holds iff any two of the random variables are equal.