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

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 $COV(X,Z)$?

• Nobody denotes covariances by $\sigma$. – Did Mar 25 '14 at 21:17
• #Did: It's updated now. – TimY Mar 26 '14 at 7:04
• @Did In the covariance matrix of the multivariate normal distribution for example... – Jimmy R. Apr 4 '14 at 8:33

## 4 Answers

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 $cov(X, Z) = 1$.

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.

• Exactly. The key piece of information you are lacking is $E[XZ]$. If you were given that, then it becomes trivial. – MentatOfDune Mar 31 '14 at 20:05

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 $$\left|\text{cov}(X,Z) - \text{cov}(X,Y)\text{cov}(Y,Z)\frac{1}{\sigma_Y^2} \right| \le \sigma_X\sigma_Z \sqrt{(1- \rho_{XY}^2)(1- \rho_{YZ}^2)}$$

Following gappy 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 $c=\sigma_x^2 \sigma_y^2 \sigma_z^2 - \sigma_x^2 \text{cov}(Y,Z) - \sigma_z^2\text{cov}(X,Y)$. This condition should be equivalent to the bound shown by gappy.