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 $COV(X,Z)$?

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