Given three random variables $X,Y, Z$, if $\text{cov}(X,Y) = 0.1$ and $\text{cov}(Y,Z) = 0.8$, is it possible that $\text{cov}(X,Z) = 0.7$?
I know how to solve the question if correlation is used instead of covariance (all the three variances are equal to 1). Any help would be greatly appreciated!
Yes, it is possible. A covariance matrix of interest is
$$ \begin{bmatrix} 1 & 0.1 & 0.7 \\ 0.1 & 1 & 0.8\\ 0.7 & 0.8 & \alpha \\ \end{bmatrix} \quad $$
for $\alpha \ge 1.03$.
For $\alpha = 1.03$, the eigenvalues are
$[2.12820779, \, 0.900905553, \, 8.86656963 \text{e-04}],$
which are positive.
For $\alpha = 1.02$, they are
$[ 2.12338262, \, 0.90090392,\, -0.00428654]$;
hence, the matrix with $\alpha = 1.02$ is not positive semi-definite (one of its eigenvalues is negative) and cannot be a covariance matrix.
For given $a,b,c$ (the covariances), the matrix
$$ \begin{bmatrix} x & a & b \\ a & y & c\\ b & c & z \\ \end{bmatrix} \quad $$
is a covariance matrix iff $x, y, z$ (the variances) satisfy the following conditions:
$$x,y,z \ge 0$$ $$xy\ge a^2,yz\ge c^2, xz\ge b^2$$ $$x(yz-c^2)-a(az-cb)+b(ac-by)\ge 0.$$
Moreover, when $a, b$ are given, you can find the minimum and maximum values of $W$, $\text{cov}(X,Z)$, for which the matrix
$$\begin{bmatrix} x & a & b \\ a & y & W\\ b & W & z \\ \end{bmatrix} \quad$$
remains a covariance matrix using semi-definite programming:
$$\text{min/max} \, W$$ subject to
$$ \begin{bmatrix} x & a & b \\ a & y & W\\ b & W & z \\ \end{bmatrix} \succeq \bf{0} \quad $$ $$ W \in \mathbb R. $$