Minimum covariance of 2 random variables given the covariance of each with a third random variable If we have 3 random variables, with the first two with a covariance of 0.1 and the second two with a covariance of 0.1, what is the minimum covariance of the first and third? Is there a generalized solution on bounds on covariance matrices?
 A: What (I believe) you want to know is for which values of $x$ your covariance matrix remains positive semi-definite (see property two here). I take your matrix to be a correlation matrix for simplicity, so that correlations are 1:
$$
V=\begin{pmatrix}
1&0.1&x\\
0.1&1&0.1\\
x&0.1&1
\end{pmatrix}
$$
By Sylvester's criterion, we need to know if the determinants of all principal minors are nonnegative. Now, 1 and $1-0.1^2$ obviously are, independent of $x$.
For the determinant, I asked Wolfram, which gave the figure below. So $x\in[-49/50,1]$.

A: (The answer has been rewritten)
Provided that you mean exactly what you are asking (i. e. such information is given about the covariances, not the correlations between the random variables), the answer is as follows.
Firstly, the covariance of the first and third random variable may not exist. Secondly, even if it does exist, there is no minimum covariance of the first and third random variable, it can be arbitrarily low.
As an example of non-existence, consider random variables $\xi_{1}  \sim F(x): \rho(x) = |x|^{-3}, |x|>1, \xi_{3} = \xi_{1}$.
Since such distribution is symmetric with respect to zero, the expectations of these random variables are zero. Due to their dependence, $cov(\xi_{1}, \xi_{3}) = D(\xi_{1}) = \infty$. However, I can find a degree $b$ such that $\xi_{2}=\xi_{1}^{\frac{1}{b}}: cov(\xi_{2}, \xi_{1}) = cov(\xi_{2}, \xi_{3}) = 0.1$, and $E\xi_{2}=0$ as well.
To instantiate the possibility of an arbitrary covariance, consider a very special case: I have three samples of mentioned random variables, say $\{x\}_{i=1}^n, \{y\}_{i=1}^n, \{z\}_{i=1}^n,$ of identical size $n$, such that $\bar{x}, \bar{y}, \bar{z} = 0$. Than their sample covariances would be simply $\frac{1}{n}\sum_{i=1}^nx_{i}\cdot y_{i}, \frac{1}{n}\sum_{i=1}^nz_{i}\cdot y_{i}, \frac{1}{n}\sum_{i=1}^nx_{i}\cdot z_{i}$, which in an orthonormal basis are respectively identical to normed scalar products of respective vectors $x, y, z$ in an N-dimensional space: $\frac{1}{n}(x, y), \frac{1}{n}(y, z), \frac{1}{n}(x, z)$, which are in turn $\frac{1}{n}|x|\cdot|y|\cdot cos(x, y) ; \frac{1}{n}|y|\cdot |z|\cdot cos(y,z); \frac{1}{n}|x|\cdot |z|\cdot cos(x,z) $ in geometric interpretation.
Now let these vectors $x, y, z$ be all in a single 3-dimensional plane, and the angles between them would be $\pi/3, \pi/3$ and $2\cdot \pi/3$ respectively, then the cosines are $0.5, 0.5, -0.5$.
In a given example, the initial conditions for covariances are satisfied whenever $|x|\cdot |y|=0.2n$ and $|z|\cdot |y|=0.2n$, and we still have a freedom to set $|x|=|z|$ to arbitrary values, setting $|y| = \frac{0.2n}{|x|}$.
Now consider a sequence of such vectors $ \{\vec{x}\}_{j=1}^\infty$, say tuples with $\pm h$ inside of dimension $j$, ($(h), (h,-h), \dots )$, then for every $j$: $|\vec{x}_{j}| = |\vec{z}_{j}|=h\cdot \sqrt{j} $ and then $cov(\vec{x_j}, \vec{z_j})=-\frac{j\cdot h}{2j}$. The last figure apparently converges to $-\frac{h}{2}$ (where $h$ is arbitrary) while the two other covariances remain stable at $0.1$. One could argue that such $x$ sample does not look like random, but here the underlying distribution is not discussed at all and in a sample case of $x_{i} \sim \{h, p = \frac{1}{2}, -h\; else\}$ the length of $\vec{x}$ would be exactly the same.
The entire considerations have been made about empirical covariances, but as long as the size of a sample goes to infinity, the Continuous mapping theorem ensures that if a theoretical covariance between the first and third variable exists, the empirical one should converge to it in probability. The example shown before shows that the theoretical $cov(x, z)$ may not even exist. By configuring another $h$ sequence, one could make the covariance converge to arbitrarily low finite numbers.
There can also be a more concise and intuitive explanation.
As covariances are expressed in squared units of measurement, one can simply adjust the units of $x$ and $z$, compensating these shifts with scaling $y$ conversely, and thus scale the covariance $cov(x,z)$ arbitrarily, so the "generalized solution" here is that in such problem setup there are actually no bounds on distinct elements of covariance matrices (in both sides).
Nevertheless, any covariance matrix should still satisfy its basic requirement: it is always a squared symmetrical positive-semidefinite matrix.
The answer above suggests a great approach but it seems to have simply changed the 'covariances' to 'correlations' which cannot be simply done since it would require scaling the covariance matrix elements differently. Posing the analogous answer about correlations leads to completely different conclusions.
