How low can the third correlation go

Idea. A correlation matrix is a Gram matrix of unit vectors. Two of the angles are fixed, so the third is at most their sum, and its cosine is at least the cosine of that sum.

Step 1: angles. Think of the standardised variables as unit vectors, with $\mathrm{corr}=\cos (\text{angle})$. Let $\theta =\arccos (0.9)$, the angle between $X$ and $Y$ and between $Y$ and $Z$. The angle between $X$ and $Z$ is at most $2\theta$, with equality when the three vectors lie in a plane with $Y$ in the middle. So $$\begin{array}{rl}\mathrm{corr}(X,Z)& \ge \cos (2\theta )=2{\cos }^2\theta -1\\ & =2(0.81)-1=\mathrm{0.62.}\end{array}$$

Step 2: the determinant check. With $c=\mathrm{corr}(X,Z)$, the determinant of $$\left(\begin{array}{ccc}1& 0.9& c\\ 0.9& 1& 0.9\\ c& 0.9& 1\end{array}\right)$$ is $1-0.81-0.81-c^2+2(0.81)c=-(c^2-1.62c+0.62)$, which is non-negative for $0.62\le c\le 1$. The leading $2\times 2$ minor is $0.19>0$, so the matrix is a valid correlation matrix exactly for $c\in [0.62,1]$.

Answer. $0.62=\frac{31}{50}$.

In general. With $\mathrm{corr}(X,Y)=a$ and $\mathrm{corr}(Y,Z)=b$, $$|\mathrm{corr}(X,Z)-ab|\le \sqrt{(1-a^2)(1-b^2)}.$$ Correlation is not transitive in general: with $a=b=0.7$, $\mathrm{corr}(X,Z)$ can be slightly negative.