Let $(X,Y,Z)^T$ be jointly normal variable with zero mean such that the correlation coefficients $\rho_{XY}=0.2, \rho_{YZ}=0.2$, what values can $\rho_{XZ}$ take?
Prove that there exists a decomposition
$Y=aN_2,$
$X=bN_1+cN_2,$
$Z=dN_1+eN_2+fN_3,$
where $N_1,N_2,N_3$ are independent stand normal variables.
I know that the covariance matrix $\Sigma$ is a semi positive definite matrix and hence admits a Cholesky decomposition $\Sigma = LL^T$. But I am not sure how to proceed
Best Answer
Consider the covariance matrix of $(X,Y,Z)^T$, $C=\mathbb E[[X,Y,Z]^T[X,Y,Z]]$. (since the means are $0$)
Compute the individual variance and standard deviation of $X,Y,Z$, call it $\sigma_X,\sigma_Y,\sigma_Z$.
Define the covariance of normalized variables $\frac{X}{\sigma_X},\frac{Z}{\sigma_Z},\frac{Z}{\sigma_Z}$, we call it $\bar C$.
For this normalized covariance matrix, it shall has this structure $$\bar C= \begin{bmatrix} 1 & 0.2 & \rho\\ 0.2 & 1 & 0.2\\ \rho & 0.2 & 1\\ \end{bmatrix} $$ Note scale and shifting of random variables do not change the pearson correlation among them. so here $\rho=\rho_{XZ}$
To make $\bar C$ a valid covariance matrix, it needs to be positive (semi-)definite. Using a generalized version of Sylvester criterion for pos-semi-definite matrices, we need to show all principal minors are non-negative (in contrast to leading ones are positive).
This criterion comes down to $$ \det(\bar C)\geq 0\\ \det(\begin{bmatrix} 1 & \rho\\ \rho & 1\\ \end{bmatrix})\geq0 $$ Which translates to $$ 1-\rho^2\geq 0\\ \frac{576}{25^2}-(\rho-\frac 1{25})^2\geq 0 $$
Then you get the range of the correlation between $X,Z$. $[-\frac{23}{25},1]$ $$ -\frac{23}{25}\leq \rho\leq 1 $$
For the last part, notice how linear transform affects the covariance of normal variables. Generally, for multivariate Gaussian $U\sim \mathcal N(0,C)$ the transform it with an invertible transform $V=AU$, $V\sim \mathcal N(0,ACA^T)$.
Thus if you have three independent standard normal variable $[N_1,N_2,N_3]^T\sim \mathcal N(0,I)$. Then if we have a Cholesky decomposition of $C=LL^T$ we can define the random variables $L[N_1,N_2,N_3]^T\sim \mathcal N(0,LL^T)=\mathcal N(0,C)$
Thus we find the transform that maps independent normal variables to $X,Y,Z$
$$ [X,Y,Z]^T=L[N_1,N_2,N_3]^T $$
Prussing, John E., The principal minor test for semidefinite matrices, J. Guid. Control Dyn. 9, 121-122 (1986). ZBL0599.15010.