# Generating correlated normal vectors with observations

Assume we are given two normally distributed random variables, $$X_1$$ and $$X_2$$, with $$X_i \sim \mathcal N (0, \sigma_{x_i}^2)$$, with correlation coefficient $$\rho_x$$. Assume further that we need to generate another two normally distributed random variables $$Y_1$$ and $$Y_2$$ with $$Y_j \sim \mathcal N (0,\sigma_{y_j}^2)$$, with correlation coefficient $$\rho_y$$ and cross correlation

$$\rho_{i,j} =\frac{\mathbb{E}\{X_i Y_j\}}{\sigma_{x_i} \sigma_{y_i}}$$

I.e., we need to generate $$[Y_1, Y_2]$$ taken into consideration the given $$[X_1, X_2]$$. I know how to generate all of them together $$[X_1,X_2, Y_1, Y_2]$$ using covariance matrix

$$M_{4\times4} = \begin{bmatrix} \sigma_{x_1}^2 &\rho_x \sigma_{x_1}\sigma_{x_2} &\rho_{1,1}\sigma_{x_1} \sigma_{y_1} &\rho_{1,2}\sigma_{x_1}\sigma_{y_2}\\ \rho_x \sigma_{x_1}\sigma_{x_2} &\sigma_{x_2}^2 &\rho_{2,1}\sigma_{x_2} \sigma_{y_1} & \rho_{2,2}\sigma_{x_2} \sigma_{y_2}\\ \rho_{1,1} \sigma_{x_1} \sigma_{y_1} & \rho_{1,2} \sigma_{x_2} \sigma_{y_1} & \sigma_{y_1}^2 & \rho_{y}\sigma_{y_1}\sigma_{y_2}\\ \rho_{1,2} \sigma_{x_1} \sigma_{y_2} & \rho_{2,2} \sigma_{x_2}\sigma_{y_2} & \rho_{y}\sigma_{y_1}\sigma_{y_2} & \sigma_{y_2}^2 \\ \end{bmatrix} = \begin{bmatrix} A_{2\times2} & B_{2\times2} \\ B^\top_{2\times2} & C_{2\times2} \\ \end{bmatrix}$$

I think we can generate $$[Y_1, Y_2]$$ by conditioning on $$[X_1, X_2]$$ and calculate the new conditional covariance, $$\tilde{C} = \begin{bmatrix} C - B^\top A^{-1}B \end{bmatrix}$$. Assuming that $$A,B$$ and $$C$$ are positive semidefinite (PSD), but not $$M$$, is it still possible to generate $$[Y_1,Y_2]$$?! or are we going to get always a non-PSD $$\tilde{C}$$.

You have a small typo: the conditional covariance is $$C - B^\top A^{-1} B$$.
If you take for granted that the formula for conditional covariance is correct, then you know it must be PSD simply because it is a covariance matrix. (Similarly, you know $$A$$ and $$C$$ must be PSD since they are covariance matrices for $$(X_1, X_2)$$ and $$(Y_1, Y_2)$$).
• I corrected the typo, it is due to the fact I was initially using $\sigma_{x_i} = \sigma_x$, i.e.,$B$ would be symmetric. – MrX Mar 10 '19 at 6:41
• But the core of my question boils down to the following: I think from the "if and only of" when $M$ is not PSD then $\tilde{C}$ is not PSD, is this correct? and thus we cannot generate $[Y_1, Y_2]$ (unless we use the generalized inverse) – MrX Mar 10 '19 at 6:43
• @MrX But why do you think $M$ is not PSD? – angryavian Mar 10 '19 at 6:48
• I am assuming it is given that $M$ is not a PSD, I was thinking if there is a work around to generate $[X_1, X_2, Y_1,Y_2]$ when $M$ is not PSD, and then thought of this iterative method (first $[X_1, X_2]$ then $[Y_1,Y_2]$)... frankly it would make sense that $\tilde{C}$ is not PSD if $M$ is not. – MrX Mar 10 '19 at 6:59