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Generating correlated normal vectors with observations

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0

$begingroup$

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

$$rho_i,j =fracmathbbEX_i Y_jsigma_x_i sigma_y_i$$

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

$$M_4times4 =
beginbmatrix
sigma_x_1^2 &rho_x sigma_x_1sigma_x_2 &rho_1,1sigma_x_1 sigma_y_1 &rho_1,2sigma_x_1sigma_y_2\
rho_x sigma_x_1sigma_x_2 &sigma_x_2^2 &rho_2,1sigma_x_2 sigma_y_1 & rho_2,2sigma_x_2 sigma_y_2\
rho_1,1 sigma_x_1 sigma_y_1 & rho_1,2 sigma_x_2 sigma_y_1 & sigma_y_1^2 & rho_ysigma_y_1sigma_y_2\
rho_1,2 sigma_x_1 sigma_y_2 & rho_2,2 sigma_x_2sigma_y_2 & rho_ysigma_y_1sigma_y_2 & sigma_y_2^2 \
endbmatrix = beginbmatrix A_2times2 & B_2times2 \ B^top_2times2 & C_2times2 \ endbmatrix $$

I think we can generate $[Y_1, Y_2]$ by conditioning on $[X_1, X_2]$ and calculate the new conditional covariance, $tildeC = beginbmatrix C - B^top A^-1B endbmatrix$. 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 $tildeC$.

statistics numerical-linear-algebra conditional-probability

share|cite|improve this question

edited 19 hours ago

MrX

asked yesterday

MrXMrX

302313

$endgroup$

add a comment | 

0

$begingroup$

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

$$rho_i,j =fracmathbbEX_i Y_jsigma_x_i sigma_y_i$$

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

$$M_4times4 =
beginbmatrix
sigma_x_1^2 &rho_x sigma_x_1sigma_x_2 &rho_1,1sigma_x_1 sigma_y_1 &rho_1,2sigma_x_1sigma_y_2\
rho_x sigma_x_1sigma_x_2 &sigma_x_2^2 &rho_2,1sigma_x_2 sigma_y_1 & rho_2,2sigma_x_2 sigma_y_2\
rho_1,1 sigma_x_1 sigma_y_1 & rho_1,2 sigma_x_2 sigma_y_1 & sigma_y_1^2 & rho_ysigma_y_1sigma_y_2\
rho_1,2 sigma_x_1 sigma_y_2 & rho_2,2 sigma_x_2sigma_y_2 & rho_ysigma_y_1sigma_y_2 & sigma_y_2^2 \
endbmatrix = beginbmatrix A_2times2 & B_2times2 \ B^top_2times2 & C_2times2 \ endbmatrix $$

I think we can generate $[Y_1, Y_2]$ by conditioning on $[X_1, X_2]$ and calculate the new conditional covariance, $tildeC = beginbmatrix C - B^top A^-1B endbmatrix$. 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 $tildeC$.

statistics numerical-linear-algebra conditional-probability

share|cite|improve this question

edited 19 hours ago

MrX

asked yesterday

MrXMrX

302313

$endgroup$

add a comment | 

$begingroup$

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

$$rho_i,j =fracmathbbEX_i Y_jsigma_x_i sigma_y_i$$

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

$$M_4times4 =
beginbmatrix
sigma_x_1^2 &rho_x sigma_x_1sigma_x_2 &rho_1,1sigma_x_1 sigma_y_1 &rho_1,2sigma_x_1sigma_y_2\
rho_x sigma_x_1sigma_x_2 &sigma_x_2^2 &rho_2,1sigma_x_2 sigma_y_1 & rho_2,2sigma_x_2 sigma_y_2\
rho_1,1 sigma_x_1 sigma_y_1 & rho_1,2 sigma_x_2 sigma_y_1 & sigma_y_1^2 & rho_ysigma_y_1sigma_y_2\
rho_1,2 sigma_x_1 sigma_y_2 & rho_2,2 sigma_x_2sigma_y_2 & rho_ysigma_y_1sigma_y_2 & sigma_y_2^2 \
endbmatrix = beginbmatrix A_2times2 & B_2times2 \ B^top_2times2 & C_2times2 \ endbmatrix $$

I think we can generate $[Y_1, Y_2]$ by conditioning on $[X_1, X_2]$ and calculate the new conditional covariance, $tildeC = beginbmatrix C - B^top A^-1B endbmatrix$. 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 $tildeC$.

statistics numerical-linear-algebra conditional-probability

share|cite|improve this question

edited 19 hours ago

MrX

asked yesterday

MrXMrX

302313

$endgroup$

share|cite|improve this question

edited 19 hours ago

MrX

asked yesterday

MrXMrX

302313

share|cite|improve this question

edited 19 hours ago

MrX

asked yesterday

MrXMrX

302313

asked yesterday

MrXMrX

302313

asked yesterday

MrXMrX

302313

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0

$begingroup$

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 $(Y_1, Y_2)$).

For general matrices that don't have any context as covariance matrices, you can refer to these results on the Schur complement.

share|cite|improve this answer

answered yesterday

angryavianangryavian

42k23381

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  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.
  $endgroup$
  – MrX
  19 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  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 $tildeC$ is not PSD, is this correct? and thus we cannot generate $[Y_1, Y_2]$ (unless we use the generalized inverse)
  $endgroup$
  – MrX
  19 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @MrX But why do you think $M$ is not PSD?
  $endgroup$
  – angryavian
  19 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  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 $tildeC$ is not PSD if $M$ is not.
  $endgroup$
  – MrX
  19 hours ago
  
  

  
  
  
  

add a comment | 

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0

$begingroup$

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 $(Y_1, Y_2)$).

For general matrices that don't have any context as covariance matrices, you can refer to these results on the Schur complement.

share|cite|improve this answer

answered yesterday

angryavianangryavian

42k23381

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  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.
  $endgroup$
  – MrX
  19 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  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 $tildeC$ is not PSD, is this correct? and thus we cannot generate $[Y_1, Y_2]$ (unless we use the generalized inverse)
  $endgroup$
  – MrX
  19 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @MrX But why do you think $M$ is not PSD?
  $endgroup$
  – angryavian
  19 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  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 $tildeC$ is not PSD if $M$ is not.
  $endgroup$
  – MrX
  19 hours ago
  
  

  
  
  
  

add a comment | 

0

$begingroup$

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 $(Y_1, Y_2)$).

For general matrices that don't have any context as covariance matrices, you can refer to these results on the Schur complement.

share|cite|improve this answer

answered yesterday

angryavianangryavian

42k23381

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  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.
  $endgroup$
  – MrX
  19 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  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 $tildeC$ is not PSD, is this correct? and thus we cannot generate $[Y_1, Y_2]$ (unless we use the generalized inverse)
  $endgroup$
  – MrX
  19 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @MrX But why do you think $M$ is not PSD?
  $endgroup$
  – angryavian
  19 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  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 $tildeC$ is not PSD if $M$ is not.
  $endgroup$
  – MrX
  19 hours ago
  
  

  
  
  
  

add a comment | 

$begingroup$

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 $(Y_1, Y_2)$).

For general matrices that don't have any context as covariance matrices, you can refer to these results on the Schur complement.

share|cite|improve this answer

answered yesterday

angryavianangryavian

42k23381

$endgroup$

share|cite|improve this answer

answered yesterday

angryavianangryavian

42k23381

answered yesterday

angryavianangryavian

42k23381

answered yesterday

angryavianangryavian

42k23381