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Suppose that X,Y and X,Z are bivariate normally distributed. We have

$E(X)=0, Var(X)=10$, $E(Y)=0, Var(Y)=6$ and $ρ_{xy}=0.87$

Moreover,

$E(X)=0, Var(X)=10$, $E(Z)=0, Var(Z)=4$ and $ρ_{xz}=0.87$

Will also Y and Z be bivariate normally distributed ? (I guess yes) If yes, which is their coefficient of correlation?

Posted also here: https://math.stackexchange.com/questions/3271570/correlation-coefficient-bivariate-normally-distributed

Added after comment of Whuber: Indicating as K the joint distribution of Y and Z,i know from the theory of the problem i'm dealing with that K is for sure a bivariate normally distributed. I expect this will pose some constraint on the value of $ρ_{yz}$

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  • $\begingroup$ The correlation question is answered at stats.stackexchange.com/questions/72790, which shows there generally is a range of correlations consistent with this information. The bivariate normal question is answered (in the negative) at stats.stackexchange.com/questions/30159. $\endgroup$ – whuber Jun 23 '19 at 13:17
  • $\begingroup$ Comment after some thought on the problem i'm dealing with. Indicating as K the joint distribution of Y and Z, I know from the theory of the problem i'm dealing with that K is for sure a bivariate normally distributed. I expect this will pose some constraint on the value of $ρ_{yz}$ $\endgroup$ – Andrea Mazzolari Jun 23 '19 at 14:55
  • $\begingroup$ Also see How to infer correlations from correlations $\endgroup$ – Glen_b -Reinstate Monica Jun 23 '19 at 16:08
  • $\begingroup$ @Glen_b thank you! that's exactly what i needed! Taking into account that K (my joint distribution) is for sure a bivariate normally distributed, is there a way to lower the interval of the inferred correlation ? $\endgroup$ – Andrea Mazzolari Jun 23 '19 at 16:52
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    $\begingroup$ The second part of my answer demonstrates that restricting the question about the possible range of correlations to bivariate Normal distributions doesn't change anything. The underlying idea is that for every possible covariance matrix, there is a (multivariate) Normal distribution whose covariance matrix is equal to it. The simplest proof proceeds from the expression of the characteristic function, which depends only on the mean and covariance. $\endgroup$ – whuber Jun 23 '19 at 18:34