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Is it possible to derive Cov (Y,Z) from Cov (X,Y), Cov(X,Z) and the corresponding Variances? All 3 variables are assumed to be normal with 0 mean.

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consider the two alternative cases where (a) Y=Z and (b) Y and Z are independent.

If X is independent to both, then Cov(X,Y)=Cov(X,Z)=0 in both cases, but Cov(Y,Z) non-zero in case (a) and zero in case (b).

So the answer is no, in general: you cannot deduce Cov(Y,Z) from Cov(X,Y), Cov(X,Z) and the variances.

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It should be noted that covariance of two variables is intrinsically independent on their expected values, as long as they actually have those; therefore the condition $E[X]=E[Y]=E[Z]=0$ is largely irrelevant.

It is a general fact that, for all vectors $v\in\Bbb R^n$ and for all $A\in\Bbb R^{n\times n}$ symmetric and positive definite (or positive semi-definite, depending on whether or not you accept the constant random variable $c$ as $\mathcal N(c,0)$), there are $n$ normal random variables $X_1,\cdots, X_n$ such that, for all $i,j$, $v_i=E[X_i]$ and $A_{ij}=\operatorname{Cov}(X_i,X_j)$. See the Wikipedia page on Gaussian vectors for a quick overview and/or the related chapter on your textbook.

If you specialise the theorem to $X_1=X$, $X_2=Z$ and $X_3=Y$, then you have at least as much freedom in assigning $\operatorname{Cov}(Y,Z)$ as you have of assigning $\alpha$ in the matrix $$A(\alpha)=\begin{pmatrix}\operatorname{Cov}(X,X) & \operatorname{Cov}(X,Z) &\operatorname{Cov}(X,Y)\\ \operatorname{Cov}(X,Z) & \operatorname{Cov}(Z,Z) & \alpha\\ \operatorname{Cov}(X,Y) &\alpha & \operatorname{Cov}(Y,Y)\end{pmatrix}$$

while preserving the condition of Sylvester's criterion: $$\begin{cases}\operatorname{Cov}(X,X)>0\\ \operatorname{Cov}(X,X)\operatorname{Cov}(Z,Z)-(\operatorname{Cov}(X,Z))^2>0\\ \det(A(\alpha))>0\end{cases}$$ This is usually enough to allow a lot of possibile values of $\alpha$. For instance, with $X,Y$ independent, $X,Z$ independent and all variables having variance $1$, all values $\alpha\in(-1,1)$ are possible.

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