# Inferring Covariance / Correlation

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.

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.

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.