Suppose $cov(X,Y)=0\;$ and $\;cov(Y,M)=0$. Does this imply $cov(X,M)=0\;$, if all distinct RV are normal?

  • $\begingroup$ Individually normal doesn't help. $\endgroup$ – André Nicolas Jun 26 '12 at 19:30
  • $\begingroup$ Calling the third random variable M instead of Z is quite odd. Any reason for that? $\endgroup$ – Did Jul 2 '12 at 8:33

In general, no. Take $X = M$ ;)

It also does not hold if $(X,Y,Z)$ follow a Multivariate Normal $(\mu,\Sigma)$.

Necessary and sufficient conditions for a matrix to be a covariance matrix $(\Sigma)$ are presented here. A matrix is a covariance matrix if and only if it is positive semi-definite.

Hence, take

$$\Sigma =\left( \begin{array}{ccc} 1 & 0 & 0 \\ 0 & 1 & 1 \\ 0 & 1 & 1 \end{array} \right)$$

and observe that $Cov(X,Y) = Cov(X,Z) = 0$ but $Cov(Y,Z)=1$.

  • 2
    $\begingroup$ +1: The question has changed so this becomes "Take $X=M$" $\endgroup$ – Henry Jun 26 '12 at 19:56

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