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

  • 0
    *Individually* normal doesn't help.2012-06-26
  • 0
    Calling the third random variable M instead of Z is quite odd. Any reason for that?2012-07-02

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