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For a normally distributed random vector $(X,Y)$, $Cov(X,Y) = 0$ implies $X$ and $Y$ being independent.

But how about other distribution?

Then for a $n\times n$ normally distributed random vector, if the the covariance matrix is a diagonal matrix, I guessed components of the random vectors are all independent of each other, is this correct?

TJT
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  • @SassatelliGiulio I agree with your first statement, but I am not sure about the second statement because I encountered this question that has proved for $normally$ distributed random vector $(X,Y)$, $X,Y$ are independent if and only if $\operatorname{Cov}(X,Y)=0$. So this make me wonder if there are other distribution that follow this in multivariate case – TJT Jan 06 '24 at 23:47
  • @RobertIsrael Partially, so why is multivariate normal so unique it has such property of $Cov(X,Y)$ implies independence between $X,Y$? is this due to its construction? – TJT Jan 06 '24 at 23:59
  • Very very easy to find examples where covariance is $0$ and the variables are not independent. Did you make any effort at all to find such an example? – Kavi Rama Murthy Jan 07 '24 at 00:13
  • @geetha290krm yes for univariate case, but I am not sure about multivariate case – TJT Jan 07 '24 at 00:20

1 Answers1

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In the Normal case yes, $Cov(X,Y) = 0$ implies $X , Y$ independent, if the covariance matrix is diagonal, we can say that they are independent.

In general $Cov(X,Y) = 0$ does not imply independence. See

$$f_{X,Y}(x,y) = \frac{1}{\pi} I(x^2 + y^2 \leq 1) $$

$Cov(X,Y) = 0$ but $f_{X}(x)f_{Y}(y) \neq f_{X,Y}(x,y)$

daniel
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