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  1. Whether there is a covariance of two random variables, which both don't have variances?
  2. Is existence of the variances of two random variables implies existence of covariance?

Thanks in advance.

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  • $\begingroup$ Hint: Cauchy Schwarz $\endgroup$ – Batman May 21 '14 at 15:40
  • $\begingroup$ I up-voted the question, and its total is now $0$, so someone downvoted it and should state their objections. $\endgroup$ – Michael Hardy May 21 '14 at 15:50
  • $\begingroup$ @MichaelHardy I didn't vote but the question's score in terms of personal input is around $-\infty$, afaiac. $\endgroup$ – Did May 21 '14 at 17:39
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The Cauchy--Schwarz inequality implies that $$ |\operatorname{cov}(X,Y)| \le \sqrt{\operatorname{var}(X)\operatorname{var}(Y)}, $$ so existence of the two variancees (meaning both are finite) implies existence of the covariance. Notice that no generality is lost by assuming $\mathbb E(X)=\mathbb E(Y)=0$, so the variances of $X$ and $Y$ are $\mathbb E(X^2)$ and $\mathbb E(Y^2)$. Cauchy--Schwarz says $\mathbb E(|XY|) \le \sqrt{(\mathbb E(X^2))(\mathbb E(Y^2))} $, and that means $\mathbb E(XY)$ exists, and that is the covariance.

As for existence of the covariance when the variances of $X$ and $Y$ are both infinite, I would let $$ X = \begin{cases} W & \text{if some even number} \le W<\text{that even number}+1, \\ 0 & \text{otherwise}, \end{cases} $$ $$ Y = \begin{cases} W & \text{if some odd number} \le W<\text{that odd number}+1, \\ 0 & \text{otherwise}, \end{cases} $$ and then see if I can choose the distribution of $W$ in such a way that $\mathbb E(X)=\mathbb E(Y)=0$ but both have infinite variance. Then the covariance between $X$ and $Y$ would be $\mathbb E(XY)=0$.

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    $\begingroup$ For the second part, consider $X$ and $Y$ independent and integrable but not square integrable, then $XY$ is integrable. $\endgroup$ – Did May 21 '14 at 17:38

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