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I need to prove or disprove the following:
There exist two random variables $X,Y$ s.t:
$Var[X]=Var[Y]=2$
$Cov(X,Y)=4$

I tried to stick to the definitions here and couldn't find any contradiction.
$E[X^2]-(E[X])^2=E[Y^2]-(E[Y])^2=2$
$E[XY]-E[X]E[Y]=4$

Nothing out of the ordinary... is it?

so.very.tired
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  • You must either prove the statement, by demonstrating that two such variables exist; or disprove the statement, by fashioning an argument to show that two such variables cannot exist.

    A hint: what is $\rm{Corr}(X,Y)$?

     – Eddie E. Feb 06 '14 at 21:31

2 Answers2

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Aren't the variances too low for the covariance? The correlation $\rho_{XY}$ has to be in $[-1,1]$.

$$ \rho_{XY} = \frac{Cov(X,Y)}{\sigma_X\sigma_Y} = \frac{4}{\sqrt{2}\sqrt{2}} = 2 > 1$$

TooTone
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Cauchy-Schwarz says that $$ \begin{align} \mathrm{Cov}(X,Y) &=\mathrm{E}\Big((X-\mathrm{E}(X))\;(Y-\mathrm{E}(Y))\Big)\\ &\le\mathrm{E}\left((X-\mathrm{E}(X))^2\right)^{1/2}\mathrm{E}\left((Y-\mathrm{E}(Y))^2\right)^{1/2}\\[2pt] &=\sigma(X)\,\sigma(Y) \end{align} $$ which verifies TooTone's asssertion.

TooTone
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robjohn
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  • I'd never thought of Cauchy-Schwartz as applying to expectations (but the expectation is an integral and the integrand is a product). – TooTone Feb 06 '14 at 23:11
  • @TooTone: math is cool, n'est-ce pas? – robjohn Feb 06 '14 at 23:35
  • oui, funny you switched language because math seems a lot like a language to me, with all the pain of learning the grammar, but also the poetry that comes with it. – TooTone Feb 06 '14 at 23:40