Let $X∼N(μ,σ^2)$ and $Y= 1−X^2$ be two real-valued r.v. How do I proof that $Cov(X,Y)=0$ but $X,Y$ not independent?
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Are you having trouble showing that covariance is zero, or dependence? There doesn’t seem much effort here. – Erick Wong Aug 11 '19 at 20:02
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It is clear by the definition of $Y$ that $X$ and $Y$ are not independent. What have you tried in terms of showing zero covariance? – Dave Aug 11 '19 at 20:10
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My Trouble is with the independence especially. – KingDingeling Aug 11 '19 at 20:24
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This should help. – RRL Aug 11 '19 at 20:45
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@RRL, what do you mean? – KingDingeling Aug 11 '19 at 20:53
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Is covariance actually zero? Quick math yielded $-2\mu \sigma^2$ for me – Makina Aug 11 '19 at 21:36
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An easy way to see that $X$ and $Y$ are independent is by considering for any $t\in\mathbb R$ the probability $$ \mathbb P(X>t,Y<1-t^2), $$ which obviously equals $\mathbb P(X>t)$ since $X>t$ implies that $Y<1-t^2$.
On the other hand, suppose that $X$ and $Y$ are independent. Then $$ \mathbb P(X>t,Y<1-t^2)=\mathbb P(X>t)\mathbb P(Y<1-t^2). $$ Comparing this with the first equation, it follows that $\mathbb P(Y<1-t^2)=1$ for all $t\in\mathbb R$, which means that $\mathbb P(Y=-\infty)=1$ which is a contradiction. Thus $X$ and $Y$ are not independent.
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