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Let $x$ and $y$ be jointly distributed numeric variables and let $z=a+by$ , where $a$ and $b$ are constants. Show that $\operatorname{cov}(x,z)=b\, \operatorname{cov}(x,y)$.

Here's what I have so far, but then I got stuck.

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Finished.

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It's not the most elegant proof, but you're almost there!

Hint: Note that $\sum_{i=1}^n x_i a = a\,n\,\mu(x)$, and $\mu(a + by) = a + b\mu(y)$. All the terms from the left that have an $a$ in them should "cancel out".

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  • $\begingroup$ Those two facts will help a lot, thanks! I couldn't figure out how to deal with the a+by inside the mu. $\endgroup$ – Dana Hill Sep 11 '17 at 0:07

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