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With correlation coefficient defined as:

$$\rho(X, Y) = \frac{\text{Cov}(X, Y)}{\sqrt{\text{Var}(X)}\sqrt{\text{Var}(Y)}}$$

can you help me prove

$$|\rho(X, Y)| = 1 \implies Y = aX + b$$

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  • $\begingroup$ Just a comment... if data is $\{(0,0),(1,1)\}$ then the correlation coefficient is 1 I imagine. Now while all the data obeys $Y=X$ it doesn't prove that this is in fact the law. I assume this is obvious to you but just pointing it out. $\endgroup$ – JP McCarthy Aug 28 '13 at 12:54
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    $\begingroup$ I believe you want to write $Y=aX+b$ almost surely. $\endgroup$ – Stefan Hansen Aug 28 '13 at 12:59
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Hint: In order for the correlation to be well-defined we must assume that $X$ and $Y$ are not degenerate ($X$ being degenerate meaning that $X=a$ almost surely). Now, show that $$ \exists a,b\in\mathbb{R}:Y=aX+b\quad\text{a.s.}\iff \exists a\in\mathbb{R}\setminus\{0\}:\,\mathrm{Var}(Y-aX)=0. $$ Then you just need to show that $$ |\rho(X,Y)|=1\;\;\Longrightarrow\;\; \exists a\in\mathbb{R}\setminus\{0\}:\,\mathrm{Var}(Y-aX)=0, $$ so start out by assuming that $\rho(X,Y)=1$ and then expand $\mathrm{Var}(Y-aX)$ and show that it is indeed zero for some $a\neq 0$.

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  • $\begingroup$ I think you meant to say the variance of $Y-aX$ not $Y-bX$. $\endgroup$ – Dilip Sarwate Aug 28 '13 at 13:37
  • $\begingroup$ @DilipSarwate: Indeed, thanks for pointing out the mistake. $\endgroup$ – Stefan Hansen Aug 28 '13 at 13:41

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