Proving $Y = aX + b$ given correlation coefficient $|\rho(X, Y)| = 1$

Question

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

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

Answer 1

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 $a\neq 0$.