Rigorous proof for if $\operatorname{Corr}(X,Y) =1$ then $X = a Y + b$ with $a>0$

Question

I know that if $X = a Y + b$ and $a>0$ then $\operatorname{Corr}=1$. it can be proved easily using the fact that $Cov(X,Y) = E(XY)-E(X)E(Y)$ and $\sigma^2_X =Var(X) = a^2 Var(Y) = a^2 \sigma^2_Y$ and $E(X) = a E(Y) + b$, and substituting $X = a Y + b$ in the covariance formula, and also the definition of correlation that $\operatorname{Corr}(X,Y) = \frac{\operatorname{Cov}(X,Y)}{\sigma_X \sigma_Y}$

But as I know, it is an "if and only if" relation (biconditional) and so we have that if $\operatorname{Corr}(X,Y) =1$ then $X = a Y + b$ with $a>0$. I can understand this intuitively but I don't know how to rigorously prove this.

Answer 1

We have that

$$0 \le E\left(\frac{X-EX}{\sigma_X} - \frac{Y-EY}{\sigma_Y}\right)^2=\frac{E(X-EX)^2}{\sigma_X^2}+\frac{E(Y-EY)^2}{\sigma_Y^2}-\frac{2E[(X-EX)(Y-EY)]}{\sigma_X\sigma_Y}=2- 2\text{Corr}(X,Y)=0,$$

since $\text{Corr}(X,Y)=1$. This implies that $$\frac{X-EX}{\sigma_X} - \frac{Y-EY}{\sigma_Y}=0 \hspace{10pt} \text{a.s}.$$

You can reorganize the last identity to obtain $$Y=aX+b \hspace{10pt} \text{a.s.},$$

with $a=\frac{\sigma_Y}{\sigma_X}$ and $b=EY-\frac{\sigma_Y}{\sigma_X}EX$.

Answer 2

I explained here that, if we identify variables of finite variance with their $Z$-scores, they become the unit vectors in an inner product space, its inner product being the correlation coefficient. By Cauchy-Schwarz, this is $1$ iff the vectors are parallel, and since it's a real space and we're dealing with unit vectors, this is equivalent to $X,\,Y$ having the same $Z$-score. But this, in turn, is equivalent to $X=aY+b$ with $a:=\frac{\sigma_X}{\sigma_Y},\,b:=\mu_X-a\mu_Y$.