proving linear relation with correlation Given $\rho(x,y) =1$ where $\rho$ denotes pearson correlation coefficient.
can you prove $y= mx+c$ where $m$ and $c$ are constants?
I can go the other way around, but haven't yet proved this. I wonder if y needs to be of this form only.
 A: Suppose that $X$ and $Y$ have variances $\sigma_X^2$ and $\sigma_Y^2$.  Then, $\frac{X}{\sigma_X}$ and $\frac{Y}{\sigma_Y}$ are unit-variance random variables.  Now, consider the variance of the random variable $\frac{X}{\sigma_X}-\frac{Y}{\sigma_Y}$.  Since
$$\operatorname{var}(aX-bY) = a^2\cdot\operatorname{var}(X) + b^2\cdot\operatorname{var}(Y) - 2ab\cdot\operatorname{cov}(X,Y),$$ we have that
\begin{align}\operatorname{var}\left(\frac{X}{\sigma_X}-\frac{Y}{\sigma_Y}\right)
&= \frac{1}{\sigma_X^2}\cdot \operatorname{var}(X) + \frac{1}{\sigma_Y^2}\cdot \operatorname{var}(Y) - 2\frac{1}{\sigma_X\sigma_Y}\cdot \rho_{X,Y}\sigma_X\sigma_Y\\
&= 1 + 1 - 2\frac{1}{\sigma_X\sigma_Y}\cdot (1\times\sigma_X\sigma_Y)\\
&= 0.
\end{align}
Now, a random variable with variance $0$ equals some constant, say  $d$, and so
$$\frac{X}{\sigma_X}-\frac{Y}{\sigma_Y} = d.
$$
 Can you take it from here and show what $Y = mX+c$ of some suitably chosen constants $m$ and $c$?
Nitpickers, please sprinkle in "with probability $1$" at various points.
