How to prove the correlation between two variables using covariance If two variables are related in this way,
$y=a+bx$
How to prove that these variables are perfectly correlated(correlation coefficient of 1) using covariance?
Need some guidance to start..
 A: The one above is right but it seems kind of strange to prove it via a linear model including assumptions. I think it is a bit easier still:
$$
Cov(x,y) = Cov(x,a+bx) = bCov(x,x) = b Var(x) $$
Which gives us for the correlation with $y=a+bx$ so $V(y) = V(a+bx)$:
$$
cor(x,y) = \frac{bV(x)}{\sqrt{V(x)}\sqrt{V(a+bx)}}\\
$$ And because the properties of variance, $a$ is out and $b$ is squared:$$
= \frac{bV(x)}{\sqrt{V(x)}\sqrt{b^2V(x)}} \\
$$ Now it's just fractions$$
= \frac{bV(x)}{\sqrt{V(x)}\sqrt{V(x)}\sqrt{b^2}} \\
= \frac{bV(x)}{\sqrt{V(x)}\sqrt{V(x)}|b|} \\
= \frac{bV(x)}{|b|V(x)} \\
=\pm 1
$$
Important here is just to use the rules of variances. No need to assume a model or error term. Another way to think of this is that the correlation shows the linear relationship. If these variables are related like in the formula, they are obviously perfectly linear dependent and correlation must be 1
A: Start with an econometric model (including error term $e$)
$y=a+bx+e$
$cov(y,x)=cov(a+bx+e,x)=cov(a,x)+cov(bx,x)+cov(e,x)=0+bvar(x)+0$
So, $cov(y,x)=bvar(x)$
or, $b=cov(y,x)/var(x)$
but, we know that $cov(y,x)=cor(x,y)*sd(y)*sd(x)$
so, $b=cor(x,y)*sd(y)/sd(x)$
If $cor(x,y)=1$ (perfect positive correlation):
$b=sd(y)/sd(x)$
If $cor(x,y)=-1$ (perfect negative correlation):
$b=-sd(y)/sd(x)$
Note: by exogenity assumption, $cov(e,x)=0$, $sd()$ means standard deviation, $var()$ means variance, $cov()$ means covariance, and $cor()$ means correlation. 
