Standard deviation of error in simple linear regression How to derive the following formula:
where $\sigma(y)$ is the standard deviation of $y$ (dependent variable), and $\rho(x,y)^2$ is correlation between $x$ and $y$ squared, $\sigma(\epsilon)$ the standard deviation of the error. 
$$ \sigma(\epsilon) = \sigma(y) \sqrt{1-\rho(x,y)^2} $$ 
if $y = a +bx + \epsilon$.
 A: Since $X$ and $\epsilon$ are supposed to be independent and so uncorrelated, you can subtract the means, square and average to give $\sigma^2_Y = b^2 \sigma^2_X + \sigma^2_\epsilon$. 
But $b^2 \sigma^2_X = \rho_{X,Y}^2 \sigma^2_Y$ so you can substitute and rearrange to get your result.
A: And here is a really explicit derivation:
For simplicity assume that $a=0$ and that your data is mean adjusted: $E[x]=0; E[y]=0$:
$$\epsilon=y-\beta x$$
$$E[\epsilon]=E[y]-E[x]\beta=0$$
So you get the formula for the variance of $\epsilon$:
$$Var[\epsilon]=E[(y-\beta x)^2]$$
$$=E[y^2-2y\beta x+\beta^2 x^2]$$
$$=E[y^2]-2\beta E[xy]+\beta^2E[x^2]$$
$$=\sigma_y^2-2\beta \sigma_{xy}+\beta^2 \sigma_x^2$$
And about the slope coefficient $\beta$ you know: $$\beta=\frac{\sigma_{xy}}{\sigma_x^2}$$
Then insert this in the last formula:
$$Var[\epsilon]=\sigma_y^2-2\frac{\sigma_{xy}}{\sigma_x^2}\sigma_{xy}+\frac{\sigma_{xy}^2}{\sigma_x^2 \sigma_x^2}\sigma_x^2$$
Then expand the last two terms with $\frac{\sigma_y^2}{\sigma_y^2}$ to get:
$$Var[\epsilon]=\sigma_y^2-2\frac{\sigma_{xy}^2}{\sigma_x^2} \frac{\sigma_y^2}{\sigma_y^2}+\frac{\sigma_{xy}^2}{\sigma_x^2 \sigma_x^2}\sigma_x^2 \frac{\sigma_y^2}{\sigma_y^2}$$
$$=\sigma_y^2-2\rho_{xy}^2\sigma_y^2+\rho_{xy}^2\sigma_y^2$$
$$=\sigma_y^2-\rho_{xy}^2\sigma_y^2=\sigma_y^2(1-\rho_{xy}^2)$$
And finally:
$$\sqrt{Var[\epsilon]}=\sigma_\epsilon=\sigma_y\sqrt{(1-\rho_{xy}^2)}$$
