Given $X$ and $Y = a+bX$, I have to prove that:
If $b \lt 0$, then $\rho = -1$. If $b \gt 0$, then $\rho = 1$.
I've gotten to the point where I have:
$ \rho = \frac{b \cdot \sigma_x }{ \sqrt{\sigma_y^2} }$
I need to find why $\sigma_y^2 = b^2 \sigma_x^2$.
Can anyone please explain to me why that is the case?
I tried rewriting $\sigma_y^2$ as: $E[(a+bX)(a+bX)] - E^2[a+bX] $. And got: $a^2 + 2abE[X] + b^2E[X^2] - a^2 - b^2E^2[X] ,$ which became: $2ab E[X] + b^2 \sigma_x^2$.
What is the extra 2abE(x) term? Is that supposed to go away somehow?
Thanks.