$Y=a-bX$, does it work the same way as $Y=a+bX$?

Question

Does $Y=a-bX$ work the same way as $Y=a+bX$ in terms of finding $\mathrm E[Y]$ and $\mathrm{var}[Y]$?

What I mean is, will $\mathrm E[Y] = a-b\mathrm E[X]$, similar to how when $Y=a+bX$, $\mathrm E[Y] = a+b\mathrm E[X]$? Does the variance of $Y$ stay the same too?

Answer 1

The variances would be the same in both cases, but the expectations would not:

$$Y=a-bX\implies\begin{cases}E[Y]=a-b E[X]\\V[Y]=b^2 V[X]\end{cases}$$

$$Y=a+bX\implies\begin{cases}E[Y]=a+b E[X]\\V[Y]=b^2 V[X]\end{cases}$$

Answer 2

Does the Variance of Y stay the same too?

Remember that $V(a)=0$ and $V(aX)=a^2V(X)$

Thus

$$V(a+bX)=V(a-bX)=b^2V(X)$$

Answer 3

$E[a - bX] = a - bE(X)$, and $E[a + bX] = a + bE(X)$. The variances would be the same.