I have to find $E[Y]$ and $Var[Y]$.
$Y=1-2X$ and the variable, X, is exponentially distributed with $\lambda=\frac12$. I know that $E[X]=\frac1\lambda$ and $Var[X]=\frac{1}{\lambda^2}$, this gets me 2 for E[X] and 4 for Var[X] but the question is asking for E[Y] and Var[Y]. I also know that if is of the form Y=aX, $\lambda$ should be replaced with $\frac{\lambda}{a}$ but I'm not sure how to apply these but here I have a one.
And finally, $f(x)=\lambda e^{-\lambda x}$ for all $x>0$. I think it's the one in the Y function that's confusing me. Any help is appreciated!
EDIT: According to the answers, E[Y] = -3 and Var[Y] = 16.
So long as you know the expected value and variance of $X$, you know the expected value and variance of $Y$ (because $Y$ is an affine linear combination of $X$).
$$ E[Y] = E[1-2X] = 1 - 2E[X] $$
$$ V[Y] = V[1-2X] = V[-2X] = 4V[X] $$