Let $X$ and $Y$ have respective cumulative distribution functions $F_x$ and $F_y$, and suppose that for some constants $a \in \mathbb{R}$ and $b > 0$,
$F_x(x) = F_y((x − a)/b)$
(a) Determine $E(X)$ in terms of $E(Y)$ and $V(X)$ in terms of $V(Y)4.
I'm not sure about $V(X)$ but I think $E(X)$ should be $(E(y)-a)/b.$ Would this be right?
For ease of notation, let's say $Y$'s CDF and density are $G(y)$ and $g(y)$ respectively. We have $f(x) = \frac{1}{b}g(\frac{x-a}{b})$, from which we have $EX = \int xf(x)dx = \int (bz+a)g(z)dz = bEY+a$.
From linearity of $E$ you can guess $Var(X) = b^2Var(Y)$. Also, it is not too difficult to go through the change of variables in the integration to evaluate $EX^2 = \int x^2f(x)dx$ as in $EX$.