I'm a bit confused with the first steps in this problem:
$F(x)=x^4$ for $0<x<1$
a) Find $E[(1/X)-1]$
b) Let $Y=(1/X)-1$. Find the support of $Y$, its pdf and CDF. Name its distribution and its parameters.
c) Compute $E(Y)$ and compare it to the result in part a.
d) Compute the variance of $Y$ (or use know formulae).
So I can tell that the CDF given is a Beta correct? With a pdf of $f(x)=4x^3$. Now to find the expected value of a) is there some sort of transformation involved here? or do I just (integrate $x \cdot ((1/X)-1)$) but that is just an equation so I don't know here I'm lost. Just looking for guidance with this problem that would be really helpful.
Hint: Suppose that the random variable $X$ has density function $f_X(x)$. Let $Y=g(X)$. Then (for reasonable $g$) we have $$E(Y)=E(g(X))=\int_{-\infty}^\infty g(x)f_X(x)\,dx.$$
For the variance of $Y^2$, one way is to calculate $E(Y^2)$ in basically the same way as you computed $E(Y)$, and use the fact that $\text{Var}(Y)=E(Y^2)-(E(Y))^2$.
For the distribution of $Y$, the support of $X$ is $(0,1)$, so $\frac{1}{X}$ ranges over the interval $(1,\infty)$, and therefore $Y$ ranges over $(0,\infty)$.
To find the distribution of $Y$, you can use a transformation. Or else you can compute $\Pr(Y\le y)$ directly, since for positive $y$ we have $\frac{1}{X}-1\le y$ if and only if $X\ge \frac{1}{1+y}$.