Show that $X=F^{-1}(U)$ Cumulative distribution $F(x)$

Question

Let $F(x)$ be a cdf, and assume that that $U$~$uniform(0,1)$ Show that then $X=F^{-1}(U)$ cumulative distribution $F(x)$.

$F(X)=1-k^{\theta}x^{-\theta}$ for $x>k$, $0$ otherwise

my initial thought is that we can let $F_U(U)$ be a cdf of $X=F(X)$ for every $U\in[0,1]$

$F_U(U)=P(X\leq U)$

I am not sure if this even the right way to do it.

Answer 1

The cdf of $X$ is $P(X\le x)=P(U\le F(x))=F(x)$, as required. In the case $F=1-(k/x)^\theta$, $x=k(1-F)^{-1/\theta}$ with $F\sim U(0,\,1)$.

Answer 2

Not sure if you're asking for the general situation or for the specific cumulative distribution that you gave, but if the cumulative density of $X$, $F_X$, is invertible then $F_X^{-1}(U)$ and $X$ have the same distribution because $$ P(F_X^{-1}(U) \leq x) = P(U \leq F_X(x)) = \int_0^{F_X(x)} 1 dx = F_X(x) = P(X \leq x) $$ so they have the same cumulative density function.