Let $X$ be the annual income to a randomly selected person in a population group. It is normal to assume that $X$ is Partero-distributed, that means that $X$ has the PDF of
$$ f_X(x)=\begin{cases}\theta k^\theta x^{-\theta -1}& \text{for} ~~x>k\\0 & \text{otherwise,}\end{cases} $$
show that the CDF is given by
$$ F_X(x)=\begin{cases}1-k^\theta x^{-\theta} & \text{for} ~~x>k\\ 0 & \text{otherwise}\end{cases} $$
and find the median of the annual income
Me:
I know that
$$ \int^x_{-\infty}{f_X(x)dx}=F_X(x) $$
since $x$ is not defined under $k$ we get $\int_k^x{f(x)}dx$ if we integrate this with respect to $x$ and apply the limits we would get
$$ F_X(x)=1-k^{\theta}x^{\theta} ~~\text{for}~~~ x>k $$
now that that's done. I have to find the median. and this is where i'm stuck. Do I find the median of the CDF or PDF?
$$ F_X(x)=\frac{1}{2}\Longrightarrow \int^M_? 1-k^{\theta}x^{\theta}=\frac{1}{2} $$
and what under limit am i supposed to use?
The idea is
$$ \int_{-\infty}^m f_X(x){\rm d}x = \frac{1}{2} = F(m) $$
So the problem becomes
$$ \frac{1}{2} = 1 - \left(\frac{k}{m}\right)^{\theta} $$
and from there
$$ \frac{k}{m} = 2^{-1/\theta} $$
So the median is
$$ m = 2^{1/\theta}k $$