How could I find a function $f: \Bbb {R}\rightarrow\Bbb {R}$ so that $Y=f(X)$ has the given probability density?

Question

Let $X$ be a random variable that is uniformly distributed on the interval $[0, 1]$.

How could I find a fucntion $f: \Bbb {R}\rightarrow\Bbb {R}$ so that $Y=f(X)$ has a probability density given by $g_Y(y)=1/y^2$ when $y\geqslant 1$ and everywhere else $g_Y(y)=0$?

I know that for a uniform distribution you get the function $f(x)=1/(\beta-\alpha)$ for $\alpha\leqslant x \leqslant \beta$. So I guess that for variable $X$, $f(x)=1$.

Answer 1

Hint: This is called Inverse transform sampling in statistics.