Transformation of a uniform distribution in order to get a random variable distributed like Y.

Question

$f(y)=\begin{cases} \frac{b}{y^2}, & y\ge b,\\ 0, & \mbox{elsewhere}\end{cases}$.

is a bona fide probability density function for a random variable, $Y$. Assuming $b$ is a known constant and $U$ has a uniform distribution on the interval $(0, 1)$, transform $U$ to obtain a random variable with the same distribution as $Y$.

I have no clue how to get started on this question. Could anyone helps me get started on this question or give some hints?

Answer 1

The target distribution is characterized by the fact that any random variable $X$ with this distribution is such that, for every $x\geqslant b$, $$ P(X\geqslant x)=\int_x^\infty f=\int_x^\infty \frac{b}{y^2}\,\mathrm dy=\frac{b}x. $$ On the other hand, if $Y=\dfrac{b}U$ with $U$ uniform on $(0,1)$, then for every $y\geqslant b$, $$ P(Y\geqslant y)=P\left(U\leqslant \frac{b}y\right)=\frac{b}y. $$ Ergo.

Answer 2

Assume $b>0$.

Let $\phi(\alpha) = p \{ y | y \le \alpha \} = \int_{-\infty}^\alpha f(y) dy = \begin{cases} 0, & \alpha <b \\ 1-{b \over \alpha}, & \alpha \ge b\end{cases}$. Note that the restricted $\phi:[b,\infty) \to [0,1)$ is a bijection, and we have $\phi^{-1}:[0,1) \to [b,\infty)$ is given by $\phi^{-1}(y) = { b \over 1-y}$.

Then $\phi^{-1}(U)$ is a random variable with distribution $\phi$.