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

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

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?

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.

• you mean $y^2$? Feb 2, 2014 at 17:42
• Do I? Really? Where?
– Did
Feb 2, 2014 at 17:53
• your y... I mean I do not quite get your logic here...Also, I am confused by what the question really wants us to do. Could you let me know what this question wants in term of the actual function. Feb 2, 2014 at 17:58
• Ach so... you do not understand the question. Which part of 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. is unclear?
– Did
Feb 2, 2014 at 18:18
• "transform U to obtain a random variable with the same distribution as Y" what are we looking for in terms of math symbols? Feb 2, 2014 at 18:19

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 $\phi$.