0
$\begingroup$

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

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?

$\endgroup$

2 Answers 2

2
$\begingroup$

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.

$\endgroup$
14
  • $\begingroup$ you mean $y^2$? $\endgroup$ Feb 2, 2014 at 17:42
  • $\begingroup$ Do I? Really? Where? $\endgroup$
    – Did
    Feb 2, 2014 at 17:53
  • $\begingroup$ 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. $\endgroup$ Feb 2, 2014 at 17:58
  • $\begingroup$ 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? $\endgroup$
    – Did
    Feb 2, 2014 at 18:18
  • $\begingroup$ "transform U to obtain a random variable with the same distribution as Y" what are we looking for in terms of math symbols? $\endgroup$ Feb 2, 2014 at 18:19
1
$\begingroup$

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 $\phi$.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .