Let X be a random variable that is uniformly distributed on $[0,1]$. What are the distribution and probability density functions of $Y$ with $Y=\frac{3X}{1-X}$?

I know that the density is the derivative of the distribution function, so if someone could help me find the distribution, I will try to find the density myself. Thanks in advance!

  • $\begingroup$ Can you figure out what values $Y$ can take on? $[0,\infty)$? $(-\infty,\infty)$? $[0,1]$? Once you have done that, can you find, for your favorite number $\alpha$, what is $P\{Y \leq \alpha\}$? How about for other, less favorite numbers? Once you have done this, you have found $P\{Y \leq y\}$ from which you can determine the density function. $\endgroup$ – Dilip Sarwate Mar 28 '15 at 16:17
  • $\begingroup$ @DilipSarwate I think $Y $ can take on values from $[0,\infty)$, but I can't really figure out how your next step should be done $\endgroup$ – JimmyP Mar 28 '15 at 16:19

If $X$ is distributed over $[0,1]$ then $Y$ is distributed over $\mathbb{R}^+$ and for any $r\in\mathbb{R}^+$ we have: $$ \mathbb{P}[Y\leq r]=\mathbb{P}\left[0\leq X \leq \frac{r}{r+3}\right]=\frac{r}{r+3}\tag{1} $$ so, by differentiation, we get that the pdf of $Y$ is given by: $$ f_Y(t) = \frac{3}{(3+t)^2}\cdot\mathbb{1}_{t\geq 0}.\tag{2}$$

  • $\begingroup$ Why do we need the indicator function in the pdf? $\endgroup$ – JimmyP Mar 28 '15 at 16:30
  • $\begingroup$ @JimmyP: because $Y$ does not take negative values. $\endgroup$ – Jack D'Aurizio Mar 28 '15 at 16:32
  • $\begingroup$ And how do you know that (1) is valid? $\endgroup$ – JimmyP Mar 28 '15 at 16:41
  • 1
    $\begingroup$ @JimmyP: because if $Y=\frac{3X}{1-X}$, then $X=\frac{Y}{Y+3}$. $\endgroup$ – Jack D'Aurizio Mar 28 '15 at 16:54

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