Friday, 7 February 2014

probability - Independent Uniformly Distributed Random Variables




I have this problem in my textbook and I just can't understand the answer provided.



It says "Let $X$ and $Y$ be independent random variables that are uniformly distributed on the interval $[0,1]$. What is the $PDF%$ of the random variable $Z=Y/X$?"



It then goes on to first find the $CDF$ of $Z$ and then differentiate to get the $PDF$. I understand that step, but I can't figure out where they're getting these answers for the $CDF$.



It says the $CDF$ of $Z$ from $[0,1]$ is $z/2$. I flat out do not understand how that is correct and would really appreciate it if someone could explain it.



Then it says the $CDF$ of $Z$ if $z>1$ is $1-1/(2z)$. I don't understand this one either...




I'm pretty sure this is supposed to be a relatively simple example, so I'm a little worried that I don't understand it. Some explanation here would be incredibly appreciated.


Answer



We have $Z\le z$ if $Y/X\le z$, i.e. $Y\le zX$. For $z\le1$, the probability for this, given $X=x$, is $zx$, and then integrating over $x$ yields



$$
\int_0^1zx\,\mathrm dx=\frac z2\;.
$$



To get the result for $z\gt1$, consider $1/Z=X/Y$ and use the result for $z\le1$.



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