Show that, for $t>1$, $P[\frac{Y}{Z}\leq t]=\frac{t-1}{t+1}$ Let the distribution of $X$ be $U(0,1)$. Let U be the length of the shorter of the intervals $(0,X)$ and $(X,1)$; that is, $Z=min(X,1-X)$ and let $Y=1-Z$ be the length of the larger part. Show that, for $t>1$, $P[\frac{Y}{Z}\leq t]=\frac{t-1}{t+1}$;Find the PDF of $Y/Z$.
I am unable to find the distribution function.I can find PDF of $Y/Z$. Please help me to find out $P[\frac{Y}{Z}\leq t]=\frac{t-1}{t+1}$

First, let's determine the behaviour of $\frac{Y}{Z}$ given $X$. We have $\frac{Y}{Z} = \frac{1-X}{X} = \frac{1}{X} - 1$ for $X \in (0,\frac{1}{2}]$ and $\frac{Y}{Z} = \frac{X}{1-X} = \frac{1}{1-X} -1$ for $X \in [\frac{1}{2}, 1)$
Now the question is, when $\frac{Y}{Z} \leq t$? For $X \in (0, \frac{1}{2}]$ it's equivalent to $\frac{1}{X} - 1 \leq t$ or $X \geq \frac{1}{t+1}$. For $X \in [\frac{1}{2}, 1)$ it is equivalent to $\frac{1}{1-X} - 1 \leq t$, which is equivalent to $1-X \geq \frac{1}{t+1}$ or $X \leq \frac{t}{t+1}$.
That means that $\frac{Y}{Z} \leq t $ if and only if $X \in [\frac{1}{t+1} , \frac{t}{t+1}]$. Since $X$ is distributed from $(0,1)$ with uniform probability, the probability it will be in desired range is equal to the length of this range, and hence we have the desired $P[\frac{Y}{Z} \leq t] = \frac{t-1}{t+1}$.
Now to get the PDF when you have the probability function, you just need to differentiate the probability over $t$. 
