Transform of a discrete random variable Ok guys, I'm having troubles with this transformation. If I have a random variable $Y \sim \mathcal{Unif}(a, b-1)$ and another random variable expressed ad a function of $Y$, $Z$, defined ad follows:
$$Z= \frac{1}{\binom{b}{Y}}$$, how do I compute the PMF of $Z$? Obviously the first distribution is a discrete uniform... I hope that somebody can help!
 A: I'll have a go at it: I assume below that both $a$ and $b$ are nonnegative integers with $b>0$ and $a<b$ so that everything is well defined. We want, for any number $x$, the probability $P_Z(x)$ of $Z$ achieving value $x$. The only possible values that $Z$ can achieve are $\frac{1}{\binom{b}{Y}}$, for $Y=0,1,\ldots,b-1$. Thus $P_Z(x)=0$ for all other numbers $x$.
Each of the values of $Y$ can occur with a probability of $\frac{1}{b-a}$ (careful counting how many elements there are in $\{a,\ldots,b-1\}$!), but we have the symmetry $\binom{n}{k}=\binom{n}{n-k}$ of binomial coefficients, so we must lump together some probabilities before we are done. I'm more comfortable identifying the unique values instead, so, for all integers $c\in \{1,2,\ldots,a-1\}$ we must have $P_Z(1/\binom{b}{c})=\frac{1}{b-a}$, and if $b$ is even, then $P_Z(1/\binom{b}{b/2})=\frac{1}{b-a}$. In the case that $a=0$, then $P_Z(1/\binom{b}{0})=\frac{1}{b-a}(=\frac{1}{b})$ as well, since $Y$ cannot equal $b$. 
For the other cases, the symmetry implies that the probability must be doubled, so $P_Z(1/\binom{b}{c})=\frac{2}{b-a}$ for $c\in\{a,a+1,\ldots,b-1\}\backslash\{b/2\}$.
Sorry if the above is a bit messy, I wrote as I thought on the problem…
Kudos for posting such a fun problem, though :)
