Can the sum be simplified? $\sum_\limits{x=y}^{\infty} {x \choose y}\left(\frac{1}{3}\right)^{x+1}$ Let:
$$f(y) = \sum_{x=y}^{\infty} {x \choose y} \left(\frac{1}{3}\right)^{x+1}$$
Can this be simplified somehow? Is it a standard probability distribution?
I can only get as far as:
$$f(y) = \frac{1}{3} \sum_{x=y}^{\infty} \frac{x!}{y!(x-y)!} \left(\frac{1}{3}\right)^{x}$$
This arose in the context of a probability problem in which X is a geometrically distributed random variable ($p=1/2$) and $Y$ is a binomially distributed random variable ($p=1/2$, $n=x$)
 A: $$f(y)=\sum_{x=y}^∞ \binom xy(\tfrac 13 )^{x+1} = {\tfrac 12}^{1+x}$$

This arose in the context of a probability problem in which X is a geometrically distributed random variable (p=1/3) and Y is a binomially distributed random variable (p=1/2, n=x)

In effect you have a sequence of trials each with three equally probable outcomes: head, tail, end and $Y$ is the count of heads before the (first) end.   $Y$ is also the count of heads before the end in all trials which don't show tails.   In any trial that doesn't show tails, there is an equal probability of instead showing heads or an end.
Thus $Y\sim\mathcal{Geom}_0(1/2)$ and hence $f_Y(y) = (\tfrac 1 2)^{y+1}$
(Didn't I answer this earlier?)
A: Use the generalized binomial theorem
$(1-z)^{-s}
=\sum_{k=0}^{\infty} \binom{k+s-1}{s-1}z^k
=\sum_{k=s-1}^{\infty} \binom{k}{s-1}z^{k-s+1}
=z^{-s+1}\sum_{k=s-1}^{\infty} \binom{k}{s-1}z^{k}
$
so that
$\sum_{k=s-1}^{\infty} \binom{k}{s-1}z^{k}
=z^{s-1}(1-z)^{-s}
$
or
$\sum_{k=u}^{\infty} \binom{k}{u}z^{k}
=z^{u}(1-z)^{-u-1}
$.
See, for example,
here:
https://en.wikipedia.org/wiki/Binomial_theorem
Then
$\begin{array}\\
f(y) 
&= \sum_{x=y}^{\infty} {x \choose y} \left(\frac{1}{3}\right)^{x+1}\\
&= \frac13\sum_{x=y}^{\infty} {x \choose y} \left(\frac{1}{3}\right)^{x}\\
&=\frac13(\frac13)^{y}(1-\frac13)^{-y-1}\\
&=(\frac13)^{y+1}(\frac23)^{-y-1}\\
&=(\frac12)^{y+1}\\
\end{array}
$
