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 ($p=1/2$, $n=x$)

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}
$
