CDF for Negative Binomial Distribution I am trying to show that the following statement is true.
$$
\sum_{x = r}^{X}\binom{x-1}{r-1}p^r(1-p)^{x-r} =
\sum_{x = r}^{X}\binom{X}{x}p^x(1-p)^{X-x}
$$
Where $X$ and $r$ and $p$ are constants, with $X \geq r$, and $ 0 \leq p \leq 1.$
How did I get there? Well, this is the story:
Consider a sequence of independent binomial trials, each one producing the result success or failure, with probabilities $p$, and $1-p$, respectively.
Let $x$ be the total number of trials which must be carried out in order to attain exactly $r$ successes.
Knowing that the probability mass function for this Negative Binomial Distribution is as follows,
$P(x=X)=\binom{X-1}{r-1}p^r(1-p)^{X-r}$, (for $X \geq r$),
I was trying to prove the following about the corresponding Cumulative Distribution Function.
$P(x \leq X)=\sum_{x=r}^{x=X}\binom{x-1}{r-1}p^r(1-p)^{x-r}=1-\sum_{x=0}^{x=r-1}\binom{X}{x}p^x(1-p)^{X-x}$     
I started out with the following:
$\sum_{x=r}^{\infty}\binom{x-1}{r-1}p^r(1-p)^{x-r}=1$, 
which can be recast as below. (Relation I)
$\sum_{x=r}^{x=X}\binom{x-1}{r-1}p^r(1-p)^{x-r}+\sum_{x=X+1}^{\infty}\binom{x-1}{r-1}p^r(1-p)^{x-r}=1$ 
In addition, from binomial theorem, we have:
$\left ( p+(1-p) \right )^X=\sum_{x=0}^{x=X}\binom{X}{x}p^x(1-p)^{X-x}=1$
Which can be restated in Relation II as below.
$\sum_{x=0}^{x=r-1}\binom{X}{x}p^x(1-p)^{X-x}+\sum_{x=r}^{x=X}\binom{X}{x}p^x(1-p)^{X-x}=1$
Comparing the relations I and II with the expression for the CDF, the proof boils down to verification of the following:
$\sum_{x=r}^{x=X}\binom{x-1}{r-1}p^r(1-p)^{x-r}=\sum_{x=r}^{x=X}\binom{X}{x}p^x(1-p)^{X-x}$
Any idea how to continue form this point onward?
Thanks.

Here is another  variation of the theme. It is convenient to  use the  coefficient of  operator  $[z^r]$  to denote the coefficient of $z^r$ of a series. This way we can write e.g.
\begin{align*}
[z^r](1+z)^t=\binom{t}{r}
\end{align*}

We observe LHS and RHS are polynomials in $p$ with lowest degree $r$ and highest degree $X$.
We prove the polynomials 
  \begin{align*}
G(p)&=\sum_{j=r}^X\binom{j-1}{r-1}p^r(1-p)^{j-r}\qquad\qquad\qquad 0\leq r \leq X, 0\leq p\leq 1\\
H(p)&=\sum_{j=r}^X\binom{X}{j}p^j(1-p)^{X-j}
\end{align*}
  are equal by showing equality of the coefficients
  \begin{align*}
[p^t]G(p)=[p^t]H(p)\qquad\qquad\qquad\qquad\qquad\qquad& r\leq t\leq X
\end{align*}

$$ $$

We obtain
  \begin{align*}
[p^t]G(p)&=[p^t]\sum_{j=r}^X\binom{j-1}{r-1}p^r(1-p)^{j-r}\\
&=\sum_{j=r}^X\binom{j-1}{r-1}[p^{t-r}]\sum_{k=0}^{j-r}\binom{j-r}{k}(-p)^k\tag{1}\\
&=\sum_{j=r}^X\binom{j-1}{r-1}\binom{j-r}{t-r}(-1)^{t-r}\tag{2}\\
&=(-1)^{t-r}\binom{t-1}{r-1}\sum_{j=t}^X\binom{j-1}{t-1}\tag{3}\\
\end{align*}

Comment:


*

*In (1) we use the linearity of the coefficient of operator and apply the rule $[z^{t-r}]A(z)=[z^t]z^rA(z)$.

*In (2) we select the coefficient of $p^{t-r}$.

*In (3) we use the binomial identity $$\binom{j-1}{r-1}\binom{j-r}{t-r}=\binom{t-1}{r-1}\binom{j-1}{t-1}$$ and we set the lower limit of the sum to  $j=t$ since otherwise $\binom{j-1}{t-1}=0$.

Since
  \begin{align*}
\sum_{j=t}^X&\binom{j-1}{t-1}=\sum_{j=0}^{X-t}\binom{t+j-1}{j}=\sum_{j=0}^{X-t}\binom{-t}{j}(-1)^j\\
&=\sum_{j=0}^{X-t}[z^j](1+z)^{-t}(-1)^j\\
&=[z^0](1+z)^{-t}\sum_{j=0}^{X-t}\left(-\frac{1}{z}\right)^j\\
&=[z^0](1+z)^{-t}\frac{1-\left(-\frac{1}{z}\right)^{X-t+1}}{1+\frac{1}{z}}\\
&=(-1)^{X-t}[z^{X-t}](1+z)^{-(t+1)}\\
&=(-1)^{X-t}\binom{-(t+1)}{X-t}\\
&=\binom{X}{t}
\end{align*}
we obtain from (3)
  \begin{align*}
[p^t]G(p)=(-1)^{t-r}\binom{X}{t}\binom{t-1}{r-1}\qquad\qquad r\leq t\leq X
\end{align*}

And now the RHS

We obtain using the same techniques as above
  \begin{align*}
[p^t]H(p)&=[p^t]\sum_{j=r}^X\binom{X}{j}p^j(1-p)^{X-j}\\
&=\sum_{j=r}^X\binom{X}{j}[p^{t-j}]\sum_{k=0}^{X-j}\binom{X-j}{k}(-p)^k\\
&=\sum_{j=r}^X\binom{X}{j}\binom{X-j}{t-j}(-1)^{t-j}\\
&=(-1)^t\binom{X}{t}\sum_{j=r}^t\binom{t}{j}(-1)^j\tag{4}
\end{align*}

Comment:


*

*In (4) we use the binomial identity 
\begin{align*}
\binom{X}{j}\binom{X-j}{t-j}=\binom{X}{t}\binom{t}{j}
\end{align*}
and we also set the upper limit of the sum to $j=t$ since otherwise $\binom{t}{j}=0$.



Since
  \begin{align*}
\sum_{j=r}^t&\binom{t}{j}(-1)j=\sum_{j=0}^{t-r}\binom{t}{j+r}(-1)^{j+r}\\
&=\sum_{j=0}^\infty[z^{j+r}](1+z)^t(-1)^{j+r}\tag{5}\\
&=[z^r](1+z)^t(-1)^r\sum_{j=0}^\infty\left(-\frac{1}{z}\right)^j\\
&=(-1)^r[z^r](1+z)^t\frac{1}{1+\frac{1}{z}}\\
&=(-1)^r[z^{r-1}](1+z)^{t-1}\\
&=(-1)^r\binom{t-1}{r-1}
\end{align*}

Comment:


*

*In (5) we set the upper limit to $\infty$ without changing anything since we are adding zeros only.

*In (6) we use the formula of the geometric series expansion.

we obtain from (4)
  \begin{align*}
[p^t]H(p)=(-1)^{t}\binom{X}{t}\binom{t-1}{r-1}\qquad\qquad r\leq t\leq X
\end{align*}

showing the coefficients of $G(p)$ and $H(p)$ are equal. We finally conclude:

The following is valid
  \begin{align*}
G(p)=H(p)=\sum_{j=r}^X(-1)^{j}\binom{X}{j}\binom{j-1}{r-1}p^j
\end{align*}

