Negative binomial distribution and negative binomial series missing $(-1)^k$ term

Question

The pmf of a negative binomial distribution is

$$p_X(x)= {x-1 \choose r-1}~ p^r~ (1-p)^{x-r}\quad x=r,r+1,\cdots$$

I want to verify that

$$\sum \limits_{x=r}^{\infty} p_X(x)= 1$$

I start with

$$\sum \limits_{x=r}^{\infty} {x-1 \choose r-1}~ p^r~ (1-p)^{x-r}$$

Now let $k=x-r\Rightarrow x=k+r$

$$ p^r ~\sum \limits_{k=0}^{\infty} {k+r-1 \choose r-1}~ (1-p)^{k+r-r}= p^r ~\sum \limits_{k=0}^{\infty} {k+r-1 \choose r-1}~ (1-p)^{k} $$

At this point, the teacher is claiming that

$$ (1- p)^{-r}= \sum \limits_{k=0}^{\infty} {k+r-1 \choose r-1}~ (1-p)^{k} $$

is a negative binomial series.

However, when I look up negative binomial serial on Wolfram, I see that it has a $(-1)^k$ term in the summation.

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• The Negative Binomial Series, arises from binomial theorem for negative integer, $-n$.

• ${-n \choose k}=(-1)^k~{n+k-1 \choose k}~~~\text{where } n>0,~ k>0$

• $(a+b)^{-n} = \sum \limits_{k=0}^{\infty}~ {-n \choose k}~a^k~b^{-(n+k)}$

  where $|a| < b$

• $(a+b)^{-n} = \sum \limits_{k=0}^{\infty} (-1)^k~{n+k-1\choose k}a^k~ b^{-(n+k)}$

  where $|a| < b$

• for $b=1$

  $(a+1)^{-n} = \sum \limits_{k=0}^{\infty} {-n \choose k}~a^k$

  $\boxed{(a+1)^{-n} =\sum \limits_{k=0}^{\infty} (-1)^k~{n+k-1\choose k}~a^k}$

  $(a+1)^{-n} = 1 - na + \frac{1}{2}n(n+1)a^2 - \frac{1}{6}n(n+1)(n+2)a^3+\cdots$

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But, anyways returning the teacher's instruction. At this point he says that the negative binomial series formula is

$$(1-y)^{-r}=\sum_{k=0}^{\infty} {k+r-1 \choose r-1}y^r$$

Therefore, our equation becomes

$$ p^r(1-(1-p))^{-r}=p^r(p)^{-r}=1 $$

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Now for my question:

What happened to the $(-1)^k$ term in the negative binomial series? Is there a mistaken in the math somewhere where it should have a $(-1)^k$?

Answer 1

Taking your boxed equation $$(a+1)^{-n} = \sum_{k=0}^\infty (-1)^k \binom{n+k-1}{k} a^k$$ and choosing $a = p-1$, $n = r$, we obtain $$p^{-n} = (1 + (p-1))^{-n} = \sum_{k=0}^\infty (-1)^k \binom{k+r-1}{k} (p-1)^k = \sum_{k=0}^\infty (-1)^k (-1)^k \binom{k+r-1}{r-1} (1-p)^k,$$ since $$\binom{k+r-1}{r-1} = \frac{(k+r-1)!}{k! (r-1)!} = \binom{k+r-1}{k},$$ and since the product $(-1)^k (-1)^k = ((-1)^2)^k = 1$, the result follows.

Answer 2

To equate your professor's formula with Wolfram Alpha's formula for a negative binomial series, plug $-y$ in for $a$ and $r$ for $n$. After that, the trick to get rid of the $(-1)^k$ can be seen in heropup's answer.

Answer 3

$$(a+1)^{-n} =\sum \limits_{k=0}^{\infty} (-1)^k~{n+k-1\choose k}~a^k$$

Now let a=-b where b >0

$$(-b +1)^{-n} =\sum \limits_{k=0}^{\infty} (-1)^k~{n+k-1\choose k}~(-b)^k$$

$$(1-b)^{-n} = \sum \limits_{k=0}^{\infty} (-1)^k (-1)^k~{n+k-1\choose k}~(b)^k$$

$$(1-b)^{-n} = \sum \limits_{k=0}^{\infty} {n+k-1\choose k}~(b)^k$$