Application of power series/ binomial theorem in inverse sampling I have posted this already in other forums. Apologies for cross posting. 
In order to establish some properties of inverse sampling, Haldane (1945) uses power series and the binomial theorem I assume. According to the inverse-sampling method (Haldane 1945) you continue sampling until m of the rare items have been found. 
Let p be the frequency of the rare item and q = 1-p. m is the number of rare items observed. n is the number of observations 
you continue sampling until m of the rare items have been found. 
Let p be the frequency of the rare item and q = 1-p. m is the number of rare items observed. n is the number of observations 
What is now the probability that exactly n observations have been made before m rare items are observed? According to Haldane (1945) this probability is 
$w_{n} = \binom{n-1}{m-1}p^{m}q^{n-m}$
Haldane then concludes that "this is the coefficient of $t^n$ in $(\dfrac{qt}{1-qt})^m$ " (Haldane 1945: 222)
He does not go more into detail here and just continues with his proof on inverse sampling. However, I just do not see how this coefficient of $t^n$ is related to the probability $w_{n}$. 
Assuming a geometric series, I could write 
$\dfrac{qt^m}{(1-qt)} = \sum_{i=0} qt^mqt^i$
Furthermore, through differentiation of the power series, we get:
$\dfrac{qt^m}{(1-qt)^m} = \sum_{i=0} \binom{m+i+1}{i}qt^mqt^i$
Here, I get stuck. What is implied? Does the coefficient of $t^n$ simply refer to $\binom{m+i+1}{i}qt^mqt^i$? If so, however, how does it link up with $w_{n} = \binom{n-1}{m-1}p^{m}q^{n-m}$? Thanks for any hints/ helps!
 A: It's convenient to use the coefficient of operator $[t^n]$ to denote the coefficient of $t^n$ in a series. We can write this way
\begin{align*}
  [t^n](1+t)^m=\binom{m}{n}
  \end{align*}

I think there's a small typo and we have instead
  \begin{align*}
  w_n=\binom{n-1}{m-1}p^mq^{n-m}=[t^n]\left(\frac{\color{blue}{p}t}{1-qt}\right)
  \end{align*}
We obtain
  \begin{align*}
[t^n]\left(\frac{pt}{1-qt}\right)^m&=[t^n]p^mt^m\frac{1}{(1-qt)^m}\\
&=p^m[t^{n-m}]\sum_{k=0}^{\infty}\binom{-m}{k}(-q)^kt^k\tag{1}\\
&=p^m[t^{n-m}]\sum_{k=0}^{\infty}\binom{m+k-1}{m-1}q^kt^k\tag{2}\\
&=\binom{n-1}{m-1}p^mq^{n-m}\tag{3}\\
\end{align*}
  and the claim follows.

Comment:


*

*In (1) we use the linearity of the coefficient of operator and the rule $$[t^{n-m}]A(t)=[t^n]t^mA(t)$$ 
We obtain      for  $0\leq m\leq n$
\begin{align*}
[t^{n-m}]A(t)&=[t^{n-m}]\sum_{k=0}^{\infty}a_kt^k=a_{n-m}\\
[t^n]t^mA(t)&=[t^n]t^m\sum_{k=0}^{\infty}a_kt^k
=[t^n]\sum_{k=0}^{\infty}a_kt^{k+m}
=a_{n-m}
\end{align*}
We also apply the binomial series expansion.

*In (2) we use the binomial identity
\begin{align*}
  \binom{-m}{k}=\binom{m+k-1}{m-1}(-1)^k
  \end{align*}

*In (3) we select the coefficient of $t^{n-m}$ and take the summand with $k=n-m$.
A: The expression $w_n$ that is described is the probability mass function of a particular parametrization of the negative binomial distribution.  If $W$ is the random number of observations until we obtain the $m^{\rm th}$ rare event, where the probability of observing a rare event is $p$, then $$\Pr[W = n] = w_n = \binom{n-1}{m-1} p^m (1-p)^{n-m}, \quad n = m, m+1, \ldots.$$  This is because there are $\binom{n-1}{m-1}$ ways to arrange $m$ rare events among $n$ total observations such that the final observation is a rare event, and the joint probability of observing any particular arrangement of $m$ rare events among $n-m$ non-rare events is $p^m (1-p)^{n-m}$.
Thus, $$1 = \sum_{n=m}^\infty \Pr[W = n] = \sum_{n=m}^\infty \binom{n-1}{m-1} p^m (1-p)^{n-m}.$$  Now consider the function $$\begin{align*} P_W(t) = \operatorname{E}[t^W] &= \sum_{n=m}^\infty t^n \Pr[W = n] \\ &= (tp)^m \sum_{n=m}^\infty \binom{n-1}{m-1} (t(1-p))^{n-m} \\ &= \frac{(tp)^m}{(1-t(1-p))^m} \sum_{n=m}^\infty \binom{n-1}{m-1} (1-t(1-p))^m (t(1-p))^{n-m} \\ &= \left(\frac{pt}{1-t(1-p)}\right)^m \sum_{n=m}^\infty \binom{n-1}{m-1} (p^*)^m (1-p^*)^{n-m} \\ &=  \left(\frac{pt}{1-t(1-p)}\right)^m, \quad 0 < t < \frac{1}{1-p}, \end{align*}$$ where $p^* = 1 - t(1-p)$, and the last summation is equal to $1$ because it is the sum over all outcomes of a modified negative binomial random variable $W^*$ with modified probability of observing a rare event $p^*$.  We conclude that $w_n$ is the coefficient of the $t^n$ term in the series expansion of the function $$P_W(t) = \left(\frac{pt}{1-t(1-p)}\right)^m.$$  Note that the numerator should be $(pt)^m$, not $(qt)^m$.
