There is this passage in a textbook Stochastics by Hans-Otto, Georgi:
An urn contains red and white balls, where the proportion of red balls is $0<p<1$. The balls are drawn with replacement. What is the distribution of the waiting time until the $r$th success, i.e., until we draw the $r$th red ball? Here, $r\in\mathbb{N}$ is a fixed number.
Since we must draw at least $r$ times, we consider the remaining waiting time after $r$ draws, or equivalently, the number of failures before the $r$th success; then the sample space is $\Omega=\mathbb{Z_+}$. Which $P$ describes the situation?
On the infinite product space $\{0,1\}^\mathbb{N}$ we could define the random variable
$$T_r(\omega)=\min\left\{k:\sum_{i=1}^{k}\omega_i=r\right\}-r$$
and obtain $P$ as the distribution of $T_r$.
My doubts are:
I understand $\min\left\{k:\sum_{i=1}^{k}\omega_i=r\right\}$ is the minimum value of draws to get the $r$th success. What I don't quite understand is why do we still need to take away $r$ in the end? Does that mean we can get a negative value for $T$? If yes, what does the negative value of $T$ represent?
What does the random variable $T$ ACTUALLY represent? Why do we have to take the minimum?
Also I don't quite understand how can the statement "the remaining waiting time after $r$ draws" to be equivalent to "the number of failures before the $r$th success". Should we erase the word "remaining"? It doesn't make sense for me.
Can anyone please help me clarify these doubts?
Thanks.
There are two commonly understood parametrizations of a negative binomial random variable. One counts the number of trials (in your example, a "trial" consists of a random draw of a ball) that have occurred when the $r^{\rm th}$ red ball is drawn. The other counts the number of failures (in your example, a "failure" consists of a random draw of a white ball) that have occurred when the $r^{\rm th}$ red ball is drawn.
Clearly, if we call the first random variable $T$ and the second one $S$, then $$S = T + r,$$ which expresses the idea that if we have observed $T$ draws of a white ball when the $r^{\rm th}$ red ball is drawn, then the total number of balls drawn at that moment must be $S = T+r$, or equivalently, $T = S - r$.
Now, $S$ is the minimum index of the sum in your notation. This sum cannot be less than $r$, since each $\omega_i \in \{0,1\}$, so $$\sum_{i=1}^k \omega_i \le k,$$ with equality occurring if and only if $\omega_i = 1$ for all $i \in \{1, 2, \ldots, k\}$. So the smallest possible $k$ such that $S = r$ is $k = r$, which guarantees that the subtraction of $r$ does not make $T$ negative. Put in words, $T$ counts the number of white balls drawn. You could conceivably draw no white balls--every single one is red--but this is another way of saying that of the first $r$ balls drawn, all of them were red; hence in such a case $S = r$ and $T = r-r = 0$. Everything works out.
Why do we take the minimum? Because if $\sum_{i=1}^k \omega_i = r$, by no means does this guarantee that $\sum_{i=1}^{k+1} \omega_i > r$. In words, the stopping rule is to count only those balls up until the $r^{\rm th}$ red ball. If we continued to draw and get a white ball, the total number of red balls we observed is still $r$. So to mathematically express this idea that we stop as soon as the $r^{\rm th}$ red ball is drawn, we take the minimum $k$ such that the sum equals $r$: this is the notion that we count the fewest possible balls necessary to observe the required $r$ red balls, and no more.
As for your last question, the book phrases it poorly in my opinion. The "waiting time" here is expressed in units of trials, not actual chronological time. That is to say, it doesn't matter how long we wait between trials: the "waiting time" the book describes is simply the number of draws we make.
That said, suppose $r = 3$ and we are counting the number of trials needed to observe the third red ball. Suppose further that our random outcome was $$\{w, r, w, w, r, w, r\}.$$ We stop drawing once the third instance of $r$ is observed. Here, we find $S = 7$ trials in total, $T = 4$ of which were white. We have thus far thought of $T$ as the number of white balls drawn until the third red ball. But you can also think of it as the number of trials after the third trial to observe the third red ball. It makes no difference whether you count failures, or you count the extra trials after the third.