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I'm reading Achim Klenke's text on probability theory, and a few different distributions (with applications) have been introduced. I'm confused by the explanation given of the negative binomial distribution: $$ b_{r,p}^- := \sum_{k=0}^\infty \binom{-r}{k}(-1)^k p^r(1-p)^k \delta_k $$ where $\delta_k : 2^\mathbb R \to \{0,1\}$ is the point measure. Klenke says that "If $r \in \mathbb N$, then one can show ... that $b_{r,p}^-$ is the distribution of the waiting time for the $r$th success in a series of random experiments". I think I'm interpreting this explanation incorrectly, because as I understand it, something doesn't make sense.

Suppose that $X_r : \Omega \to \mathbb R$ is this random variable, and $(\Omega, \mathcal A, \mathbb P)$ is our probability space. In particular, $\mathbb P_{X_r} = \mathbb P \circ X_r^{-1} = b_{r,p}^-$. In this setting, $\mathbb P(X_r=k) = b_{r,p}^-(\{k\}) = \binom{-r}k (-1)^k p^r(1-p)^k$ is the probability that the $r$th success will come at time $k$. The reason I'm confused is, practically speaking, if $r > k$, shouldn't this probability be $0$? For example, if our experiments are coin tosses with $p=1/2$, and we consider "heads" to be success, if $r=3$, the probability of getting $3$ heads on the $2$nd experiment ($k=2$) is obviously $0$, but according to this formula, the probability is $$\mathbb P(X_3 = 2) = \binom{-3}{2}(-1)^2\left(\frac 1 2 \right)^3 \left(\frac 1 2 \right)^2 = \frac 1{32} \times \frac{(-3)(-4)}{2} = \frac 3{16}$$ which seems absurd to me. Can anyone shed some light on this interpretation?

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For the negative binomial defined by $$\Pr[\mathbf X = k] = (-1)^k\binom{-r}{k} p^r (1-p)^k$$ the precise "waiting time" characterization is that $\mathbf X$ counts the number of failures seen before the $r^{\text{th}}$ success. Thus, $\mathbf X=0$ corresponds to the outcome in which there are no failures at all: the first $r$ outcomes are successes. The formula above correctly says that this has probability $p^r$.

That being said, some people do define the negative binomial to be the total number of trials before the $r^{\text{th}}$ success. Since the total number of trials is just $r$ more than the number of failures, this shifts everything over by $r$ and then the $\Pr[\mathbf X = k] $ is indeed equal to $0$ for $k < r$. In this version, the formula is $$\Pr[\mathbf X = k] = \binom{k-1}{r-1} p^r (1-p)^{k-r}.$$

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    $\begingroup$ I see, so whereas I was interpreting $k$ as the number of experiments, I should have been interpreting it as the number of failures. Thank you for the clarification! $\endgroup$ – D Ford Nov 7 '17 at 4:33

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