求零截尾负二项分布n阶矩相关的含组合数无穷级数的值

Hey there! Let's break down how to compute this infinite series, which ties directly to moments of the zero-truncated negative binomial distribution. Here's a step-by-step approach using combinatorial identities and generating functions:

Key Setup & Identity Transformations #

First, recall that the binomial coefficient can be rewritten using symmetry:
$$\binom{r+y-1}{y} = \binom{r+y-1}{r-1}$$
This standard combinatorial identity makes manipulating the series much easier.

Next, we'll use second-kind Stirling numbers ($S(n,k)$) to expand $y^n$ into a sum of combinatorial terms—this is a go-to trick for handling moments in discrete distributions:
$$y^n = \sum_{k=0}^n S(n,k) \cdot k! \cdot \binom{y}{k}$$
Stirling numbers of the second kind count the ways to partition $n$ elements into $k$ non-empty subsets, and this identity lets us convert powers of $y$ into combinations that play nicely with the binomial coefficient in our series.

Substitute & Simplify the Series #

Plug the Stirling number expansion into our original series:
$$\sum_{y=1}^\infty \binom{r+y-1}{y} y^n (1-p)^y = \sum_{y=1}^\infty \binom{r+y-1}{y} \left( \sum_{k=0}^n S(n,k) k! \binom{y}{k} \right) (1-p)^y$$

Swap the order of summation (justified by absolute convergence, since $0 < 1-p < 1$):
$$\sum_{k=0}^n S(n,k) k! \sum_{y=k}^\infty \binom{r+y-1}{y} \binom{y}{k} (1-p)^y$$

Now simplify the product of binomial coefficients using another identity:
$$\binom{r+y-1}{y} \binom{y}{k} = \binom{r+k-1}{k} \binom{r+y-1}{r+k-1}$$
Let $m = y - k$ (so $y = m + k$, $m \geq 0$) to reindex the inner sum:
$$\sum_{m=0}^\infty \binom{r+k-1}{k} \binom{r+m+k-1}{m} (1-p)^{m+k}$$

Factor out constants and recognize the remaining sum as a convergent negative binomial series (valid because $|1-p| < 1$):
$$\binom{r+k-1}{k} (1-p)^k \sum_{m=0}^\infty \binom{(r+k)+m-1}{m} (1-p)^m = \binom{r+k-1}{k} (1-p)^k \cdot \frac{1}{p^{r+k}}$$

Final Closed-Form Expression #

Combine all these pieces back together. Note that $k! \cdot \binom{r+k-1}{k} = (r)_k$, where $(r)_k = r(r+1)\dots(r+k-1)$ is the rising factorial (Pochhammer symbol). This simplifies the series to:

$$\boxed{\frac{1}{p^r} \sum_{k=0}^n S(n,k) \cdot (r)_k \cdot \left( \frac{1-p}{p} \right)^k}$$

Alternatively, expanding the rising factorial explicitly gives:
$$\frac{1}{p^r} \sum_{k=0}^n S(n,k) \cdot r(r+1)\dots(r+k-1) \cdot \left( \frac{1-p}{p} \right)^k$$

Quick Verification with Small n #

Let's test this with $n=1$ (first moment):

• $S(1,1)=1$, $(r)_1=r$
• Result: $\frac{1}{p^r} \cdot 1 \cdot r \cdot \frac{1-p}{p} = \frac{r(1-p)}{p^{r+1}}$
• This matches the expected value: the standard negative binomial distribution has mean $\frac{r(1-p)}{p}$, so dividing by $p^r$ (the constant term from the distribution's PMF) gives exactly this result.

For $n=2$ (second moment):

• $S(2,1)=1$, $S(2,2)=1$; $(r)_1=r$, $(r)_2=r(r+1)$
• Result: $\frac{1}{p^r} \left( r \cdot \frac{1-p}{p} + r(r+1) \cdot \left( \frac{1-p}{p} \right)^2 \right)$
• This aligns with the second moment of the standard negative binomial distribution scaled by $1/p^r$.

内容的提问来源于stack exchange，提问作者EllipticalInitial