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I would like to know the standard form of the negative multinomial expansion i.e. $(x_1 + x_2 + \ldots + x_p)^{-n}$.

I understand that I can probably derive something by applying the negative binomial expansion recursively but it would be nice to know the standard expansion.

Omar Shehab
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    I've shown the derivation of PGF for Negative Multinomial distribution here: https://math.stackexchange.com/questions/1263942/probability-generating-function-of-a-negative-multinomial-distribution. See if it helps. – Vectorizer Oct 21 '20 at 17:10
  • @Vectorizer I am afraid I am asking a stupid question. How is the PGF related to the coefficients of the terms in the series/expansion? – Omar Shehab Oct 21 '20 at 17:17
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    The "probability" tag of your question made me think you are after the negative multinomial distribution. Otherwise, here is a similar question: https://math.stackexchange.com/questions/2724780/negative-multinomial-theorem – Vectorizer Oct 21 '20 at 17:25
  • @Vectorizer, it seems like the question you have referred to (answered by yourself) has an additional constraint of "where the k-th occurrence of the event with the probability $p_{0}$ stops the trials." Hence, I am not sure whether it will be directly applicable. – Omar Shehab Oct 22 '20 at 15:04
  • @Vectorizer, btw, I have found the formula for the PGF in Definition 8.1 of this chapter (https://onlinelibrary.wiley.com/doi/abs/10.1002/9781118445112.stat01250). It is as follows. $$G_x(\boldsymbol{t}) = \left(\mathcal{Q} - \sum^k_{i=1} P_{it_i}\right)^{-n}$$, where $n > 0$, $P_i > 0(i = i, \ldots, k)$, and $\mathcal{Q} - \sum^k_{i=1} P_{i} = 1$. I am still confused what $\mathcal{Q}$ and $P_i$ mean. – Omar Shehab Oct 22 '20 at 15:25
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    @OmarShebah : "where the k-th occurrence of the event with the probability p0 stops the trials." It follows directly from the negative multinomial distribution's definition. – Vectorizer Oct 22 '20 at 15:42
  • @Vectorizer, am I correct to understand that $k$ in your definition correspond to $n$ in $G_x(\boldsymbol{t}) = \left(\mathcal{Q} - \sum^k_{i=1} P_{it_i}\right)^{-n}$ used in my comment above? – Omar Shehab Oct 22 '20 at 17:50

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