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Derive the probability generating function (pfg) of a negative multinomial distribution with parameters $(k; p_{0}, p_{1}, ..., p_{r})$ where the k-th occurrence of the event with the probability $p_{0}$ stops the trials.

My approach: Find the pgf of the event that is stopped by the first occurrence of the event associated with $p_{0}$ then raise that expression to the k-th power.

This elementary event is a collection of multinomial event (excluding the stopping event) sequence of length $0 \rightarrow \infty$, followed by the stopping event:

$g_{elementary}(s_{1},...,s_{r}) = p_{0}(\sum_{j=0}^\infty (g_{multinomial}(s_{1},...,s_{r}))^j =\\ p_{0} / (1-g_{multinomial}(s_{1},...,s_{r}))$

where $g_{multinomial}(s_{1},...,s_{r})$ is the pgf of the multinomial event sequence of length $1$ with r possible outcomes, i.e. $\sum_{i=1}^r p_{i}s_{i}$

Raising $g_{elementary}(s_{1},...,s_{r})$ to the k-th power results in:

$g(s_{1},...,s_{r}) = p_{0}^k (1-\sum_{i=1}^r p_{i}s_{i})^{-k}$

While this result is correct, I am not sure about my reasoning.

1 Answers1

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As I mentioned in my original post, I was not comfortable with the reasoning.

We can think of negative-multinomial as a multinomial sequence of length M $(M:0\rightarrow \infty)$ mixed with $k-1$ occurences of type $0$ event, which can happen in $\binom{M+k-1}{k-1}$ ways, followed by the last occurrence of the type $0$ event; which terminates the trials. Thus the distribution function of negative-multinomial $(k, p_{0};p_{1},...,p_{r})$: $$ \sum_{M=0}^{\infty} p_{0}^{k-1} \binom{M+k-1}{k-1} (\sum_{j_{1}+...+j_{r} = M}\frac{M!}{j_1!...j_r!}\prod_{q=1}^r p_{q}^{j_q}) p_0$$ $$\phi(s_1,...,s_r) = \sum_{j_1=0}^\infty...\sum_{j_r=0}^\infty{\{\sum_{M=0}^{\infty} p_{0}^{k} \binom{M+k-1}{k-1} (\sum_{j_{1}+...+j_{r} = M}\frac{M!}{j_1!...j_r!}\prod_{q=1}^r p_{q}^{j_q}s_q^{j_q})\}}$$ $$\phi(s_1,...,s_r) = \sum_{M=0}^\infty p_0^k \binom{M+k-1}{k-1}(s_1p_1+...+s_rp_r)^M$$ $$\phi(s_1,...,s_r) = \frac{p_0^k}{(1-\sum_{j=1}^r s_j p_j)^k}$$