The probability mass function of the negative multinomial distribution is: \begin{align*} \mathbb{P}(\boldsymbol{\rm{X}}=\boldsymbol{\rm{x}}|\mathbf{p})=\frac{\Gamma\left(x_0+\sum_{i=1}^{m}x_{i}\right)}{\Gamma\left(x_0\right)}p_0^{x_0} \prod_{i=1}^{m}\frac{p_{i}^{x_i}}{x_i!}, \end{align*}

where $\mathbf{p} \equiv (p_1,...,p_m)\in\mathbb R_+^m$ and $p_0 \equiv 1-\sum_{i=1}^m p_i$. Taking $x_0$ to be fixed as the number of successful draws from the "first" group, the mean vector and variance matrix of the distribution of the remaining $m$ values are given respectively by:

\begin{align*} \boldsymbol{\rm{\mu}}=\frac{x_0}{p_{0}}\boldsymbol{\rm{p}} \quad \quad \quad \quad \quad {\boldsymbol {\Sigma }}=\frac{x_0}{p_{0}^{2}}\boldsymbol{\rm{p}}\boldsymbol{\rm{p'}}+\frac{x_0}{p_{0}}{\rm{diag}(\boldsymbol{\rm{p}})}. \end{align*}

I want to show that the determinant of the variance matrix is:

\begin{align*} {\displaystyle |{\boldsymbol {\Sigma }}|={\frac {1}{p_{0}}}\prod _{i=1}^{m}{\mu _{i}}}, \end{align*}

where the values in the product are elements of the mean vector. How can I derive this result?

  • $\begingroup$ There is this equation on Wikipedia(en.m.wikipedia.org/wiki/Negative_multinomial_distribution), but I don't know how they got it $\endgroup$
    – d d
    Jun 3, 2022 at 9:47
  • $\begingroup$ It's crucial to specify that $p_0 = 1 - (p_1+\cdots+p_m),$ for otherwise this result is not generally true. One way to derive it is via row reduction to relate the case $m$ to the case $m-1,$ then proceed inductively. $\endgroup$
    – whuber
    Jun 3, 2022 at 13:37

1 Answer 1


As a preliminary step, we note the result of the following quadratic form:

$$\mathbf{p}' \text{diag} (\mathbf{p})^{-1} \mathbf{p} = \mathbf{p}' \mathbf{1} = \sum_{i=1}^m p_i = 1-p_0.$$

Now, using the matrix determinant lemma, we have:

$$\begin{align} |\boldsymbol{\Sigma}| &= \bigg| \frac{x_0}{p_{0}^{2}} \mathbf{p} \mathbf{p}' + \frac{x_0}{p_{0}}\text{diag} (\mathbf{p}) \bigg| \\[6pt] &= \bigg( 1 + \frac{\mathbf{p}' \text{diag} ( \mathbf{p})^{-1} \mathbf{p}}{p_0} \bigg) \bigg| \frac{x_0}{p_{0}} \text{diag} (\mathbf{p}) \bigg| \\[6pt] &= \bigg( 1 + \frac{1-p_0}{p_0} \bigg) \prod_{i=1}^m \frac{x_0}{p_0} p_i \\[6pt] &= \frac{1}{p_0} \prod_{i=1}^m \frac{x_0}{p_0} p_i \\[6pt] &= \frac{1}{p_0} \prod_{i=1}^m \mu_i. \\[6pt] \end{align}$$

As you can see, the result you are interested in is a direct result of the matrix determinant lemma; if you are interested in a proof of the latter you can find it here.


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