Negative multinomial distribution

If m-dimensional x is partitioned as follows

${\displaystyle \mathbf {X} ={\begin{bmatrix}\mathbf {X} ^{(1)}\\\mathbf {X} ^{(2)}\end{bmatrix}}{\text{ with sizes }}{\begin{bmatrix}n\times 1\\(m-n)\times 1\end{bmatrix}}}$

and accordingly ${\displaystyle {\boldsymbol {p}}}$

${\displaystyle {\boldsymbol {p}}={\begin{bmatrix}{\boldsymbol {p}}^{(1)}\\{\boldsymbol {p}}^{(2)}\end{bmatrix}}{\text{ with sizes }}{\begin{bmatrix}n\times 1\\(m-n)\times 1\end{bmatrix}}}$

and let

${\displaystyle q=1-\sum _{i}p_{i}^{(2)}=p_{0}+\sum _{i}p_{i}^{(1)}}$

The marginal distribution of ${\displaystyle {\boldsymbol {X}}^{(1)}}$ is ${\displaystyle \mathrm {NM} (x_{0},p_{0}/q,{\boldsymbol {p}}^{(1)}/q)}$. That is the marginal distribution is also negative multinomial with the ${\displaystyle {\boldsymbol {p}}^{(2)}}$ removed and the remaining p's properly scaled so as to add to one.

The univariate marginal ${\displaystyle m=1}$ is the negative binomial distribution.

If ${\displaystyle \mathbf {X} _{1}\sim \mathrm {NM} (r_{1},\mathbf {p} )}$ and If ${\displaystyle \mathbf {X} _{2}\sim \mathrm {NM} (r_{2},\mathbf {p} )}$ are independent, then ${\displaystyle \mathbf {X} _{1}+\mathbf {X} _{2}\sim \mathrm {NM} (r_{1}+r_{2},\mathbf {p} )}$. Similarly and conversely, it is easy to see from the characteristic function that the negative multinomial is infinitely divisible.

If

${\displaystyle \mathbf {X} =(X_{1},\ldots ,X_{m})\sim \operatorname {NM} (x_{0},(p_{1},\ldots ,p_{m}))}$

then, if the random variables with subscripts i and j are dropped from the vector and replaced by their sum,

${\displaystyle \mathbf {X} '=(X_{1},\ldots ,X_{i}+X_{j},\ldots ,X_{m})\sim \operatorname {NM} (x_{0},(p_{1},\ldots ,p_{i}+p_{j},\ldots ,p_{m})).}$

This aggregation property may be used to derive the marginal distribution of ${\displaystyle X_{i}}$ mentioned above.

The entries of the correlation matrix are

${\displaystyle \rho (X_{i},X_{i})=1.}$

${\displaystyle \rho (X_{i},X_{j})={\frac {\operatorname {cov} (X_{i},X_{j})}{\sqrt {\operatorname {var} (X_{i})\operatorname {var} (X_{j})}}}={\sqrt {\frac {p_{i}p_{j}}{(p_{0}+p_{i})(p_{0}+p_{j})}}}.}$

If we let the mean vector of the negative multinomial be

${\displaystyle {\boldsymbol {\mu }}={\frac {x_{0}}{p_{0}}}\mathbf {p} }$

and covariance matrix

${\displaystyle {\boldsymbol {\Sigma }}={\tfrac {x_{0}}{p_{0}^{2}}}\,\mathbf {p} \mathbf {p} '+{\tfrac {x_{0}}{p_{0}}}\,\operatorname {diag} (\mathbf {p} )}$,

then it is easy to show through properties of determinants that ${\displaystyle |{\boldsymbol {\Sigma }}|={\frac {1}{p_{0}}}\prod _{i=1}^{m}{\mu _{i}}}$. From this, it can be shown that

${\displaystyle x_{0}={\frac {\sum {\mu _{i}}\prod {\mu _{i}}}{|{\boldsymbol {\Sigma }}|-\prod {\mu _{i}}}}}$

and

${\displaystyle \mathbf {p} ={\frac {|{\boldsymbol {\Sigma }}|-\prod {\mu _{i}}}{|{\boldsymbol {\Sigma }}|\sum {\mu _{i}}}}{\boldsymbol {\mu }}.}$

Substituting sample moments yields the method of moments estimates

${\displaystyle {\hat {x}}_{0}={\frac {(\sum _{i=1}^{m}{{\bar {x_{i}}})}\prod _{i=1}^{m}{\bar {x_{i}}}}{|\mathbf {S} |-\prod _{i=1}^{m}{\bar {x_{i}}}}}}$

and

${\displaystyle {\hat {\mathbf {p} }}=\left({\frac {|{\boldsymbol {S}}|-\prod _{i=1}^{m}{{\bar {x}}_{i}}}{|{\boldsymbol {S}}|\sum _{i=1}^{m}{{\bar {x}}_{i}}}}\right){\boldsymbol {\bar {x}}}}$