# Multinomial transformation for matrices

Suppose we have a vector of probabilities $\mathbf{p}=(p_1,...,p_n)$, where $p_i>0$ for $i=1,...n$ and $\sum p_i=1$. Define new vector $\mathbf{r}=(r_1,...,r_{n-1})$ in a following way:

$r_i=\log(p_i/p_n)$

This defines the transformation $T:(0,1)^n\to\mathbb{R}^{n-1}$, $\mathbf{r}=T\mathbf{p}$. This transformation can be called multinomial transformation (or to be more precise inverse multinomial transformation), since similar formula is used in http://en.wikipedia.org/wiki/Multinomial_logit>multinomial logit model.

This transformation is useful for modelling, since resulting $r_i$ can be any real number, and there is an easy way to transform $r_i$ back to probabilities:

$p_n=\dfrac{1}{1+\sum \exp(r_i)},$

$p_i=\exp(r_i)p_n$.

My question is whether there exists a similar transformation for matrices. Suppose we have two probability vectors $\mathbf{p}=(p_1,...,p_n)$, $\mathbf{q}=(q_1,...q_m)$ and $n\times m$ matrix $P=(p_{ij})$, satisfying

$\sum p_i=1$, $\sum q_i=1$

$\sum_{j=1}^m p_{ij}=p_i$, for each $i=1,...,n$, (1)

$\sum_{i=1}^n p_{ij}=q_j$, for each $j=1,...,m$, (2)

(what we actualy have is a bivariate discrete probability distribution with given marginal distributions).

Now what I am looking for is a transformation which transforms $p_{ij}$ to unbounded real numbers, but such that the inverse would satisfy constraints (1) and (2). In effect I am looking for the bijection from subset of $(0,1)^{nm}$ to $R^{k}$, where $k$ should be $(n-1)(m-1)$.

I suspect that maybe copulas can be involved here, or some properties of stochastic matrices. If somebody could give me any pointers I would be very grateful.

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Constraints 1 and 2 are not enough to describe your bijection. Marginals can be represented as two vectors of log-odds, n+m-2 parameters total, then you can get the joint satisfying the constraints by multiplying the marginals –  Yaroslav Bulatov Aug 20 '10 at 16:36