invert multinomial logit link with three unknown I am attempting to invert the multinomial logit link with three variables.  I can do it with two variables, but I do not know how to do it with three.

A multinomial logit function for three states, i.e., three probabilities, $a, b, c$ is written as follows:

$a = \frac{e^{x}} {(1 + e^{x} + e^{y})}$

$b = \frac{e^{y}} {(1 + e^{x} + e^{y})}$

$c = 1 - a - b$

These three probabilities are defined by the parameters $x$ and $y$.  If we know $x$ and $y$ we can obtain $a$, $b$ and $c$.

However, given $a, b, c$, how do we obtain $x$ and $y$?  One way is to use multinomial logistic regression.  However, there should be a closed form solution in which $x$ and $y$ are obtained using basic algebra.  I can obtain the closed form solution for two parameters, $x$ and $y$:

$x = \log(\frac{a  (1 - b) + (a  b)}{ (1 - a)  (1 - b) - a  b})$

$y = \log(\frac{b  (1 - a) + (b  a) }{ (1 - b)  (1 - a) - b  a})$

Which simplifies to:

$x = \log(\frac{a}{1 - a - b})$

$y = \log(\frac{b}{1 - a - b})$

How can I obtain the closed form solution when there are three parameters $x, y, z\;$?

$a = \frac{e^{x}} {(1 + e^{x} + e^{y} + e^{z})}$

$b = \frac{e^{y}} {(1 + e^{x} + e^{y} + e^{z})}$

$c = \frac{e^{z}} {(1 + e^{x} + e^{y} + e^{z})}$

$d = 1 - a - b - c$
For every fixed number of variables, you are considering, for every $i$, 


  $$a_i=\frac{e^{x_i}}{1+s}$$ 


where 


  $$s=\sum_ie^{x_i}$$ 


and 


  $$z=\frac1{1+s}$$ 


and you are asking how to invert this system, that is, how to deduce the collection $(x_i)$ from the collection $(a_i)$ and $z$, or even, from $(a_i)$ only. 

To solve this, consider that $$z+\sum_ia_i=1$$ hence, for every $i$, $$e^{x_i}=a_i\cdot(1+s)=\frac{a_i}z$$ that is,


  $$x_i=\log a_i-\log z=\log\left(\frac{a_i}{1-\sum\limits_ka_k}\right)$$
Solve each variable $x$, $y$ and $z$ as a function of the other two.

$x = \log(\frac{(a + a  e^{y} + a  e^{z})}  {(1 - a)})$

$y = \log(\frac{(b + b  e^{x} + b  e^{z})}  {(1 - b)})$

$z = \log(\frac{(c + c  e^{x} + c  e^{y})}  {(1 - c)})$

Then: 

Substitute $x$ into the equation for $y$.

Substitute $x$ into the equation for $z$.

Substitute $y$ into the equation for $x$.

Substitute $y$ into the equation for $z$.

Substitute $z$ into the equation for $x$. 

And substitute $z$ into the equation for $y$.

Each parameter $x$, $y$ and $z$ is now expressed as a function of just one of the other two.

To express each parameter as a function of itself, substitute, for example, the equation expressing $x$ as a function of $y$ into the function expressing $y$ as a function of $x$.

Once these three equations are simplified we have:

$x = \log(-(a * c - a) / (c^2 + (b+a-2)*c + (a - 1)*b - a + 1 - a * b))$

$y = \log(-(b * c - b) / (c^2 + (b+a-2)*c + (a - 1)*b - a + 1 - a * b))$

$z = \log(((1-b)*c) / (((b+a-1)*c+b^2+(a-2)*b-a+1) - (a*c)))$