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 $x, y, z\;$?
$a = \frac{e^{x}} {(1 + e^{x} + e^{y} + e^{z})}$
$b = \frac{e^{y}} {(1 + e^{x} + e^{y} + e^{z})}$
$c = \frac{e^{z}} {(1 + e^{x} + e^{y} + e^{z})}$
$d = 1 - a - b - c$
 A: 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,

$$x_i=\log a_i-\log z=\log\left(\frac{a_i}{1-\sum\limits_ka_k}\right)$$

A: 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:
$x = \log(-(a * c - a) / (c^2 + (b+a-2)*c + (a - 1)*b - a + 1 - a * b))$
$y = \log(-(b * c - b) / (c^2 + (b+a-2)*c + (a - 1)*b - a + 1 - a * b))$
$z = \log(((1-b)*c) / (((b+a-1)*c+b^2+(a-2)*b-a+1) - (a*c)))$
A: If you want to solve
$$\begin{eqnarray}
a_1 &=& \frac{e^{x_1}} {1 + \sum_{i=1}^n e^{x_i}}\\
&\ldots&  \tag{1}\\
a_n &=& \frac{e^{x_n}} {1 + \sum_{i=1}^n e^{x_i}}\\
\end{eqnarray}
$$
with constants $$a_1,\ldots,a_n$$
and variables $$x_1,\ldots,x_n$$
then set 
$$
\begin{eqnarray}
u_1&=&e^{x_1} \\
&\ldots&  \tag{2}\\
u_n&=&e^{x_n} 
\end{eqnarray}
$$
and substitute in $(1)$ and multiply each equation by its denominator and you have
$$\begin{eqnarray}
a_1 + \sum_{i=1}^n a_1 u_i&=& u_1\\
&\ldots&  \tag{3}\\
a_n + \sum_{i=1}^n a_1 u_i&=& u_n\\
\end{eqnarray}
$$
This is a linear equation with variables $u_i.$ You can solve this numerically or algebraically with the well known methods for linear equations. 
Then you can so the back substitution to get the $x_i.$
$$
\begin{eqnarray}
x_1&=&\log{u_1} \\
&\ldots&  \tag{4}\\
x_n&=&\log{u_n} 
\end{eqnarray}
$$
