<s>[INST] 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 $x, y, z\;$?
$a = \frac{e^{x}} {(1 + e^{x} + e^{y} + e^{z})}$
$b = \frac{e^{y}} {(1 + e^{x} + e^{y} + e^{z})}$
$c = \frac{e^{z}} {(1 + e^{x} + e^{y} + e^{z})}$
$d = 1 - a - b - c$
 [/INST] 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:
$x = \log(-(a * c - a) / (c^2 + (b+a-2)*c + (a - 1)*b - a + 1 - a * b))$
$y = \log(-(b * c - b) / (c^2 + (b+a-2)*c + (a - 1)*b - a + 1 - a * b))$
$z = \log(((1-b)*c) / (((b+a-1)*c+b^2+(a-2)*b-a+1) - (a*c)))$
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