I am trying to solve the following system of non linear equations:
$$\pi_1=\frac{k_2k_3(1-(1-x_2)(1-x_3))}{k_2k_3(1-(1-x_2)(1-x_3))+k_1k_3(1-(1-x_1)x_3)+k_1k_2(1-x_1x_2)}\\\pi_2=\frac{k_1k_3(1-(1-x_1)x_3)}{k_2k_3(1-(1-x_2)(1-x_3))+k_1k_3(1-(1-x_1)x_3)+k_1k_2(1-x_1x_2)}\\\pi_3=\frac{k_1k_2(1-x_1x_2)}{k_2k_3(1-(1-x_2)(1-x_3))+k_1k_3(1-(1-x_1)x_3)+k_1k_2(1-x_1x_2)}$$
with real constants $k_1,k_2,k_3>0$ and $0\leq\pi_1\leq1,0\leq\pi_2\leq1,0\leq\pi_3\leq1$. And the unknowns are real and $\vec{x}\in [0,1]$.
If I try to solve with e.g. wolfram alpha, I do not get the solution I expect to find if I set $k_1,k_2,k_3$ and $\pi_1,\pi_2,\pi_3$.
Can anybody help me to understand the correct way to solve this system?
@mvw as well
This problem has a fundamental indeterminacy. I show it using your example.
Let us take the following simplified notations:
$$\begin{cases}N_1 &=& k_2 k_3(1 - (1 - x_2)(1 - x_3))\\ N_2 &=& k_1 k_3(1 - (1 - x_1)x_3)\\ N_3 &=& k_1 k_2(1 - x_1 x_2)\\ D &=& N_1 + N_2 + N_3\end{cases}$$
Instead of defining $\pi_1$, $\pi_2$ and $\pi_3$ by
$$\pi_1 = \dfrac{N_1}{D}, \pi_2 = \dfrac{N_2}{D} , \pi_3 = \dfrac{N_3}{D} \ \ \ (1)$$
we are interested by the error vector $E$ defined as:
$$E=(e_1,e_2,e_3)= (\pi_1 - \dfrac{N_1}{D}, \pi_2 - \dfrac{N_2}{D} , \pi_3 - \dfrac{N_3}{D})$$
Let us take the test values you have provided
$\pi_1 = 0.4039, \pi_2 = 0.1962, \pi_3 = 0.3998, k_1 = 0.6278, k_2 = 0.9675, k_3 = 0.5654$
considered as fixed for all the coming tests.
If we first take the values you propose
$$x_1 = 0.2895, x_2 = 0.7586, x_3 = 0.4848 \ \ \text{we obtain} \ \ E=(-0.000078, -0.000051, 0.000029)$$
which displays a very good agreement. But with :
$$x_1 = 0.7255, x_2 = 0.1764, x_3 = 0.9745 \ \ \text{we obtain} \ \ E=(-0.0002, -0.000001, 0.00011)$$
we get an almost as good agreement, though with very different values of $x_1,x_2,x_3$. And now, with this set of values:
$$x_1 = 0.5160, x_2 = 0.4561, x_3 = 0.7391 \ \ \text{we obtain} \ \ E=(-0.0001, -0.00002, 0.00004)$$
we have a very small discrepancy too.
What is the conclusion ? That, intrinsicaly, there are an infinite set of solutions in a very wide range : it is much more than instability. We need at least supplementary constraints...
Remark 1: Probabilistic interpretation. You have not stressed the fact that your question originates "probably" from a Markovian context. Thus, the following question: is this indeterminacy connected with the fact that, in a markovian interpretation, you can't retrieve the past ?
Remark 2: Relationships (1) can be fruitfully interpreted as barycentrical coordinates. This is by using them that I have realized the indeterminacy of the problem, but it would be out of context to spend time here to discuss about this very interesting tool.