# Solve system of non linear equations

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?

• Looks very markovian... – Jean Marie Jun 15 '16 at 11:26
• It is false : there are explicit solutions. I just did it with mathematica. But the expressions are awfully complicated. – Jean Marie Jun 15 '16 at 11:32
• Intersecting each of the first three equations with the plane of the fourth gives simpler curves. Found no common solutions yet, though. – mvw Jun 15 '16 at 11:37
• @JeanMarie I know I get solutions from mathematica but if I plug in e.g. $k_1=0.6278,k_2=0.9675,k_3=0.5654$ and $\pi_1=0.4039,\pi_2=0.1962,\pi_3=0.3998$ you should get $x_1=0.2895,x_2=0.7586,x_3=0.4848$. Can you get this from the solutions you find in mathematica? I could not. – JFNJr Jun 15 '16 at 11:43
• @JFNJr Well the question is now substantially different - I may look at it again later. – almagest Jun 15 '16 at 12:37

@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.