I have a nonlinear system of equations. I would like your help to show that this system has a unique solution.
Some useful notation:
Let us fix a positive integer $T$.
Define the matrix $A$ of size $2^T\times T$, stacking all possible $1\times T$ rows of ones and zeros that can be formed. Any order of the rows is fine.
For each $j$-th row of $A$, let $\mathcal{R}^j_1$ denotes the total number of ones, let $\mathcal{R}^j_0$ denotes the total number of zeros.
For example, if $T=3$, then $$ A=\begin{pmatrix} 1 & 1 & 1\\ 1 & 1 & 0\\ 1 & 0 & 1\\ 1 & 0 & 0\\ 0 & 1 & 1\\ 0 & 1 & 0\\ 0 & 0 & 1\\ 0 & 0 & 0\\ \end{pmatrix} $$ $$ \mathcal{R}^1_1=3,\mathcal{R}^2_1=\mathcal{R}^3_1=\mathcal{R}^5_1=2, \mathcal{R}^4_1=\mathcal{R}^6_1=\mathcal{R}^7_1=1,\mathcal{R}^8_1=0 $$ $$ \mathcal{R}^1_0=0,\mathcal{R}^2_0=\mathcal{R}^3_0=\mathcal{R}^5_0=1, \mathcal{R}^4_0=\mathcal{R}^6_0=\mathcal{R}^7_0=2,\mathcal{R}^8_0=3$$
My question:
Let $T=3$ and consider the following system of $4\times 2^T $ equations (i.e., 32 equations) $$ \begin{cases} \frac{\alpha^{\mathcal{R}^j_1} (1-\alpha)^{\mathcal{R}^j_0}p_i}{\alpha^{\mathcal{R}^j_1} (1-\alpha)^{\mathcal{R}^j_0}p_i+\beta^{\mathcal{R}^j_1} (1-\beta)^{\mathcal{R}^j_0}(1-p_i)}=B_{i,j} & i\in \{1,...,4\} \text{ and } j\in \{1,...,2^T\} \end{cases} $$ where
- $\alpha, p_1,p_2,p_3,p_4$ are unknowns in $[0,1]$
- $\{B_{i,j}\}$ are known real numbers in $[0,1]$
- The sets $\{\mathcal{R}^j_*\}$ are pre-computed using the matrix $A$ above.
I want to show that this system has a unique solution with respect to $\alpha, p_1,p_2,p_3,p_4$. I have a proof (below) which makes use of 7 equations out of 32. I would like your help to understand if we can show uniqueness using less than 7 equations.
My proof:
Note that, for the system to have a solution, it is necessary that $$ B_{i,2}=B_{i,3}=B_{i,5} \text{ for }i\in \{1,2,3,4\}\\ B_{i,4}=B_{i,6}=B_{i,7} \text{ for }i\in \{1,2,3,4\}\\ $$ I assume that these conditions hold
Consider the following $4$ equations of the system $$ \begin{cases} (1) \text{ }\alpha^3p_1=B_{1,1}\\ (2) \text{ }(1-\alpha)^3p_1=B_{1,8}\\ (3) \text{ }\alpha^2(1-\alpha)p_1=B_{1,2}\\ (4) \text{ }\alpha (1-\alpha)^2p_1=B_{1,4}\\ \end{cases} $$
Put (1) in (2),(3) and get $$ \begin{cases} (2)^* \text{ }p_1-B_{1,1}-3\alpha p_1+3\alpha^2p_1=B_{1,8}\\ (3)^* \text{ }p_1\alpha^2=B_{1,1}+B_{1,2}\\ \end{cases} $$
Put (1) and (3)* in (4) and get $$ (4)^* \text{ }\alpha p_1 =- B_{1,1}+2(B_{1,1}+B_{1,2})+B_{1,4}\\ $$
Put (4)* and (3)* in (2)* and get $p_1$
Replace such $p_1$ in (1) and get $\alpha$ from $(\frac{B_{1,1}}{D_1})^{1/3}$
Obtain $p_2, p_3, p_4$ from the following 3 equations of the system
$$ \begin{cases} (5) \text{ }\alpha^3p_2=B_{2,1}\\ (6) \text{ }\alpha^3p_3=B_{3,1}\\ (7) \text{ }\alpha^3p_4=B_{4,1}\\ \end{cases} $$