Checking and proving unicity of solution of a system of equations Consider the following system of equations:
$$\prod_{j=1}^K\alpha_j^{R_j} p_i+\prod_{j=1}^K(1-\alpha_j)^{R_j}(1-p_i)=y_{i,(R_1,\cdots,R_K)}$$
for each $i\in\{1,\cdots,I\}$ and each $(R_1,R_2,\cdots,R_K)\in\{0,1,\cdots,T\}^K$ satisfying $\sum\limits_{k=1}^K R_k=T$.
Here:

*

*$I\geq 2$, $K\geq 2$, $T\geq 2$  are integers


*the unknowns are $\{\alpha_j\}_{j=1}^K$, $\{p_i\}_{i=1}^I$


*$0<\alpha_j<1$ for each $j=1,\cdots,K$ and $\sum\limits_{j=1}^K \alpha_j=1$


*$0<p_i<1$ for each $i=1,\cdots,I$


*the number of equations is $I×\binom{K+T-1}{K-1}$


*the quantities on the right-hand-side are known, potentially different across equations (this is why they have that complicated subscript)
Question: does the system has a unique solution? If yes, can we show it?

Further discussion:
The system above is a generalised version of a system that I have studied in a simplified setting where unicity holds. For example, for $I=2$, $K=2$, $T=2$, the system is:
$$
\begin{cases}
\alpha^2_1 p_1+(1-\alpha_1)^2(1-p_1)]=y_{1,(2,0)}\\
\alpha^2_2 p_1+(1-\alpha_2)^2(1-p_1)]=y_{1,(0,2)}\\
\alpha_1 \alpha_2 p_1+(1-\alpha_1)(1-\alpha_2)(1-p_1)]=y_{1,(1,1)}\\
{}^{\underline{\hphantom{\Huge------------}}}\\
\alpha^2_1 p_2+(1-\alpha_1)^2(1-p_2)]=y_{2,(2,0)}\\
\alpha^2_2 p_2+(1-\alpha_2)^2(1-p_2)] =y_{2,(0,2)}\\
\alpha_1 \alpha_2 p_2+(1-\alpha_1)(1-\alpha_2)(1-p_2)]=y_{2,(1,1)}
\end{cases}
$$
which can be shown to have a unique solution with respect to $\alpha_1, \alpha_2, p_1, p_2$ if $\alpha_1>1/2$. Just recalling that $\alpha_1=1-\alpha_2$, derivations are easy. I am unable to generalise such derivations though. Can you see some patterns/properties?
 A: $\def\paren#1{\left(#1\right)}\def\Rsum{\sum_{\substack{R_1, \cdots, R_K \geqslant 0\\R_1 + \cdots + R_K = T}} }$Firstly, $p_i$'s can be solved uniquely: Since\begin{align*}
&\mathrel{\phantom=} \Rsum \frac{T!}{R_1! \cdots R_K!} y_{i, (R_1, \cdots, R_K)}\\
&= \Rsum \frac{T!}{R_1! \cdots R_K!} \paren{ p_i \prod_{j = 1}^K α_j^{R_j} + (1 - p_i) \prod_{j = 1}^K (1 - α_j)^{R_j} }\\
&= p_i \Rsum \frac{T!}{R_1! \cdots R_K!} \prod_{j = 1}^K α_j^{R_j}  + (1 - p_i) \Rsum \frac{T!}{R_1! \cdots R_K!} \prod_{j = 1}^K (1 - α_j)^{R_j}\\
&= p_i \paren{ \sum_{j = 1}^K α_j }^T + (1 - p_i) \paren{ \sum_{j = 1}^K (1 - α_j) }^T\\
&= p_i + (K - 1)^T (1 - p_i),
\end{align*}
then$$
p_i = \frac{1}{(K - 1)^T - 1} \Biggl( (K - 1)^T - \Rsum \frac{T!}{R_1! \cdots R_K!} y_{i, (R_1, \cdots, R_K)} \Biggr).
$$
Now consider a fixed $i_0$. For each $j$, there is\begin{gather*}
α_j^T p_{i_0} + (1 - α_j)^T (1 - p_{i_0}) = y_{i_0, (0, \cdots, 0, T, 0, \cdots 0)}. \tag{1}
\end{gather*}
Since$$
\frac{\partial}{\partial α_j}(α_j^T p_{i_0} + (1 - α_j)^T (1 - p_{i_0})) = T (α_j^{T - 1} p_{i_0} - (1 - α_j)^{T - 1} (1 - p_{i_0})),
$$
then $α_j^T p_{i_0} + (1 - α_j)^T (1 - p_{i_0})$ is strictly decreasing with respect to $α_j$ for $α_j \in (0, α^*)$ and strictly increasing for $α_j \in (α^*, 1)$, where $α^* = \dfrac{(1 - p_{i_0})^{\frac{1}{T - 1}}}{p_{i_0}^{\frac{1}{T - 1}} + (1 - p_{i_0})^{\frac{1}{T - 1}}}$. Thus (1) implies that there are at most two values of $α_j$.
Furthur more, if one of $α_j$'s is known, say $α_1$, then for each $j ≠ 1$, there is$$
α_1^{T - 1} α_j p_{i_0} + (1 - α_1)^{T - 1} (1 - α_j) (1 - p_{i_0}) = y_{i_0, (T - 1, 0, \cdots, 0, 1, 0, \cdots 0)},
$$
which implies that\begin{gather*}
(p_{i_0} α_1^{T - 1} - (1 - p_{i_0}) (1 - α_1)^{T - 1}) α_j = y_{i_0, (T - 1, 0, \cdots, 0, 1, 0, \cdots 0)} - (1 - p_{i_0}) (1 - α_1)^{T - 1}. \tag{2}
\end{gather*}
If $p_{i_0} α_1^{T - 1} - (1 - p_{i_0}) (1 - α_1)^{T - 1} ≠ 0$, i.e. $α_1 ≠ α^*$, then (2) shows that$$
α_j = \frac{y_{i_0, (T - 1, 0, \cdots, 0, 1, 0, \cdots 0)} - (1 - p_{i_0}) (1 - α_1)^{T - 1}}{p_{i_0} α_1^{T - 1} - (1 - p_{i_0}) (1 - α_1)^{T - 1}}.
$$
