Checking and proving unicity of solution of a system of equations

Question

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?

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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?

Answer 1

The answer is no.

If $K=2$ and $T$ is even and there are two integers $i_1, i_2$ such that $$y_{i_1,(T/2, T/2)}\not=y_{i_2,(T/2, T/2)}\qquad\text{and}\qquad 1\le i_1\lt i_2\le I$$ then the system has no solution.

Proof :

For $(R_1,R_2)=(T/2,T/2)$, we have $I$ equations $$\alpha_1^{T/2}\alpha_2^{T/2}p_i+(1-\alpha_1)^{T/2}(1-\alpha_2)^{T/2}(1-p_i)=y_{i,(T/2, T/2)}$$ which can be written as $$\alpha_1^{T/2}\alpha_2^{T/2}=y_{i,(T/2, T/2)}$$ for every $i=1,2,\cdots, I$ since $\alpha_1+\alpha_2=1$.

So, if there are two integers $i_1, i_2$ such that $$y_{i_1,(T/2, T/2)}\not=y_{i_2,(T/2, T/2)}\qquad\text{and}\qquad 1\le i_1\lt i_2\le I$$ then the system has no solution.

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Added 1 : I'm going to add a generalization of "If $K=T=I=2$ and $\alpha_1\gt\frac 12$, then the system has at most one solution".

Claim 1 : If $K=2$ and $T$ is even and $\alpha_1\gt \frac 12$, then the system has at most one solution.

Proof :

For $(R_1,R_2)=(T,0)$, we have $$p_i\alpha_1^{T}+(1-p_i)(1-\alpha_1)^{T}=y_{i,(T,0)}$$ which can be written as $$p_i(\alpha_1^{T}-(1-\alpha_1)^{T})=y_{i,(T,0)}-(1-\alpha_1)^{T}$$ for $i=1,2,\cdots, I$.

Here, note that $$\alpha_1^{T}-(1-\alpha_1)^{T}=0\iff \alpha_1^{T}=(1-\alpha_1)^{T}\iff \alpha_1=\frac 12$$ from which we can say that if $\alpha_1\not=\frac 12$, then $p_i\ (i=1,2,\cdots, I)$ can be solved uniquely.

For $(R_1,R_2)=(T/2,T/2)$ with $i=1$, we have $$\begin{align}&p_1\alpha_1^{T/2}\alpha_2^{T/2}+(1-p_1)(1-\alpha_1)^{T/2}(1-\alpha_2)^{T/2}=y_{1,(T/2,T/2)} \\\\&\iff p_1\alpha_1^{T/2}\alpha_2^{T/2}+(1-p_1)\alpha_1^{T/2}\alpha_2^{T/2}=y_{1,(T/2,T/2)} \\\\&\iff \alpha_1^{T/2}\alpha_2^{T/2}=y_{1,(T/2,T/2)} \\\\&\iff \alpha_1\alpha_2=y_{1,(T/2,T/2)}^{2/T} \\\\&\iff\alpha_1^2-\alpha_1+y_{1,(T/2,T/2)}^{2/T}=0 \\\\&\iff\alpha_1=\frac 12\pm\frac{\sqrt{1-4y_{1,(T/2,T/2)}^{2/T}}}{2}\end{align}$$

Therefore, we can see that if $\alpha_1\gt \frac 12$, then the system has at most one solution.$\quad\blacksquare$

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Added 2 : I'm going to add another generalization for any $K\ge 2$.

Claim 2 : If $$\begin{cases}\alpha_1\not=\dfrac 12 \\\\b\ge 2^{-T} \\\\\max\bigg(p,\dfrac{-b K T +bK + 2 b T - b + C T}{T(C+b)}\bigg)\lt\alpha_1\lt \min\bigg(q,\dfrac{b(T-1)}{T(C+b)}\bigg)\end{cases}$$where $p,q\ (p\lt q)$ are the roots of $$K(1-2 T) \alpha_1^2 + (K T + 2 T - 2) \alpha_1+1- T=0$$with $$b=y_{1,(T,0,0,\cdots, 0)},\qquad C=\displaystyle\sum_{j=2}^{K}y_{1,(T-1,0,0,\cdots, 0,R_j=1,0,0,\cdots, 0)}$$ then the system has at most one solution.

Proof :

For $(R_1,R_2,\cdots, R_K)=(T,0,0,\cdots, 0)$, we have $$p_i\alpha_1^{T} +(1-p_i)(1-\alpha_1)^{T}=y_{i,(T,0,0,\cdots, 0)}$$ which can be written as $$p_i(\alpha_1^T-(1-\alpha_1)^{T})=y_{i,(T,0,0,\cdots, 0)}-(1-\alpha_1)^{T}\tag1$$ Here, note that $$\alpha_1^T-(1-\alpha_1)^{T}=0\iff \alpha_1^T=(1-\alpha_1)^{T}\iff \alpha_1=\frac 12$$ So, if $\alpha_1\not=\frac 12$, then $p_i\ (i=1,2,\cdots, I)$ can be solved uniquely.

For $i=1$ and $(R_1,R_2,\cdots, R_K)=(T-1,0,0,\cdots, 0,1,0,0,\cdots, 0)$ where $R_j=1$ with $j\not=1$, we have

$$p_1\alpha_1^{T-1}\alpha_j+(1-p_1)(1-\alpha_1)^{T-1}(1-\alpha_j)=y_{1,(T-1,0,0,\cdots, 0,1,0,0,\cdots, 0)}$$ which can be written as $$\alpha_j(p_1\alpha_1^{T-1}-(1-p_1)(1-\alpha_1)^{T-1})=y_{1,(T-1,0,0,\cdots, 0,1,0,0,\cdots, 0)}-(1-p_1)(1-\alpha_1)^{T-1}$$

Here, note that $$\begin{align}&p_1\alpha_1^{T-1}-(1-p_1)(1-\alpha_1)^{T-1}=0 \\\\&\implies p_1(\alpha_1^{T-1}+(1-\alpha_1)^{T-1})=(1-\alpha_1)^{T-1} \\\\&\implies p_1(\alpha_1^T-(1-\alpha_1)^{T})(\alpha_1^{T-1}+(1-\alpha_1)^{T-1})=(1-\alpha_1)^{T-1}(\alpha_1^T-(1-\alpha_1)^{T}) \\\\&\implies (b-(1-\alpha_1)^{T})(\alpha_1^{T-1}+(1-\alpha_1)^{T-1})=(1-\alpha_1)^{T-1}(\alpha_1^T-(1-\alpha_1)^{T}) \\\\&\implies b(\alpha_1^{T-1}+(1-\alpha_1)^{T-1})= \alpha_1^{T-1}(1-\alpha_1)^{T-1} \\\\&\implies \frac{\alpha_1^{T-1}(1-\alpha_1)^{T-1}}{\alpha_1^{T-1}+(1-\alpha_1)^{T-1}}=b\end{align}$$ where $b:=y_{1,(T,0,0,\cdots, 0)}$.

So, if $$\frac{\alpha_1^{T-1}(1-\alpha_1)^{T-1}}{\alpha_1^{T-1}+(1-\alpha_1)^{T-1}}\not=b$$ then $\alpha_j\ (j=2,3,\cdots, K)$ can be solved uniquely.

Using $\displaystyle\sum_{j=1}^{K}\alpha_j=1$, we have

$$\alpha_1+\sum_{j=2}^{K}\frac{y_{1,(T-1,0,0,\cdots, 0,R_j=1,0,0,\cdots, 0)}-(1-p_1)(1-\alpha_1)^{T-1}}{p_1\alpha_1^{T-1}-(1-p_1)(1-\alpha_1)^{T-1}}=1$$ which can be written as

$$(1-\alpha_1)(p_1\alpha_1^{T-1}-(1-p_1)(1-\alpha_1)^{T-1})=C-(K-1)(1-p_1)(1-\alpha_1)^{T-1},$$ i.e. $$p_1(1-\alpha_1)\alpha_1^{T-1}-C=(1-p_1)(1-\alpha_1)^{T-1}(2-\alpha_1-K)$$ where $C:=\displaystyle\sum_{j=2}^{K}y_{1,(T-1,0,0,\cdots, 0,R_j=1,0,0,\cdots, 0)}$.

Multiplying the both sides by $(\alpha_1^T-(1-\alpha_1)^{T})$ and using $(1)$ for $i=1$ give

$$(b-(1-\alpha_1)^{T})(1-\alpha_1)\alpha_1^{T-1}-C(\alpha_1^T-(1-\alpha_1)^{T})+(\alpha_1^T-b)(1-\alpha_1)^{T-1}(\alpha_1+K-2)=0\tag2$$

Let $f(\alpha_1)$ be the LHS of $(2)$.

Then, we have

$$f'(\alpha_1) =\alpha_1^{T-2}(1-\alpha_1)^{T-2}\bigg( K(1-2 T) \alpha_1^2 + (k T + 2 T - 2) \alpha_1+1- T\bigg)+(1-\alpha_1)^{T-2}\bigg( T(C + b )\alpha_1+b K T -bK - 2 b T + b - C T\bigg)+\alpha_1^{T-2}\bigg(-T(C+b)\alpha_1 + Tb - b \bigg) ​$$

Therefore, if

$$K(1-2 T) \alpha_1^2 + (K T + 2 T - 2) \alpha_1+1- T\gt 0\tag3$$ $$T(C + b )\alpha_1+b K T -bK - 2 b T + b - C T \gt 0\tag4$$ and $$-T(C+b)\alpha_1 + Tb - b\gt 0\tag5$$

then we get $f'(\alpha_1)\gt 0$, so $(2)$ has at most one solution.

Since $$(4)(5)\iff \frac{-b K T +bK + 2 b T - b + C T}{T(C+b)}\lt\alpha_1\lt \frac{b(T-1)}{T(C+b)}$$ letting $p,q(p\lt q)$ be the roots of $$K(1-2 T) \alpha_1^2 + (K T + 2 T - 2) \alpha_1+1- T=0$$ we have $$(3)(4)(5)\iff \max\bigg(p,\frac{-b K T +bK + 2 b T - b + C T}{T(C+b)}\bigg)\lt\alpha_1\lt \min\bigg(q,\frac{b(T-1)}{T(C+b)}\bigg)$$

Also, let $g(\alpha_1)=\dfrac{\alpha_1^{T-1}(1-\alpha_1)^{T-1}}{\alpha_1^{T-1}+(1-\alpha_1)^{T-1}}$. Then, we have $$\dfrac{\partial g(\alpha_1)}{\partial \alpha_1}=\frac{(T-1)\alpha_1^{T-2}(1-\alpha_1)^{T-2}((1-\alpha_1)^{T-1}-\alpha_1^{T-1})}{((1-\alpha_1)^{T-1}+\alpha_1^{T-1})^2}$$ so $g(\alpha_1)$ is increasing for $\alpha_1\lt 1/2$ and decreasing for $\alpha_1\gt 1/2$. Therefore, we can say that if $g(1/2)=2^{-T}\le b$, then $g(\alpha_1)\not=b$. This means that we can say that if $b\ge 2^{-T}$, then $\alpha_j\ (j=2,3,\cdots, K)$ can be solved uniquely.

In conclusion, if $$\begin{cases}\alpha_1\not=\dfrac 12 \\\\b\ge 2^{-T} \\\\\max\bigg(p,\dfrac{-b K T +bK + 2 b T - b + C T}{T(C+b)}\bigg)\lt\alpha_1\lt \min\bigg(q,\dfrac{b(T-1)}{T(C+b)}\bigg)\end{cases}$$where $p,q\ (p\lt q)$ are the roots of $$K(1-2 T) \alpha_1^2 + (K T + 2 T - 2) \alpha_1+1- T=0$$with $$b=y_{1,(T,0,0,\cdots, 0)},\qquad C=\displaystyle\sum_{j=2}^{K}y_{1,(T-1,0,0,\cdots, 0,R_j=1,0,0,\cdots, 0)}$$ then the system has at most one solution.$\quad\blacksquare$.

Answer 2

$\def\paren#1{\left(#1\right)}\def\Rsum{\sum_{\substack{R_1, \cdots, R_K \geqslant 0\\R_1 + \cdots + R_K = T}} }$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$$ α_j = \frac{y_{i_0, (T - 1, 0, \cdots, 0, 1, 0, \cdots 0)} - (1 - p_{i_0}) (1 - α_1)^{T - 1}}{p_{i_0} α_1^{T - 1} - (1 - p_{i_0}) (1 - α_1)^{T - 1}}. $$