For $\pi$ in the interior of the probability simplex, when is the vector $v$ that solves $v \cdot \pi = 1$ uniquely the vector of ones?

Question

Let $\pi \in \mathbb{R}^K$ be a known element of the interior of the probability simplex: \begin{equation*} \pi_k > 0, \qquad \sum_{k=1}^K \pi_k = 1. \end{equation*} Let $a \in \mathbb{R}_{++}^K$ be known. The question is: Under what conditions is the vector $b \in \mathbb{R}_{++}^K$ that solves \begin{equation*} 1 = \sum_k \frac{b_k}{a_k} \pi_k \end{equation*} unique?

Clearly $b=a$ is a solution, so another way to state the question is when is $b=a$ the only solution?

One knife-edge case is easy: if $\pi_k = \pi_{\ell} = 1/K$ for all $k, \ell$, then any $b$ that satisfies $K = \sum_k b_k/a_k$ would suffice. But what about for a generic $\pi$?

Answer 1

We can define the following vector:

$$v := \left[\frac{\pi_1}{a_1} \cdots \frac{\pi_K}{a_K}\right]$$

This is clearly known, since $\pi$ and $a$ are known.

Then the equation can be rewritten as

$$1 = b\cdot v$$

Now the question is, how many $b \in \mathbb{R}^K_{++}$ vectors satisfy this equation?

• Exactly $0$, if $v = 0$.
• Exactly $1$, if $K = 1$, namely $b = \frac{1}{v}= \frac{a}{\pi} = a$.
• Let's say now $K \ge 2$. If there is even one non-zero element in $v$, then there are infinitely many solutions. The zero elements can be multiplied with anything, and at least two non-zero elements can be weighted infinitely many ways to obtain $1$:

For example if $v = [2\ \ 3]$, then $2\cdot \frac{1}{2} + 3\cdot 0 =1$, or $2\cdot 1 + 3\cdot \left(-\frac{1}{3}\right)=1$, or in general, $2c+3\left(\frac{1-2c}{3}\right)=1$, for any $c \in \mathbb{R}$.

Since $c$ changes continuously and the original solution $(b=a)$ is inside the simplex, uncountably infinitely many solutions can also be guaranteed to be inside the simplex.

So $b=a$ is the only solution exactly if $K=1$ (and in this case, $v$ has to be $\ne 0$, which is satisfied by your conditions).