How to prove that equation over probabilities has unique solultion or find counterexample?

Question

Given equations: $$ \prod_{i=1}^n p_i = \prod_{i=1}^n (1-p_i)= \frac{1}{2^n} $$ where $p_i\in (0,1), i=\overline{1,n}$.

Is it true that this system has unique solution $p_1=p_2=\ldots=p_n=\frac12$ for all $n\in\mathbb{N}$?

In other words, if we have (discrete uniform distributed) function $f\colon X\to \{0,1\}^n$ so that $\Pr[f(x)=y]=\frac{1}{2^n} \;\forall x\in X, y\in \{0,1\}^n$ then is it true that $\Pr[y_i=0]=\frac{1}{2} \forall i\in[n]$ where $f(x)=y=(y_1,y_2,\ldots,y_n), y_j\in\{0,1\}, x\in X$?

Answer 1

From your assumption it follows that $$\prod_{i=1}^n\bigl(p_i(1-p_i)\bigr)={1\over 4^n}\ .\tag{1}$$ If $p$ is a probability then $$p(1-p)={1\over4}-\left(p-{1\over2}\right)^2$$ implies that $p(1-p)\leq{1\over4}$ with equality only if $p={1\over2}$. Therefore $(1)$ can only hold if $$p_i={1\over2}\qquad(1\leq i\leq n)\ .$$

Answer 2

Let $n=1$.

The equation:

$$\prod_{i=1}^1 p_i = \prod_{i=1}^1 (1-p_i)= \frac{1}{2}$$

does necessarily imply that $p_1=\frac{1}{2}$ is the unique solution.

Let $p_i=\frac{1}{2}$ for all $i=1, \ldots, n-1$.

Then by the same logic as the above:

$$\prod_{i=1}^n p_i = \prod_{i=1}^n (1-p_i)= \frac{1}{2^n}$$

implies $p_n=\frac{1}{2}$ (divide off the product of the $n-1$ other terms which we know by hypothesis reduces this to the $n=1$ step). Thus $p_i = \frac{1}{2}$ for each $i=1, \ldots n$ is the unique solution.