The Problem of the n-Liars

The title here is the title of the 11th challenge problem in Professor Nahin’s book [1]; I translate his problem into the following problem statement: Suppose that $X_1,\ X_2,\ \ldots,\ X_n$ are i.i.d. and

$$X_1\sim \left( \begin{array}{rr} -1 & 1\\ 1-p & p \end{array} \right), $$

where $0\le p\le 1$. Define random variable

$$ Y:=\prod_{i=1}^n X_i. $$

What is the distribution of $Y$?

Easily, we have that

$$Y\sim \left( \begin{array}{rr} -1 & 1\\ b & a \end{array} \right), $$

where

$$ a=\hbox{Pr}(\hbox{even number of}\ X_i\hbox{'s}\ \hbox{taking}\ -1), $$

and

$$ b=\hbox{Pr}(\hbox{odd number of}\ X_i\hbox{'s}\ \hbox{taking}\ -1). $$

So what are the values of $a$ and $b$?

In “A coin-flipping problem” section, Professor Nahin shows a way how to derive

$$ a = \frac{1}{2} + \frac{1}{2}(2p-1)^n. $$

Inspired by Professor Nahin's work, next, I will show a new way deriving $a$ and $b$. The key is the following system of equations:

$$ \left\{ \begin{array}{cl} a+b & = & 1,\\ a-b & = & [-(1-p)+p]^n = (2p-1)^n, \end{array} \right. $$

from which we can find

$$ a = \frac{1}{2}[1+(2p-1)^n], $$

and

$$ b = \frac{1}{2}[1-(2p-1)^n]. $$

Remark: Using the definitions of $a$ and $b$, we can directly verify that

$$ a-b = [-(1-p)+p]^n. $$

Nice!

References:

[1] Nahin, Paul J. (2014). Will You Be Alive 10 Years From Now? pp. 18-20, 87-89.