Show that the distribution is the uniform distribution if and only if coin tosses are fair and mutually independent.

Question

I am paraphrasing a question once again from Professor Ery Arias-Castro's Principles of Statistical Analysis.

Suppose we toss a fair coin $n$ times. Define the event $A_i$ as the event that the $i$th toss results in heads, and show that $A_1, A_2, \dots, A_n$ are mutually independent, that is,

$$ \mathbb{P}(A_{i_1} \cap A_{i_2} \cap \cdots \cap A_{i_k}) = \mathbb{P}(A_{i_1})\mathbb{P}(A_{i_2}) \cdots \mathbb{P}(A_{i_k}) $$

for any $k$-tuple $1 \leq i_1 < \cdots < i_k \leq n$. In fact, show that the distribution is the uniform distribution $(\mathbb{U}(A) = |A|/|\Omega|$ for any $A \subseteq \Omega$) if and only if the tosses are fair and mutually independent.

I edited the question for clarity (the question depended on several other exercises in the text). Mutual independence follows from the law of multiplication $$ \mathbb{P}(A_1 \cap A_2 \cap \cdots \cap A_n) = \prod_{k = 1}^n \mathbb{P}(A_k | A_1 \cap A_2 \cap \cdots \cap A_{k - 1}). $$ For the if and only if, I believe the forward direction follows since for $A_i$ we have

\begin{align} \mathbb{U}(A_i) &= \frac{|A_i|}{|\Omega|}\\ &= \frac{2^{n - 1}}{2^n}\\ &= \frac{1}{2}, \end{align}

proving fairness. Additionally, for events $1 \leq i_1 < \cdots < i_k \leq n$, we have

\begin{align} \mathbb{U}(A_{i_1} \cap A_{i_2} \cap \cdots \cap A_{i_k}) &= \frac{1}{2^k}\\ &= \mathbb{U}(A_{i_1})\mathbb{U}(A_{i_2}) \cdots \mathbb{U}(A_{i_k}), \end{align}

proving mutual independence.

I am not quite sure how to approach the backwards direction. We need to show that for any $A \subseteq \Omega$ we have $\mathbb{P}(A) = |A| / |\Omega|$, and we know that the coin is fair (I am assuming this means $\mathbb{P}(A_i) = 1/2$ since no rigorous definition was provided) and that the events $A_1, A_2, \dots, A_n$ are mutually independent. We know that $\cup_{i = 1}^n A_i = \Omega$, so maybe there's something there to get any $A \subseteq \Omega$ from the $A_i$s. I would appreciate any help.

Answer 1

The fact that $A_1,\ldots, A_n$ are mutually independent yields $$\begin{aligned} P(A_1^c\cap A_2\cap\ldots\cap A_n)&=P((A_2\cap\ldots\cap A_n)\backslash (A_1\cap A_2\cap\ldots\cap A_n))\\ &=P(A_2\cap\ldots\cap A_n)-P(A_1\cap A_2\cap\ldots\cap A_n)\\ &=P(A_2)\cdot\ldots\cdot P(A_n) -P(A_1)\cdot P(A_2)\cdot\ldots\cdot P(A_n)\\ &=(1-P(A_1))\cdot P(A_2)\cdot\ldots\cdot P(A_n)\\ &=P(A_1^c)\cdot P(A_2)\cdot\ldots\cdot P(A_n). \end{aligned}$$

This means that $A_1^c,A_2,\ldots,A_n$ are mutually independent.

In the next step you can show that both $A_1,A_2^c,\ldots,A_n$ and $A_1^c,A_2^c,\ldots,A_n$ are mutually independent. If you iterate this argument, you can show that $B_1,\ldots, B_n$ are mutually independent, where $B_i\in\{A_i,A_i^c\}$ for $i=1,\ldots,n$. Moreover, it holds $$P(B_i)=\begin{cases}P(A_i)&&=\frac{1}{2}&\text{ if }B_i=A_i\\ P(A_i^c)&=1-P(A_i)&=\frac{1}{2} &\text{ if }B_i=A_i^c\end{cases},$$ for $i=1,\ldots,n$, since the coin is fair.

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Pick any $\omega\in\Omega$. Then, there exists $B_i\in\{A_i,A_i^c\}$ for any $i=1,\ldots,n$, such that $\{\omega\}= B_1\cap\ldots\cap B_n$ and

$$P(\{\omega\})=P(B_1\cap\ldots\cap B_n)= P(B_1)\cdot \ldots \cdot P( B_n)=\frac{1}{2^n}=\frac{1}{|\Omega|}.$$

Ultimately, since $A$ can be written as the disjoint union $\bigcup_{\omega\in A}\{\omega\}$ and $A$ contains $|A|$ elements, you find

$$P(A)=P(\bigcup_{\omega\in A}\{\omega\})=\sum_{\omega\in A}P(\{\omega\})=\sum_{\omega\in A}\frac{1}{|\Omega|}=\frac{|A|}{|\Omega|}.$$