Suppose you toss a fair coin $n$ times. The sample space $\Omega$ is the space of (ordered) binary vectors of length $n$. A $1$ in the $i$-th position of such a binary vector indicates Heads at the $i$-th toss. The probability measure $\Bbb{P}$ on $\Omega$ is defined as $\Bbb{P}(A)=\frac{|A|}{|\Omega|}$ for all $A \subset \Omega$. If $A_i$ is the event that the $i$-th toss is Heads, $i=1, 2, ..., n$, prove using the aforementioned $\Bbb{P}$ that events $A_1, ..., A_n$ are mutually independent.
My attempt:
I have found that:
- $A_i=\{(x_1, x_2, ..., x_n ): x_i=1, x_j=0,1 \forall j\ne i\}$.
- $|\Omega|=2^n$
- $|A_i|=2^{n-1}$
- So, $\Bbb{P}(A)=\frac{2^{n-1}}{2^n}=\frac{1}{2}$
By the definition of mutual independence, I need to prove that for any $k$ and any choice of distinct indices $i_1, ..., i_k$, $\Bbb{P}(A_{i_1}\cap A_{i_2} \cap ... \cap A_{i_k})=\Bbb{P}(A_{i_1})\Bbb{P}(A_{i_2})...\Bbb{P}(A_{i_k})$.
However, I am unsure of how to go from here. Please help!!
The intersection of $k$ events $A_{i_1}\cap\ldots \cap A_{i_k}$ is the set of all binary vectors of length $n$ such that $x_{i_1}=1, \ldots, x_{i_k}=1$. There are $2^{n-k}$ of such vectors since there are $k$ coordinates with $1$ and $n-k$ coordinates which can be either $0$ or $1$. So $$ \mathbf P(A_{i_1}\cap\ldots \cap A_{i_k}) = \frac{|A_{i_1}\cap\ldots \cap A_{i_k}|}{|\Omega|}=\frac{2^{n-k}}{2^n}=\frac{1}{2^k}. $$ Compare it with the product of probabilities and conclude.