Proving $\ A_i $ and $\ A_j $ are independent events

Question

I toss a fair coin $\ 10 $ times. $\ A_i \ (i = 1,2,\dots,9) \ $ is the event of the same result in the $\ i $ and $\ i + 1 $ toss. How do I prove that for every $\ i \not = j $ the events $\ A_i, A_j $ are independent?

I understand that in order to prove the independent of those two events I need to prove that $$\ P(A_i | A_j) = P(A_i) \ P(A_j) $$

I guess I can conut on the fact that $\ i, i+1 $ are indepedent events because it is a fair coin. so $\ P(A_i) = P(A_j) = \frac{1}{2} $ because I can have whatever result I want in the first toss and then I need the same results in the second. so

$$\ P(A_j|A_i) = \frac{P(A_j\cap A_i)}{P(A_i)} = \frac{\frac{1}{2}^3}{\frac{1}{2}} = \frac{1}{4} = P(A_i)P(A_j) $$

But what if for example $\ i =1 , j = 2 $ it means I should have the same result only in the first, second and third toss and then $\ P(A_j | A_i) \not = P(A_i)P(A_j) $

am I correct?

Answer 1

$P(A_1\cap A_2)=\frac 1 8 +\frac 1 8=\frac 1 4$ and $P(A_1)=P(A_2)=\frac 1 2$ so $A_1$ and $A_2$ are independent.

Answer 2

Two events are independent if \begin{equation} P(A\cap B) = P(A)P(B). \end{equation}

Alterativelly if: \begin{equation} P(A| B) = P(A). \end{equation} Also $P(A_j\cap A_i)= \frac{1}{4} \iff i+1=j$ .

Answer 3

You are asked to prove that for $A_1,\dots,A_{10}$ are pairwise independent.

In this answer I go for a stronger result: $A_1,\dots,A_{10}$ are mutually independent.

See here for the difference.

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Work in probability space $(\Omega,\wp(\Omega),P)$ where: $$\Omega=\{H,T\}^{10}=\{(\omega_1,\dots,\omega_{10})\mid \omega_i\in\{H,T\}\text{ for every }i\in\{1,\dots,10\}\}$$and $P(\{\omega\})=2^{-10}$ for every $\omega\in\Omega$.

Let $E_i\in\{A_i,A_i^{\complement}\}$ for $i=1,\dots,9$.

Every $\omega=(\omega_1,\dots,\omega_{10})\in E_1\cap\cdots\cap E_9$ is completely determined by $\omega_1$ so that $E_1\cap\cdots\cap E_9$ contains exactly $2$ elements of $\Omega$.

So $P(E_1\cap\cdots\cap E_9)=2\cdot2^{-10}=2^{-9}=P(E_1)\times\cdots\times P(E_{9})$

This allows the conclusion that $A_1,\dots,A_9$ are mutually independent.

Consequently $A_1,\dots,A_9$ are also pairwise independent.