Cfrgtkky

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

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

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?

$P(A_1cap A_2)=frac 1 8 +frac 1 8=frac 1 4$ and $P(A_1)=P(A_2)=frac 1 2$ so $A_1$ and $A_2$ are independent.

Two events are independent if
begin{equation}
P(Acap 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.

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.

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

Let $E_iin{A_i,A_i^{complement}}$ for $i=1,dots,9$.

Every $omega=(omega_1,dots,omega_{10})in E_1capcdotscap E_9$ is completely determined by $omega_1$ so that $E_1capcdotscap E_9$ contains exactly $2$ elements of $Omega$.

So $P(E_1capcdotscap E_9)=2cdot2^{-10}=2^{-9}=P(E_1)timescdotstimes P(E_{9})$

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

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