$ v(\omega) : = \inf \{ n \geq 2 : \omega \in A_n \} $ , proving that $v$ is a random variable

Question

Let $ (X_n)_{n \geq 1} $ be a sequence of Bernoulli random variables of parameter $p$

$A_n : = \{ \omega \in \Omega : X_n(\omega) \neq X_{n-1}(\omega) \} $

$ v(\omega) : = \inf \{ n \geq 2 : \omega \in A_n \} $ with $ \inf \{ \emptyset \} = - \infty $

We want to prove that $ v$ is a random variable, find its law, and prove that $P(v = +\infty) = 0$

The way I started thinking about it was

$ v(\omega) : = \inf \{ n \geq 2 : \omega \in A_n \} $

$ v(\omega) : = \inf \{ n \geq 2 : X_n(\omega) \neq X_{n-1}(\omega) \} $

I was hoping I'd apply some result of a decreasing or increasing sequence of sets, but I didn't get to a pertinent result, the solution starts by saying :

For all $n \geq 2 $ , we have that $ \{ v = n \} = \cap^{n-2}_{i=1} \{ X_i = X_{i+1} \} \cap \{ X_{n-1}\neq X_n \}$

Which I don't see how it came, which is what this question is about.

For more context, the solution then it affirms that :

since the $X_n$ are random variables , the sets $\{ X_i = X_{i+1} \}$ , $i\leq n $ and $\{ X_{n-1} \neq X_{n} \}$ are elements of the $\sigma$ - algebra $F$ (with which we endowed $\Omega$ ). So it's also the case of $\{ v = n \}$ which shows that $v$ is a random variable.

Then it goes on saying that :

For all $n \geq 2 $ , we have that

$P( v = n) = P(X_1 = X_2 \dots = X_{n-1} = 1)P(X_n = 0)$ $ + P(X_1 = X_2 \dots = X_{n-1} = 1)P(X_n = 1)$ $ = (1-p)^{n-1}p + p^{n-1}(1-p) $ which caracterises the law of $v$

Let's calculate the probability of the complementary event $\{ v < + \infty \}$

$P(v < \infty) = P( \cup^{\infty}_{n=2} \{ v= n \})$

$= \sum^\infty_{n=2} ( (1-p)^{n-1}p + p^{n-1}(1-p) ) = 1 $

From which we conclude that $ P(v=\infty)=0 $

Answer 1

For an arbitrary $\omega\in\Omega$ and a fixed integer $n\geq2$ the following statements are equivalent:

• $\omega\in\{v=n\}$
• $v(\omega)=n$
• $\inf(\{k\geq2\mid \omega\in A_k\})=n$
• $\omega\in A_n\wedge \omega\notin A_k$ for every $k<n$
• $X_n(\omega)\neq X_{n-1}(\omega)$ and $X_k(\omega)= X_{k-1}(\omega)$ for every $k<n$.
• $\omega\in\{X_n\neq X_{n-1}\}$ and $\omega\in\{X_k=X_{k-1}\}$ for every $k<n$.
• $\omega\in\{X_n\neq X_{n-1}\}\cap\bigcap_{k<n}\{X_k=X_{k-1}\}$

This justifies the conclusion that: $$\{v=n\}=\{X_n\neq X_{n-1}\}\cap\bigcap_{k<n}\{X_k=X_{k-1}\}$$