0
$\begingroup$

The following is a passage from Durrett's book on probability.

''If $X_n \in \{0,1\}$ are independent with $P(X_n=1)=a_n \to 0$ and $\sum a_n = \infty$, then $X_n \to 0$ in probability, but if we let $N(n)=\text{inf}\{m\geq n: X_m=1\}$, then $X_{N(n)}=1$ a.s.''

The statement itself is not important, but I'm having trouble understanding the sequence $X_{N(n)}$ in a rigorous way. Can someone explain how the random variables $\{X_{N(n)}\}$ look like as measurable functions and also as a subsequence of the original sequence $\{X_n\}$?

$\endgroup$
0
$\begingroup$

It's all complicated in the beginning. I personally feel that it is much easier to understand it in terms of measure theory.

Recall that $X_n$ is just a measurable function on some probability space (about which we usually don't care). But let us fix this probability space $(\Omega, \mathcal{F}, \mathbb{P}).$ So $X_n:(\Omega, \mathcal{F}, \mathbb{P})\to \{0, 1\}$ is a measurable function with some properties (but I will ignore them as you said the exact statement is not important).

Now if you have another measurable function $N:(\Omega, \mathcal{F}, \mathbb{P})\to \mathbb{N},$ then you can define a compostie function say $Y:(\Omega, \mathcal{F}, \mathbb{P})\to \{0, 1\}$ by $Y(\omega)=X_{N(\omega)}(\omega).$ This is what we are talking about.

Probabilistically, think of the situation as follows: Think of $X_n$ as being representing head/tail in a coin toss. But let us say you are interested in some particular $X_n,$ for example, you might be interested in what happens in the $10$th toss, that is, you are interested in $X_{10}.$ This is simple. But you might be interested in something more complicated. Imagine that you are playing a game where you win if you see $5$ consecutive zeroes and then $1.$ Naturally, if you look at the time of your first win, it is a random time, you can call it $N.$

So fundamentally, at the level of measure theory, we are just talking about the composition of measurable functions. In terms of probability, we have a sequence but we are now chosing one of the term in the sequence randomly.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.