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I know that $\lim \sup A_n$ is always a tail event. E.g. from here.

But this was asked in an exam a few years ago -

Prove of disprove: Let $X_1, X_2, ... $ be a series of positive R.V.; Let $S_n = X_1 + ... + X_n$ and $a>0$. Then $A=\lim\sup\{\frac{S_n}{c_n}>a\} $ is a tail event if $c_n$ is a series of real numbers going to infinity.

The official answer claim this is not true. And gives the following explanation:

Choose $X_1$ to get 1 or 2 with equal probability, and $X_2, ... $ to get always 1. Define $c_n=n$ and let $a=1$. Then the event $\{\frac{S_n}{c_n}>a\}$ is not in the tail sigma-algebra, as it depends on the outcome of $X_1$: if $X_1=1$ it doesn't occur, and if $X_1=2$ it does occur.

I think that this explanation is wrong. Am I wrong and the official answer is correct? If so, can you please add more intuition behind this.

Maverick Meerkat
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1 Answers1

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I think the official answer is wrong. Even without the $\lim\sup$ the event in the "counter" example is actually a tail-event. So obviously with $\lim\sup$ it will remain a tail event. Here is my disproving of the "counter" example:

Denote the event $A_n = \{\omega: \frac{X_1+...+X_n}{n}>1\}$

We can associate it with a new R.V. $X_n^*(\omega)=\chi_{\frac{X_1+...+X_n}{n}>1}(\omega)$.

If we look now on some $\sigma(X_n^*)$ we will see it only depends on partition of the sample space made by $X_1,...,X_n$. So $\sigma(X_n^*)=\sigma(X_1,...,X_n)$ But all the R.V. except the 1st are constants, so they don't divide the sample space, hence $=\sigma(X_1)$.

If we now look at the $\sigma$-algebra of all the new R.V. from $n$ onwards: $\mathcal F'_n = \sigma (X_n^*,X_{n+1}^* ,...) =\sigma(X_1,...,X_n,X_1,...,X_{n+1},...)=\sigma(X_1)$. So if we look at the tail $\sigma$-algebra, $\mathcal T=\bigcap_{n\geqslant 1} \mathcal F'_n=\sigma(X_1)$.

The event will happen or not, depending on $X_1$, which is also the tail-$\sigma$ algebra, so the event is a tail event.

Maverick Meerkat
  • 1,230