I seem to have misunderstood something fundamental about $\sigma$-tail algebras, as I cannot resolve the following problem:
Given a series of real random variables $(X_n)_{n\in\mathbb{N}}$. Then
$X^*=\displaystyle \limsup_{n\rightarrow\infty}\frac{1}{n}\sum_{i=1}^n X_i$ is in the $\sigma$-tail algebra $\mathcal{T}((X_n)_{n\in\mathbb{N}})$
I can show this using the following arguments:
First step: $\displaystyle \frac{1}{n}\sum_{i=1}^n X_i$ is $\sigma(X_1,…,X_n)$ measurable, since $X_i$ is $\sigma(X_i)$ measurable for all $i=1,…,n$ and the fact that the sum of measurable maps (and multiplying by a constant) is again measurable, particularly measurable in the smallest $\sigma$-algebra generated by $X_1,…,X_n$, being $\sigma(X_1,…,X_n)$.
Second step: It follows that $\displaystyle \limsup_{n\rightarrow\infty}\frac{1}{n}\sum_{i=1}^n X_i$ is $\sigma((X_n)_{n\in\mathbb{N}})$ measurable, because the limit is measurable for measurable maps.
Third step: Since $\displaystyle \limsup_{n\rightarrow\infty}\frac{1}{n}\sum_{i=1}^n X_i= \limsup_{n\rightarrow\infty}\frac{1}{n}\sum_{i=k}^n X_i$
it follows that $X^*$ is $\sigma(X_n,\ n\ge k)$ measurable for every $k\in\mathbb{N}$ which means that $X^*$ is measurable with respect to $\mathcal{T}((X_n)_{n\in\mathbb{N}})$
This is correct so far, right?
Now lets look at $S_n=\sum_{i=1}^nX_i$ for i.i.d random variables $X_i$. Is $\limsup_{n\rightarrow\infty}S_n$ measurable w.r.t. $\mathcal{T}((S_n)_{n\in\mathbb{N}})$?
Here comes my fallacy: I would just argue that, since $\sigma((X_n)_{n\in\mathbb{N}})=\sigma(X_1)$, because $X_i$ are i.i.d, and again by the argument that the sum and limit of measurable functions are again measurable, we immediately have that $\limsup_{n\rightarrow\infty}S_n$ is $\mathcal{T}((S_n)_{n\in\mathbb{N}})$ measurable.
However, someone told me this is not correct. Where is my mistake?
Best Answer
It is far from truth that if $X$ and $Y$ are i.i.d. then $\sigma (X,Y)=\sigma (X)$. I fact these two sigma algebras cannot be equal unless $Y$ is a constant! Proof: if this is true then $Y$ is independent of $X$ and also measurable w.r.t. $\sigma (X)$ which implies that $Y$ is independent of itself, hence a constant almost surely.