I am stuck on understanding the statement of Lévy which is used in the proof of the CLT.
Lévy statement:
Given a sequence of random variables $\textbf{X}_n$ with characteristic functions $\phi_n$, if $$\phi_n(t)\rightarrow\phi(t) \text{ for all } t, \text{ then } \exists \textbf{ X} \text{ s.t. } \textbf{ X} \xrightarrow{d} \textbf{X},$$
where the characteristic function of $\textbf{X}$ is $\phi(t)$.
The proof of the CLT shows that the characteristic functions convergence to $e^{-t^2/2}$ and then claims that this random variable is the standard normal. However, the Lévy theorem only concludes that their exists some random variable $\textbf{X}$ and not anything about the uniqueness. Can their be another random variable with the same characteristic function?
I know characteristic functions uniquely determine the law / distribution functions but cannot see how this relates.
Best Answer
You can have two independent random variables with the same distribution. If they are independent, they obviously can't be identical.
Convergence in distribution just means the distribution converges. So if $X_n \xrightarrow d X$ and $Y$ is identically distributed to $X$ then $X_n \xrightarrow d Y$ as well. Weak convergence only depends on distributions and you can wildly change the functions as long as the distribution is the same.
For another example, if $X \sim \mathcal{N}(0,1)$ then $-X \sim \mathcal{N}(0,1)$ so if $X_n \xrightarrow d X$ then $X_n \xrightarrow d -X$. This is something that just doesn't happen for $L^1$ or almost-sure convergence unless $X = 0$.
We might say something like "the weak convergence topology is non-Hausdorff" meaning sequences can have more than one limit. Of course, that statement doesn't make any sense because the random variables need not share the same probability space, but that's what the quotation marks are for.