Let $X_n$ be a sequence of random variables such that
$$P(X_n = 0 \text{ eventually}) = 1 $$
Does this imply that
$$P(\sum_{i=1}^nX_{n,i} = 0 \text{ eventually}) = 1 $$
where $X_{n,1},X_{n,2},...$ is i.i.d. and has the same distribution as $X_n$ for all $n$.
Do you agree or disagree with this result. Any comment is appreciated.
The implication is false. Here's a counterexample. Consider the standard probability space, $\left([0,1], \mathcal{B}, P=\lambda\right)$. For every $n \in \{1, 2, \dots\}$ define $A_n := \left[0, \frac{1}{n}\right]$. Then $A_1\supseteq A_2 \supseteq \cdots$ is a decreasing sequence of events, such that, for every $n \in \{1, 2, \dots\}$, $P(A_n) = \frac{1}{n}$. For every $n \in \{1, 2, \dots\}$ define the random variable $X_n := \mathbb{1}_{A_n}$. Then, on the one hand, $P\left(X_n = 0\text{ eventually}\right) = 1$. On the other hand, since, for every $n \in \{1, 2, \dots\}$, $X_n \sim \mathrm{Bin}\left(1, \frac{1}{n}\right)$, then, given i.i.d. random variables $X_{n, 1}, X_{n, 2}, \dots, X_{n, n}$ that are equally distributed like $X_n$, we have $S_n := X_{n, 1} + \cdots + X_{n, n} \sim \mathrm{Bin}\left(n, \frac{1}{n}\right)$, so that $P(S_n = 0) = \left(1-\frac{1}{n}\right)^n\rightarrow e^{-1}$. We may assume, w.l.g., that the random variables $S_n$, $n \in \{1, 2, \dots\}$, are independent. Then, since $$ \sum_{n = 1}^\infty P(S_n > 0) = \sum_{n = 1}^\infty \left(1-\left(1-\frac{1}{n}\right)^n\right) = \infty, $$ the Borel-Cantelli lemma yields that $P(S_n > 0\text{ infinitely often}) = 1$, so that $P\left(S_n = 0\text{ eventually}\right) = 0$.