variance tends to zero implies uniformly bounded by a random variable

Question

Given random variables $X_1,...,X_n$ with $E(X_n) =0$ and $Var(X_n)\to 0$ does it follow that $X_n$ is uniformly bounded by a random variable $X$, i.e.,

$P(|X_n| > t) \leq P(|X|>t) \,\,\forall n$ for any $t$?

Answer 1

This question doesn't make much sense: all probabilities are bounded by 1. Namely, taking $X = \infty$ a.s. yields $\mathbb{P}(|X| > t) = 1$ for all $0 \leq t < \infty$.

Regardless, it might be helpful point out that Chebyshev's inequality yields $$ \mathbb{P}(\left|X_{n}\right|>t)=\mathbb{P}(\left|X_{n}-\mathbb{E}X_n\right|>t)\leq\text{Var}(X_{n})/t^{2}. $$ Maybe the above bound is more useful for whatever you are trying to do?