variance tends to zero implies boundedness in probability

Question

Given a sequence of random variables $X_n$ with $Var(X_n) \to 0$ as $n\to \infty$.

Does ist follow that $X_n$ is bounded in probability, i.e., $X_n = O_P(1)$?

Answer 1

No. Take $X_n=n$ for all $n$ for a counterexample. However if you assume that $EX_n=0$ for all $n$ the result is true and it follows by Chebychev's inequality.

Answer 2

This is true if the means of $\{X_n\}$ are uniformly bounded - this is a consequence of Chebyshev's inequality. Let $\sigma_n$ denote the standard deviation of $X_n$. For any $\varepsilon > 0$, there exists $M^2 = \frac{1}{\varepsilon}$ such that

$$ \mathbb{P}\left( \lvert X_n - \mu_n \rvert > M\sigma_n \right) \leq \frac{\mathbb{E}(X_n - \mu_n)^2}{M^2 \cdot \sigma^2_n} = \varepsilon$$

Otherwise, it is not true by the counterexample in the other answer.