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Assume that I have a sequence of random variables $X_1, X_2, \dots$ with means $\mu_1, \mu_2, \dots$ such that $\lim_{n \to \infty} \operatorname{Var}(X_n) = 0$.

Can I claim that for large enough $n$ the respective $X_n$ will be close to $\mu_n$ with high probability? How can I can formalize such a statement?

Here is an example of a result that would make me happy:

For any $\varepsilon > 0$ and $\delta > 0$ there exists an $N$, such that for all $n \ge N$ it holds that $\Pr( |X_n - \mu_n| \ge \delta) \le \varepsilon$.

However, I'm not sure what tools I can use to prove it, or if there exists some similar known inequality.

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I just realized that this follows trivially from the Chebyshev's inequality.

Since $\operatorname{Var}(X_n) \overset{n \to \infty}{\to} 0$, we can pick $N$ such that for all $n \ge N$ we have $\operatorname{Var}(X_n) \le \varepsilon \delta^2$. Then by Chebyshev's inequality $$\Pr(|X_n - \mu_n| \ge \delta) \le \frac{\operatorname{Var}(X_n)}{\delta^2} \le \varepsilon,$$ which completes the proof.

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