I'm having trouble improving the upper bound on Markov/Chebyshev's inequality in this particular example:

Show that $$\lim_{n\to\infty} n\mathbb{P}(|X_1|\geq \epsilon\sqrt{n})=0$$

Clearly, Markov's inequality yields the upper bound $n\mathbb{E}[|X_1|^2]/\epsilon^2$ which is not good enough. I'd really appreciate any hints or direction.

  • 3
    $\begingroup$ Why would you need to have a sequence of variables if you need to prove something about $X_1$? $\endgroup$
    – GReyes
    Dec 4, 2023 at 2:20
  • $\begingroup$ Alright, edited $\endgroup$
    – mtcicero
    Dec 9, 2023 at 23:22

1 Answer 1


Note that $nP(|X_1| > \sqrt{n}) = E[n1(X_1^2 > n)]\leq E[X_1^21(X_1^2 > n)]$. The quantity inside the expectation is bounded by $X_1^2$ and converges to zero almost surely. Now use the dominated convergence theorem.


You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .