If $\text P\left[|X_n|>n^{-\alpha}\right]\to0$ as $n\to\infty$ for some $\alpha>0$, does $(X_n)_{n\in\mathbb N}$ converge in probability?

Question

Let $(X_n)_{n\in\mathbb N}$ be a sequence of real-valued random variables on a probability space $(\Omega,\mathcal A,\operatorname P)$ and $\alpha>0$. Is there some relation between convergence in probability, i.e. $$\operatorname P\left[|X_n|>\varepsilon\right]\xrightarrow{n\to\infty}0\;\;\;\text{for all }\varepsilon>0\tag1$$ and $$\operatorname P\left[|X_n|>\frac1{n^\alpha}\right]\xrightarrow{n\to\infty}0?\tag2$$ It seems like neither implies the other. If so, can we deduce any other mode of convergence from $(2)$?

Answer 1

• $(2)\Rightarrow(1)$: Given $\epsilon>0$, then for any $n$ large enough so $n^{-\alpha}<\epsilon$, we have $$ P(|X_n|>\epsilon)<P(|X_n|>n^{-\alpha})\to 0. $$

• $(1)\;\require{cancel}{\nRightarrow}\; (2)$. Fix $\beta$ so $0<\beta<\alpha$, and let $X_n=n^{-\beta}$ deterministically.

• $(2)$ does not imply almost sure convergence. Let $X_n$ be independent Bernoulli random variables with $P(X_n=1)=1/n$, then apply Borel-Cantelli.

• $(2)$ does not implie $L_1$ convergence. On the probability space $[0,1]$ with Lebesgue measure, let $X_n=n^\alpha{\bf 1}_{(0,n^{-\alpha})}$. Then $P(|X_n|>n^{-\alpha})=n^{-\alpha}\to 0$, yet $E[|X_n|]=1\not\to 0$.