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If a sequence of $X_n \to_{\mathbb{P}} 0$, do it follow that the Cesaro sum $ \frac{1}{n} \sum_{k=1}^n X_k \to 0$ in probability?

Here is a related question.

2nd part of my question: What if $X_n$ is independent and uniformly bounded?

Edit: The answer is already answered by the accepted answer and comments in it.

Hamilton
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1 Answers1

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No. Take the standard example of a sequence that converges in probability to $0$ but not almost surely. If you multiply the terms of the sequence with a sufficiently fast increasing sequence of numbers, the sequence will still converge in probability, but the Cesaro-sum can even diverge to $\infty$ everywhere.

Michael Greinecker
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  • What if $X_n$ is independent and uniformly bounded? the result will change? I just edited the question. – Hamilton Oct 23 '22 at 03:25
  • @Beginner: If $X_n$ are independent (uniformly bounded if you want) and $X_n\rightarrow0$ in probability, then $X_n=0$ a.s. for all $n$. – Mittens Oct 24 '22 at 21:05
  • @OliverDíaz not sure that's true? Take independent events $A_n$ with $\mathbb P(A_n)=1/n$, the indicators $X_n=\mathbf1_{A_n}$ converge to $0$ in probability but not a.s. (eg by Borel Cantelli) – jlammy Oct 24 '22 at 21:14
  • @jlammy: Sure, I was thinking i.i.d ... – Mittens Oct 24 '22 at 21:16
  • @Beginner If $X_n$ converges to $0$ in probability and they are uniformly bounded (independence is unnecessary), then $\mathbb{E}|X_n|$ converges to $0$, so $\mathbb{E}\left|\frac1n\sum_{k=1}^nX_k\right| \leq \frac1n\sum_{k=1}^n \mathbb{E}|X_k|$ converges to $0$, which means $\frac1n\sum_{k=1}^n X_k$ converges to $0$ in probability. (you could probably show this directly too) – Brian Moehring Oct 24 '22 at 21:26
  • On what grounds, did you conclude directly that "$\mathbb E \lvert X_n \rvert$" converges to 0? Generally, it's not true since convergence in probability doesn't imply $L^1$ convergence. – Hamilton Oct 24 '22 at 21:56
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    @Beginner That's where you need the sequence to be uniformly bounded. The reason why convergence in probability doesn't by itself imply $L^1$ convergence is that you have no control over how big the sequence can become on "small" events, but if the sequence is uniformly bounded, you have precisely that control. The proof only uses elementary tools, so you should prove it to yourself. – Brian Moehring Oct 24 '22 at 22:10
  • @BrianMoehring Thank you I got it. – Hamilton Oct 25 '22 at 00:29