2

I saw in this question that for random variables $\{X_n\}$ converging to 0 in probability it is not necessarily true that their Cesaro sum $\frac{X_1+X_2+\ldots X_n}{n}$ converge in probability to 0. I am interested in sort of the converse of this. That is, if the Cesaro sum converge in probability to zero, then does the sequence $\{ (\frac{X_n}{n})_{n \geq 1} \}$ converge in probability to 0?

abc
  • 429
  • 3
  • 13

1 Answers1

3

If you let $S_n = X_1 + \dots + X_n$ then $$\frac{X_n}{n} = \frac{S_n}{n} - \frac{n-1}{n}\cdot\frac{S_{n-1}}{n-1}$$ Since $\frac{n-1}{n} \to 1$ you get that $\frac{X_n}{n} \to 0$ in probability.

George Giapitzakis
  • 7,097
  • Thanks! Can we see this directly from the definition of convergence in probability? I think we will have to show $P(\bigg|\frac{X_n}{n}\bigg| \geq \epsilon)=P(\bigg|\frac{S-n}{n}-\frac{n-1}{n}.\frac{S_{n-1}}{n-}\bigg| \geq \epsilon) \to 0$. How do I show the last equality? Should I use the inequality $|a-b| \geq |a|-|b|$ – abc Nov 13 '23 at 16:56
  • @abc See here for example. – George Giapitzakis Nov 13 '23 at 17:01
  • @abc I think you'll have to use $|a-b| \leq |a|+|b|$. – user196574 Nov 13 '23 at 17:19