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Consider the random variable $X_n$, not necessarily iid. If $X_n\rightarrow 0$ almost surely, then the Cesaro means $\frac 1n\sum_{k=1}^nX_n$ converge almost surely to 0. This cannot be weakened to convergence in probability, as has been discussed here before by considering an independent sequence $\{X_n\}$, such that $X_n=2^n,$ with probability $1/n$,and $X_n=0$ with probability $1-1/n$.

My question is, if $|X_n|\leq 1$ surely, does convergence in probability of $X_n$ to zero imply the almost sure convergence of $\frac 1n\sum_{k=1}^nX_k$ to zero?

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Consider a sequence of random variables $(Y_i)_{i\geqslant 1}$ such that $0\leqslant Y_i\leqslant 1$ and $Y_i\to 0$ in probability but not almost surely (for example, $Y_i=\mathbf 1(A_i)$, where $(A_i)_{i\geqslant 1}$ is a sequence of independent events). Now consider an increasing sequence of positive integers $(n_j)_{j\geqslant 0}$ such that $$\forall j\geqslant 1, \frac{n_j-n_{j-1} } {n_j} \geqslant \frac 12, n_0=1$$ and define $$X_j=Y_i\mbox{ if }n_{i-1} \leqslant j\lt n_{i}.$$
Then $X_j\to 0$ in probability, $|X_j|\leqslant 1$ and $$\frac 1{n_\ell}\sum_{j=1}^{n_\ell}X_j=\frac 1{n_\ell}\sum_{i=1}^\ell(n_i-n_{i-1})Y_i\geqslant \frac{n_\ell-n_{\ell-1}}{n_\ell}Y_\ell\geqslant \frac{Y_\ell}2,$$ hence the sequence $1/n\sum_{j=1}^n X_j$ cannot converge to $0$ almost surely.

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  • $\begingroup$ If we let $X_j$ uniformly bounded and independent and converge to 0 in probability, $1/n\sum_{j=1}^n X_j$. $\endgroup$
    – Hamilton
    Oct 23, 2022 at 20:59
  • $\begingroup$ and it should be $n_{i-1} \leqslant j\lt n_{i}$ not $n_{j-1} \leqslant i\lt n_{j}$ $\endgroup$
    – Hamilton
    Oct 23, 2022 at 21:56
  • $\begingroup$ @Beginner I agree for your second comment and I have edited accordingly : thanks you for that. However, I fail to understand the first one. $\endgroup$ Oct 23, 2022 at 22:52
  • $\begingroup$ Thank you, I was going to ask, does that guarantee $1/n \sum_{j=1}^{n}X_j$ converges? you can check my linked question to this question. $\endgroup$
    – Hamilton
    Oct 23, 2022 at 23:00
  • $\begingroup$ what about the converse? That is if the Cesaro sum converges in probability to 0 does $\frac{X_n}{n}$ converge to 0 as well? $\endgroup$
    – abc
    Nov 13, 2023 at 6:06

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