Let $X_1,X_2, \cdots$ be a sequence of random variables that converges to random variable $X$ almost surely OR in $L_p$ norm OR in probability OR in distribution. That is $$X_n\stackrel{a.s.}{\longrightarrow} X$$ OR $$X_n\stackrel{L_p}{\longrightarrow}X$$ OR $$X_n\stackrel{\mathscr{P}}{\rightarrow} X$$ OR $$X_n\stackrel{\mathscr{D}}{\rightarrow} X$$

Let $Y=\frac{1}{n}\sum_{i=1}^{n}X_i$, I am interested in the convergence of the average sequence $\{Y_n\}$ in different cases. i.e. Which of the following conjectures is true?

$$X_n\stackrel{a.s.}{\longrightarrow} X \Rightarrow Y_n\stackrel{a.s.}{\longrightarrow} X$$

$$X_n\stackrel{L_p}{\longrightarrow}X\Rightarrow Y_n\stackrel{L_p}{\longrightarrow} X$$

$$X_n\stackrel{\mathscr{P}}{\rightarrow} X\Rightarrow Y_n\stackrel{\mathscr{P}}{\longrightarrow} X$$

$$X_n\stackrel{\mathscr{D}}{\rightarrow} X\Rightarrow Y_n\stackrel{\mathscr{D}}{\longrightarrow} X$$

If any one of them is incorrect, please give a specific counter example where the limit random variable $X$ is a non-degenerate variable, i.e. $X$ is not a real number(if it is possible). If there is not an $X$ could be non-degenerate, why?

(Maybe we can add a non-degenerate $X$ to the existing counter example? Namely, if $X$ is non-degenerate, then let $X_n^{\prime}=X_n-X$. Thus, we have $X_n^{\prime}\rightarrow 0$ in different modes. By the additivity of limit, it is make sense for non-degenrate case. Would it be all right? )

Many thanks.

Update: Now we know that for the arithmetic mean, case 1 and case 2 are true, meanwhile case 3 and case 4 are false.

Further, I want to know can the conclusion be extended to geometric mean, harmonic mean and quadratic mean? By the continuous mapping theorem, I think, for case 1, so does geometric mean, harmonic mean and quadratic mean as well. But I am not sure about the case 2. Moreover, for the other forms of mean, will the conclusions of the other two cases change?

  • 1
    $\begingroup$ The first one stems from Stolz-Cesaro. The second one can be proved with a similar argument. 4. is wrong, because you can choose $U$ a Gaussian random variable, and set $X_{2n+1}=U$, $X_{2n}=U$ so that $X_n$ converges in law to $U$ while $Y_n$ converges as to $0$. $\endgroup$ – Mindlack Oct 19 '19 at 11:13
  • $\begingroup$ 4. is even wrong in the iid case because of the law of large numbers. $\endgroup$ – Mindlack Oct 19 '19 at 11:30
  • $\begingroup$ @Mindlack In your first comment you wanted to type $U$ in one place and $-U$ in the other right? $\endgroup$ – Kavi Rama Murthy Oct 19 '19 at 11:43
  • $\begingroup$ @Kabo Murphy: right, thanks for pointing it out. $\endgroup$ – Mindlack Oct 19 '19 at 12:47
  • $\begingroup$ @Mindlack Thanks! I think case 2 also can be prove by Minkowski inequality, right? And I wonder what about case 3. (in probability)--in particular, a counter example can be constructed when X follows a degenerate distribution. However, for the non-degenerate case, I can't find a good proof. $\endgroup$ – Skylin Chern Oct 19 '19 at 13:49
  1. and 2. follow from the fact that if a sequence $(a_n)$ of real numbers converges to $0$, so does $\left(n^{-1}\sum_{i=1}^na_i\right)_{n\geqslant 1}$ as well.

For 3., let $(A_i)$ be a sequence of independent events such that for $2^N+1\leqslant i\leqslant 2^{N+1}$, $\Pr(A_i)=2^{-N}$. Let $X_i=i\mathbf 1_{A_i}$. Then the following inclusion holds: $$ \left\{ \frac 1{2^{N+1}}\sum_{i=1}^{2^{N+1}}X_i\geqslant \frac 12 \right\} \supset\bigcup_{i=2^N+1}^{2^{N+1}}A_i$$ and a lower bound for $\Pr\left(\bigcup_{i=2^N+1}^{2^{N+1}}A_i\right)$ can be obtained by Bonferroni's inequality, namely, $$ \Pr\left(\bigcup_{i=2^N+1}^{2^{N+1}}A_i\right)\geqslant \sum_{i=2^N+1}^{2^{N+1}}\Pr(A_i)- \sum_{2^{N}+1\leqslant i\leqslant j\leqslant 2^{N+1}}\Pr(A_i\cap A_j) $$ and using independence and the fact that $\Pr(A_i)=2^{-N}$, we get $$ \Pr\left\{ \frac 1{2^{N+1}}\sum_{i=1}^{2^{N+1}}X_i\geqslant \frac 12 \right\}\geqslant 1-\frac 12\left(1-2^{-N}\right). $$

For 4., let $X_{2i}=U$ and $X_{2i+1}=-U$, where $U$ takes the values $1$ and $-1$ with probability $1/2$.

  • $\begingroup$ Thanks a lot! For 3 and 4, is there any other counter example when the limit random variable X is non-degenerate? Namely, X is not a real number, such as 0, but a non-degenerate random variable. $\endgroup$ – Skylin Chern Oct 20 '19 at 3:12
  • $\begingroup$ For 3., just take $Y+X_n$ where $Y$ is a non-degenerated random variable. For 4. let $X_{3i}=X_{3i+1}=U$ and $X_{3i+2}=-U$. $\endgroup$ – Davide Giraudo Oct 20 '19 at 7:12
  • $\begingroup$ Well, I think so! (I updated my idea and problem just four hours ago)So many thanks! $\endgroup$ – Skylin Chern Oct 20 '19 at 8:05

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