# Average of sequence random variables convergence

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?

• 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 $0$. – Mindlack Oct 19 '19 at 11:13
• 4. is even wrong in the iid case because of the law of large numbers. – Mindlack Oct 19 '19 at 11:30
• @Mindlack In your first comment you wanted to type $U$ in one place and $-U$ in the other right? – Kavi Rama Murthy Oct 19 '19 at 11:43
• @Kabo Murphy: right, thanks for pointing it out. – Mindlack Oct 19 '19 at 12:47
• @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. – 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 $$1/2$$.
• 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 $X_{3i+2}=-U$. – Davide Giraudo Oct 20 '19 at 7:12