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?
 A: *

*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 $1/2$.
