# Weak convergence of Cesaro sums

Suppose $\{X_n\}_{n \geq 1}$ is a sequence of random variables which converges weakly to some random variable: $$X_n \overset{w}{\longrightarrow} X$$ Question: what happens to the Cesaro sums of $X_n$ (assuming all moments exist)? $$A_n := \frac{1}{n} \sum_{i=1}^n X_n \overset{w}{\longrightarrow} \; ?$$ At first I thought maybe $A_n \to \mathbb{E}X$ by writing $X_n = (X_n-X)+X$, but then something seemed fishy. I can't seem to grasp the intuition for whether this should converge to a constant (as opposed to random) limit.

• Well why it could not converge to a constant r.v.? Think at what the law of large numbers says (although you need to check the assumptions required by that theorem as opppsed to the assumptions in your problem). – Kolmo Nov 9 '15 at 15:27
• @Kolmo: well, it definitely can in certain cases, but I guess I'm wondering if there are cases where it doesn't. – gogurt Nov 9 '15 at 16:28

• The sequence $(A_n)_{n\geqslant 1}$ may even be not tight. For example, define $X_n:=2^n\mathbf 1(E_n)$, where the sequence $(E_n)_{n\geqslant 1}$ is a sequence of independent events such that $\mu(E_n)=1/n$. Then for each $R$, $$\limsup_{n\to +\infty}\mu\{|A_n|\gt R\}\geqslant \limsup_{n\to +\infty}\mu\{2^{2n}/(2n)\bigcup_{i=n+1}^{2n}E_i \gt R\} \geqslant\limsup_{n\to +\infty}\mu\left(\bigcup_{i=n+1}^{2n}E_i\right)\geqslant 1/2$$ while $X_n\to 0$ in probability.
• If $X$ is a random variable with symmetric distribution, then for each sequence $(\varepsilon_n)_{n\geqslant 1}$ of numbers among $-1$ and $1$, the sequence $\left(\varepsilon_n X=:X_n\right)_{n\geqslant 1}$ converges to $X$ in distribution. With some good choices of $\varepsilon_n$, we may have $A_n\to \alpha X$ in distribution, where $\alpha$ is a rational number in $[-1,1]$.