Does $L^1$ boundedness and convergence in probability imply convergence in probability of the Cesaro sums?

Question

Let $X_n$ be a sequence of random variables with uniformly bounded $L^1$ norm. Suppose $X_n$ converges in probability to $X \in L^1$.

Is it true that the Cesaro sums $Y_n := \frac{1}{n} \sum_{i = 1}^n X_i$ converge in probability to $X$?

Answer 1

Consider an independent sequence of events $\left(A_i\right)_{i\geqslant 1}$ such that if $2^N+1\leqslant i\leqslant 2^{N+1}$, $\mathbb P(A_i)=2^{-N}$. Define for $2^N+1\leqslant i\leqslant 2^{N+1}$ the random variable $X_i$ by $X_i=2^N\mathbf{1}_{A_i}$. Then $\lVert X_i\rVert_1=1$ and $X_i\to 0$ in probability. If $(Y_n)$ was convergent to $0$ in probability then so would be $Y_{2^{N+1}}$. Since $$ Y_{2^{N+1}}=\frac 1{2^{N+1}}\sum_{i=1}^{2^{N+1}}X_i\geqslant \frac 1{2^{N+1}}\sum_{i=2^N+1}^{2^{N+1}}X_i\geqslant \frac 12\sum_{i=2^N+1}^{2^{N+1}}\mathbf{1}_{A_i}\geqslant \frac 12\mathbf{1}_{\bigcup_{i=2^N+1}^{2^{N+1}}A_i}=:Z_N, $$ we would derive that $Z_N\to 0$ in probability. This would imply that $\mathbb P\left(\bigcup_{i=2^N+1}^{2^{N+1}}A_i\right)\to 0$, but $$ \mathbb P\left(\bigcup_{i=2^N+1}^{2^{N+1}}A_i\right)=1-\mathbb P\left(\bigcap_{i=2^N+1}^{2^{N+1}}A_i^c\right)=1-\left(1-2^N\right)^{2^N}\to 1-e^{-1}. $$