Convergence in probability of Cesaro sums

Question

I saw in this question that for an independent sequence $\{X_n\}$, such that $X_n=2^n,$ with probability $1/n$,and $X_n=0$ with probability $1-1/n$ the sequence converges in probability to zero but the cesaro sum $\frac{X_1+X_2+\ldots X_n}{n}$ does not converge in probability. I see that the sequence itself converges to 0 because $$\lim_{n \to \infty} P(|X_n| \geq \epsilon)=\lim_{n \to \infty}\frac{1}{n}=0$$ But I dont understand how to get the same for the Cesaro sum. Can anyone please help?

Answer 1

Let $S_n=X_1+\dots+X_n$. In order to prove that $S_n/n$ does not converge in probability, we shall show that $S_n/n$ is not Cauchy in probability. Specifically, we show that $$ P(|S_n/n-S_{2n}/2n| \ge 1) \not \to 0\qquad \text{as }n\to\infty $$ To this end, suppose that there exists an index $k$ such that

• $k\ge n+1+\log_2 n$, and
• $X_k=2^k$.

I claim, in this case, that $|S_n/n-S_{2n}/2n|\ge 1$. Proof: $$ \begin{align} |S_n/n-S_{2n}/n| &\ge |S_{2n}/2n|-|S_n/n| \\&\ge |2^k/2n|-|2^{n+1}/n| \\&=\frac1n(2^{k-1}-2^n) \\&\ge \frac1n(2^{n+\log_2 n}-2^n) \\&=2^n(1-1/n) \\&\ge 1 \tag{for $n\ge 2$} \end{align} $$ Therefore, we can lower bound the probability of $|S_n/n-S_{2n}/2n|\ge 1$ by the probability of the existence of such a $k$. Each of the variables in the list $$ X_{n+1+\log_2 n},\dots,X_{2n} $$ is zero with probability at most $1-1/2n$. Therefore, $$ P(|S_n/n-S_{2n}/2n|\ge 1) \ge 1 - \left(1-\frac1{2n}\right)^{n-\log_2 n-1} $$ Since $$\lim_{n\to\infty}1 - \left(1-\frac1{2n}\right)^{n-\log_2 n-1}=1-e^{-1/2},$$ we conclude that $S_n/n$ is not Cauchy in probability.