A sequence $(x_k)$ is statistically convergent to $L$ if
$\displaystyle \lim_{n\to\infty}\frac{1}{n}|\{k\leq n:|x_k-L|\geq \epsilon\}|=0.$ (1)
Also let replace $n$ with $\log n$ as follows:
$\displaystyle \lim_{n\to\infty}\frac{1}{\log n}|\{k\leq \log n:|x_k-L|\geq \epsilon\}|=0.$ (2)
(Here, | | represent the cardinality of the set enclosed.)
My Question: I am not sure if the following is true:
(1) is satisfied if and only if (2) is satisfied.
Yes, they are equivalent. Use the following facts:
For any $N$ there exists a unique $n$ such that $\log n < N\leq log (n+1)$; as $N \to \infty$, we have $n \to \infty$ and $\frac {\log {(n+1)}} {\log n} \to 1$. $\{k \leq \log n: |x_k-L| \geq \epsilon\} \subset \{k \leq N: |x_k-L| \geq \epsilon\} \subset \{k \leq \log (n+1): |x_k-L| \geq \epsilon\} $.