I have a two statements. One says:

$$-\frac{1}{T} \sum_{t=1}^T X_t \rightarrow a $$ in probability as $T \rightarrow \infty$.

The other:

$$-\frac{1}{g(T)} \sum_{t=1}^{g(T)} X_t \rightarrow a $$ in probability as $T \rightarrow \infty$ for every function $g: N \rightarrow N$ such that $ g(T) \rightarrow \infty$ as $T \rightarrow \infty$.

Could you please tell me why are these two statements equivalent for a random sequence $X_t$?


1 Answer 1


The question is actually about the equivalence of $Y_n\to 0$ in probability and $Y_{g(n)}\to 0$ in probability for each function $g\colon \mathbb N\to\mathbb N$ going to infinity as $n$ goes to infinity (it has nothing to do with averages).

Of course, the direction $\Leftarrow$ is trivial (take $g(n)=n$ for each $n$). For the converse, fix $g$ such that $g(n)\to \infty$. Fix $\varepsilon$ and $\delta\gt 0$: there is $N$ such that $\mathbb P(|Y_n|>\varepsilon)\lt \delta$ if $n\geqslant N$. Now consider $N'$ such that $g(n)\geqslant N$ if $n\geqslant N'$ (using the fact that $g(n)\to \infty$). We obtain that $\mathbb P(|Y_{g(n)}|>\varepsilon)\lt \delta$ if $n\geqslant N'$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.