Central limit theorem for "not the whole sum" in construction of Wiener measure.

Question

I am trying to understand the proof of existence of Wiener measure and I came across this step that was not clarified further:

Let $Y_i$, $i \in \mathbb{N}$ be i.i.d., $E[Y_1]=0$, $Var[Y_1]=1$ and $0<s<t$. Does the sum $$\frac{1}{\sqrt{n}}\sum_{[ns]+1}^{[nt]}Y_i$$ converge in distribution to a normal distributed random variable? And what is then the variance of this normal distributed random variable?

I know that one has to use the standard central limit theorem, but I do not see how to estimate the sum

Thanks a lot in advance!

Answer 1

Since $\{Y_i\}$ is i.i.d $\sum_{[ns]+1}^{[nt]} Y_i$ has the same distribution as $\sum_{1}^{[nt]-[ns]} Y_i$. If you divide the sum by $\sqrt {[nt]-[ns]}$ then it converges to standard normal distribution. Note that $\frac 1 {\sqrt n} = \frac 1 {\sqrt {[nt]-[ns]}} c_n$ with $c_n \to \sqrt{t-s}$. Hence the limiting distribution os $N(0,t-s)$.