assume some i.i.d. random variables $(X_i)_{i \geq 1}$ with mean 0 and variance 1. Due to Donsker it holds that $$\left(\frac{1}{\sqrt{n}}\sum\limits_{i=1}^{nt}{X_i}\right)_t \rightarrow (W_t)_t$$ with a Standard Brownian motion $W_t$. Changing this expression a bit I guess that it holds $$\left(\frac{1}{\sqrt{n}}\sum\limits_{i=1}^{n^a t}{X_i}\right)_t \rightarrow 0$$ for $0<a<1/2$. Is this statement true and if so, why?
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Define $W_n(t):=\sum_{i=1}^{nt}X_i$. We have that $\left(n^{-\alpha/2}W_{n^\alpha}(t)\right)_{n\geqslant 0}\to (W_t)_{t\in [0,1]}$ since $\left(n^{-1/2}W_{n}(t)\right)_{n\geqslant 0}\to (W_t)_{t\in [0,1]}$. Now the equality $$n^{-1/2}W_{n^\alpha}(t)=n^{(\alpha-1)/2}n^{-\alpha/2}W_{n^\alpha}(t)$$ gives the wanted convergence.